tag:blogger.com,1999:blog-91012991483611091842024-03-11T07:58:02.216-07:00 Mymathware Shafii's math academy Mymathware is a mathematics tutorial blog that posts solutions on questions, tutor students on various topics, download useful resource materials and also write articles on mathematics that keeps people updated on happenings in the mathematics world. Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.comBlogger221125tag:blogger.com,1999:blog-9101299148361109184.post-73249900395940707412023-06-12T03:45:00.002-07:002023-06-12T05:39:10.278-07:00What is a Polynomial Ring? It's properties and Examples <div><b>What is a Polynomial Ring? It's properties and Examples </b></div><div><b><div class="separator" style="clear: both; text-align: center;">
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</div><br></b></div><div><br></div><div>Polynomial rings are fundamental mathematical structures that allow us to study polynomials and their properties in a systematic way. They find applications in various branches of mathematics, including algebra, number theory, and algebraic geometry. In this article, we will delve into the concept of polynomial rings, discuss their properties, provide examples, and explore two important theorems related to these rings.</div><div><br></div><div>Let's begin by defining a polynomial ring. Given a commutative ring $R$ with identity, the polynomial ring over $R$, denoted as $R[x]$, is the set of all polynomials in the indeterminate $x$ with coefficients in $R$. A polynomial in $R[x]$ has the form:</div><div><br></div><div>\[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \]</div><div><br></div><div>Here, $n$ is a non-negative integer, $a_n, a_{n-1}, \ldots, a_1, a_0$ are elements of the ring $R$, and $x$ is the indeterminate.</div><div><br></div><div><b>Properties</b></div><div><br></div><div>Polynomial rings possess several important properties that make them a powerful tool in mathematical analysis. Let's discuss a few of these properties:</div><div><br></div><div><b>1) Closure</b>: Polynomial rings are closed under addition and multiplication. If $f(x)$ and $g(x)$ are polynomials in $R[x]$, then $f(x) + g(x)$ and $f(x) \cdot g(x)$ are also in $R[x]$.</div><div> </div><div><b> 2. Degree: </b>The degree of a polynomial is the highest power of $x$ with a non-zero coefficient. For example, the polynomial $f(x) = 3x^3 + 2x^2 - 5x + 1$ has a degree of 3.</div><div> </div><div><b> 3. Integral Domains: </b>Polynomial rings are integral domains if the base ring $R$ is an integral domain. An integral domain is a commutative ring with no zero divisors, meaning that the product of two non-zero elements is always non-zero. This property is crucial for studying factorization of polynomials and solving equations.</div><div><br></div><div><b>Examples</b></div><div><br></div><div>To better understand polynomial rings, let's consider a few examples.</div><main class="relative h-full w-full transition-width flex flex-col overflow-auto items-stretch flex-1"><div class="flex-1 overflow-hidden"><div class="react-scroll-to-bottom--css-wwknh-79elbk h-full dark:bg-gray-800"><div class="react-scroll-to-bottom--css-wwknh-1n7m0yu"><div class="flex flex-col text-sm dark:bg-gray-800"><div class="group w-full text-gray-800 dark:text-gray-100 border-b border-black/10 dark:border-gray-900/50 bg-gray-50 dark:bg-[#444654]"><div class="flex p-4 gap-4 text-base md:gap-6 md:max-w-2xl lg:max-w-[38rem] xl:max-w-3xl md:py-6 lg:px-0 m-auto"><div class="relative flex w-[calc(100%-50px)] flex-col gap-1 md:gap-3 lg:w-[calc(100%-115px)]"><div class="flex flex-grow flex-col gap-3"><div class="min-h-[20px] flex flex-col items-start gap-4 whitespace-pre-wrap break-words"><div class="markdown prose w-full break-words dark:prose-invert light"><ol><li><p>$\mathbb{R}[x]$: The polynomial ring over the real numbers includes all polynomials with real coefficients. For instance, $h(x) = 2x^3 + 4x^2 - 3x + 1$ and $k(x) = x^4 - 2x^2 + 1$ belong to $\mathbb{R}[x]$.</p></li><li><p>$\mathbb{C}[x]$: The polynomial ring over the complex numbers consists of all polynomials with complex coefficients. An example of an element in $\mathbb{C}[x]$ is $p(x) = (1 + i)x^2 + (2 - i)x + (3 + 2i)$.</p></li><li><p>$\mathbb{F}_2[x]$: The polynomial ring over the field $\mathbb{F}_2$, which contains only the elements 0 and 1, represents polynomials with coefficients from $\mathbb{F}_2$. For example, $q(x) = x^3 + x^2 + x + 1$ and $r(x) = x^4 + 1$ are elements of $\mathbb{F}_2[x]$.</p></li></ol></div></div></div></div></div></div></div></div></div></div></main>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-52521361902485180252023-06-07T11:43:00.002-07:002023-06-07T11:46:33.622-07:00Exploring the Fascinating World of Graph Symmetry<b>Exploring the Fascinating World of Graph Symmetry</b><div><div class="separator" style="clear: both; text-align: center;">
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</div><br></div><div>When we think about symmetry, images of perfectly balanced objects or reflections often come to mind. However, symmetry is not limited to the visual world alone. In the realm of mathematics, symmetry plays a crucial role in various branches, including graph theory. Graph symmetry offers a captivating perspective on the structure and properties of graphs, unraveling intriguing patterns and connections. In this blog post, we will embark on a journey to explore the captivating world of graph symmetry and uncover its significance.</div><div><br></div><div><b>Understanding Graph Symmetry</b>:</div><div><br></div><div>In graph theory, a graph is an abstract representation of a set of objects, known as vertices, connected by lines called edges. Graph symmetry refers to a property where certain transformations or operations leave the structure of a graph unchanged. These transformations may include rotations, reflections, or combinations of both.</div><div><br></div><div>Types of Graph Symmetry:</div><div><br></div><div>1. <b>Reflection Symmetry</b>: Also known as bilateral symmetry, reflection symmetry occurs when a graph remains unchanged when mirrored along a particular axis. Imagine a graph as a visual representation of objects or patterns, and reflection symmetry is akin to folding the graph along a line and observing identical structures on both sides.</div><div><br></div><div>2. <b>Rotational Symmetry</b>: Unlike reflection symmetry, rotational symmetry involves rotating a graph by a specific angle around a central point. If the graph appears the same after the rotation, it exhibits rotational symmetry. The order of rotational symmetry of a graph is the number of distinct angles at which the graph appears unchanged.</div><div><br></div><div>3. <b>Translational Symmetry</b>: Translational symmetry involves shifting or sliding a graph along a particular direction without altering its overall structure. If the graph aligns perfectly with its original form after translation, it possesses translational symmetry. This type of symmetry is more commonly associated with periodic structures in graphs.</div><div><br></div><div><b>Applications of Graph Symmetry</b></div><div><br></div><div>Graph symmetry is a powerful concept with practical applications in various fields:</div><div><br></div><div>1. Chemistry: Graphs are extensively used to model molecular structures, and symmetry analysis helps identify identical atoms and predict chemical properties. Symmetry considerations play a significant role in understanding molecular vibrations, chirality, and electronic configurations.</div><div><br></div><div>2. Computer Vision: Symmetry detection algorithms based on graph symmetry aid in identifying objects or patterns in images. By analyzing the symmetrical properties of an image, these algorithms can recognize shapes, objects, and even detect anomalies.</div><div><br></div><div>3. Network Analysis: The study of symmetric properties in networks and graphs provides insights into the structure and organization of complex systems. Detecting and analyzing symmetrical patterns in social networks, transportation networks, and biological networks helps uncover hidden relationships and optimize network efficiency.</div><div><br></div><div>4. Graph Algorithms: Graph symmetry can guide the development of efficient algorithms for solving various graph problems. Symmetry breaking techniques help reduce the search space and improve the computational complexity of algorithms in domains such as graph coloring, graph isomorphism, and graph matching.</div><div><br></div><div>Graph symmetry offers a captivating lens through which we can explore the intricate patterns and connections within graphs. Whether it's uncovering the hidden symmetries in molecular structures, enhancing computer vision algorithms, understanding complex networks, or optimizing graph algorithms, the study of graph symmetry has far-reaching applications.</div><div><br></div><div>As we delve deeper into the realm of graph theory and symmetry, we unlock a deeper understanding of the underlying structures in our world. Graph symmetry not only enhances our knowledge but also sparks curiosity, leading to further advancements in a variety of disciplines. So, let us embrace the beauty and significance of graph symmetry as we continue to unravel the mysteries of the mathematical world.</div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-56098658979170316582023-06-06T03:15:00.091-07:002023-06-06T12:03:08.799-07:00Ring Divisors of Zero<div>In ring theory, a ring divisor of zero refers to an element of a ring that, when multiplied by another element, yields zero. These divisors of zero play a significant role in understanding the algebraic structure of rings. In this blog post, we will explore the concept of ring divisors of zero, discuss their properties, and examine some examples.<br></div><div><div class="separator" style="clear: both; text-align: center;">
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</div><br></div><div><br></div><div>Let $R$ be a ring and $a \in R$ be a nonzero element. We say that $a$ is a <i>divisor of zero</i> in $R$ if there exists a nonzero element $b \in R$ such that $ab = 0$. In other words, $a$ is a divisor of zero if it annihilates some nonzero element of $R$ under multiplication.</div><div><br></div><div><b>Properties of Ring Divisors of Zero </b></div><div><br></div><div>Ring divisors of zero possess several interesting properties that are worth noting:</div><div><br></div><div><b><i>1. Closure under Multiplication</i></b></div><div><br></div><div>If $a$ is a divisor of zero in $R$, and $c$ is any element of $R$, then $ac$ is also a divisor of zero. This property arises from the fact that if $ab = 0$ for some nonzero $b \in R$, then $(ac)b = a(cb) = 0$, implying that $ac$ is also a divisor of zero.</div><div><br></div><div><b><i>2. Zero Divisors in Commutative Rings</i></b></div><div><br></div><div>In a commutative ring, every nonzero divisor of zero is a zero divisor. This means that if $a$ is a divisor of zero in a commutative ring $R$, then there exists a nonzero element $b \in R$ such that $ab = ba = 0$. The proof of this property relies on the commutativity of multiplication in the ring.</div><div><br></div><div><b><i>3. Non-Existence of Multiplicative Inverses</i></b></div><div><br></div><div>Divisors of zero are never units or have multiplicative inverses. If $a$ is a divisor of zero in $R$, then there is no element $b \in R$ such that $ab = ba = 1$. This property can be easily proven by contradiction, assuming the existence of a multiplicative inverse for a divisor of zero and showing that it leads to a contradiction.</div><div><br></div><div><b>Examples</b></div><div><br></div><div>Let's look at some examples of divisors of zero in different rings:</div><div><br></div><div><b><i>Integers Modulo $n$</i></b></div><div><br></div><div>Consider the ring $\mathbb{Z}_6$ of integers modulo $6$. In this ring, both $2$ and $3$ are divisors of zero since $2 \cdot 3 = 3 \cdot 2 = 0 \pmod{6}$. These elements demonstrate that divisors of zero need not be zero themselves.</div><div><br></div><div><b><i>Matrix Rings</i></b></div><div><br></div><div><br></div><div>Let $R$ be a ring and $a \in R$ be a nonzero element. We say that $a$ is a \textit{divisor of zero} in $R$ if there exists a nonzero element $b \in R$ such that $ab = 0$. In other words, $a$ is a divisor of zero if it annihilates some nonzero element of $R$ under multiplication.</div><div><br></div><div><b>Properties</b></div><div><br></div><div>Ring divisors of zero possess several interesting properties that are worth noting:</div><div><br></div><div><b>Closure under Multiplication</b></div><div><br></div><div>If $a$ is a divisor of zero in $R$, and $c$ is any element of $R$, then $ac$ is also a divisor of zero. This property arises from the fact that if $ab = 0$ for some nonzero $b \in R$, then $(ac)b = a(cb) = 0$, implying that $ac$ is also a divisor of zero.</div><div><br></div><div><b>Zero Divisors in Commutative Rings</b></div><div><br></div><div>In a commutative ring, every nonzero divisor of zero is a zero divisor. This means that if $a$ is a divisor of zero in a commutative ring $R$, then there exists a nonzero element $b \in R$ such that $ab = ba = 0$. The proof of this property relies on the commutativity of multiplication in the ring.</div><div><b><br></b></div><div><b>Non-Existence of Multiplicative Inverses</b></div><div><br></div><div>Divisors of zero are never units or have multiplicative inverses. If $a$ is a divisor of zero in $R$, then there is no element $b \in R$ such that $ab = ba = 1$. This property can be easily proven by contradiction, assuming the existence of a multiplicative inverse for a divisor of zero and showing that it leads to a contradiction.</div><div><br></div><div><b>Examples</b></div><div><br></div><div>Let's look at some examples of divisors of zero in different rings:</div><div><br></div><div><b>Integers Modulo $n$</b></div><div><br></div><div>Consider the ring $\mathbb{Z}_6$ of integers modulo $6$. In this ring, both $2$ and $3$ are divisors of zero since $2 \cdot 3 = 3 \cdot 2 = 0 \pmod{6}$. These elements demonstrate that divisors of zero need not be zero themselves.</div><div><br></div><div><b>Matrix Rings</b></div><div><br></div><div>Matrix rings provide another interesting context to explore ring divisors of zero. Let's consider the ring of $2 \times 2$ matrices over a field $\mathbb{F}$, denoted as $\text{Mat}(2, \mathbb{F})$. In this ring, a matrix $A$ is a divisor of zero if there exists a nonzero matrix $B$ such that $AB = 0$.</div><div><br></div><div>$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ </div><div><br></div><div>To check whether $A$ is a divisor of zero, we can multiply $A$ by another nonzero matrix $B$ and see if the result is the zero matrix. Let's take</div><div>$B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$</div><div>Now, let's compute the product $AB$:</div><div><div class="separator" style="clear: both; text-align: center;">
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</div>$ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ $\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$</div><div>$=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}$</div><div><br></div><div>which is the zero matrix.</div><div><br></div><div>Therefore, in the ring $(2, \mathbb{F})$, the matrix $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ is a divisor of zero, as it annihilates the nonzero matrix $B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.</div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-18563388522769004332023-04-17T00:17:00.018-07:002023-04-18T01:12:37.306-07:00Accessing Machine Learning Impact In Mathematics - Trends and Application<div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgoMb39pZGfs8Rz6FfCnfwp3TsBGkW8tunJUYbvijhLIGWbOqP6fxfBPFLI77Z60Kk3B3m3Mj46qgUA6DX6RtdfHyfgUZYysqibgu-GoYPiaS3gs4pUBsfpwc_0UAbvd1byd0i_TDQaJOBpcfU_H9hBoNmHiY-c8LU9NZNDAWbMAgLc4I9y3TqkkHsqw/s447/ML.PNG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="361" data-original-width="447" height="258" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgoMb39pZGfs8Rz6FfCnfwp3TsBGkW8tunJUYbvijhLIGWbOqP6fxfBPFLI77Z60Kk3B3m3Mj46qgUA6DX6RtdfHyfgUZYysqibgu-GoYPiaS3gs4pUBsfpwc_0UAbvd1byd0i_TDQaJOBpcfU_H9hBoNmHiY-c8LU9NZNDAWbMAgLc4I9y3TqkkHsqw/s320/ML.PNG" width="320" /></a></div><br />Accessing Machine Learning Impact In Mathematics - Trends and Application</div><div><br /></div><div>Machine learning has revolutionized the field of mathematics, enabling mathematicians to develop highly effective models for data prediction. By applying machine learning algorithms to large datasets, mathematicians can now identify patterns and relationships that would be difficult or impossible to detect using traditional mathematical techniques.</div><div><br /></div><div>One area where machine learning has had a significant impact is in the field of statistics. In the past, statisticians relied on simple linear regression models to predict future trends based on historical data. However, these models often failed to capture the complexity and nuance of real-world data, resulting in inaccurate predictions.</div><div><br /></div><div>With the advent of machine learning, statisticians now have access to more sophisticated models that can account for multiple variables and nonlinear relationships. For example, support vector machines (SVMs) and neural networks are powerful machine learning algorithms that can be trained to recognize complex patterns in data and make highly accurate predictions.</div><div><br /></div><div>Another area where machine learning has been transformative is in the field of data analysis. By using unsupervised learning algorithms such as clustering and dimensionality reduction, mathematicians can now identify hidden patterns and relationships in large datasets that would be difficult to discern using traditional statistical techniques.</div><div><br /></div><div>In addition, machine learning has enabled mathematicians to develop more accurate and reliable models for image and signal processing. For example, deep learning algorithms such as convolutional neural networks (CNNs) can be trained to recognize patterns and features in images and signals, allowing mathematicians to develop highly effective models for tasks such as object recognition, speech recognition, and natural language processing.</div><div><br /></div><div>Machine learning has also enabled mathematicians to make significant strides in the field of optimization. By using reinforcement learning algorithms, mathematicians can now develop highly efficient algorithms for solving complex optimization problems, such as the traveling salesman problem or the knapsack problem.</div><div><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0Pretoria, South Africa-25.7478676 28.2292712-56.203816440191716 -6.9269788000000005 4.7080812401917171 63.3855212tag:blogger.com,1999:blog-9101299148361109184.post-405467904683760632022-02-28T07:14:00.002-08:002022-02-28T07:42:36.660-08:00From UDUS to KMUTT: The journey of a thousand miles ends with a PhD in Mathematics by Abubakar Muhammad ShafiiFrom UDUS to KMUTT: The journey of a thousand miles ends with a PhD in Mathematics by Abubakar Muhammad Shafii<div><br><div><div class="separator" style="clear: both; text-align: center;">
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</div><div><i>Dr. Ibrahim Hassan Abdulkareem</i></div><div><i><br></i></div><div>An anonymous speaker once said </div><div>"Excellence is never an accident; it is the result of high intention, sincere effort, intelligent direction, skillful execution and the vision to see obstacles as opportunities", in our world today, attaining excellence especially in academics can be quite challenging because of the less attention and reward attached to it, this I think should necessitate the need to celebrate people of excellence in our society at every little opportunity we have, people who have against all odds beaten the hurdles and come out victorious in their pursuit of academics, they are the real heroes and role models of the upcoming generation.</div><div><div class="separator" style="clear: both; text-align: center;">
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</div><i>P</i></a><i>hoto of Dr. Abdulkareem during one of his presentations</i></div><br></div><div>Ibrahim Hassan Abdulkareem also known as Akareem is a bonafide product of department of mathematics, Usmanu Danfodiyo University, Sokoto where he obtained his bachelor's degree under the mentorship and supervision of Dr. Aremu Kazeem publishing two papers in the process in the years 2017 and 2018. His undergraduate research was focused on Pattern Popularity of Gamma-1 non-deranged permutations. </div><div><br></div><div>Akareem, the newest PhD holder in town, has made the 105 family really proud being the first to attain this academic milestone. </div><div>Many who knew him before now are aware of his quite brilliant history, right from his primary, secondary to university education, Akareem secured more than 3 awards of academic excellence and 2 scholarship awards. Among his notable awards is the Bronze medal, National Mathematics competition for University students in 2017. He also secured the PTDF scholarship award in 2016 and the Petchra Pra Jom Klao PhD scholarship award at the king Mongkut University of Technology Thailand (KMUTT) in 2018.</div><div>Akareem's PhD research is based on numerical optimization with applications in image processing, supervised by the erudite Prof. Poom Kumam who is the head of Center of Excellence in Theoretical and Computational studies (TaCS-CoE) and KMUTT Fixed Point Theory and Applications Research Group.</div><div><div class="separator" style="clear: both; text-align: center;">
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</div><i>Dr.Abdulkareem and Prof. Poom kumam</i></div><div><br></div><div>Prof. Poom was ranked among the top 1000 researchers globally with top citations and h-index in February 2022 by researcher.com</div><div>One can easily see that Akareem has been lucky to pass through some of the best mentorships. His mentorship can not be complete without mentioning his pioneer mentor Prof. Taofeek Ibrahim who happens to be his father. </div><div>Prof. Ibrahim is an erudite professor of health science and a former university Vice-Chancellor in Nigeria. </div><div><br></div><div>At 28 years old, Akareem is one of the few Nigerians to obtain a PhD below the age of 30 making him stand in the halls of the pioneer ABU Vice-Chancellor Prof. Iya Abubakar who was also a mathematician.</div><div><div class="separator" style="clear: both; text-align: center;">
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</div><i>Photo of Dr. Abdulkareem at University Novi Sad,Serbia.</i></div><div><br></div><div>Akareem's PhD adventure saw him visit Erzurum Technical University, Erzurum, Turkey as a visiting graduate student and University of Novi Sad, Serbia under the hospitality and host of Prof. Dr. Sanja Rapajic .</div><div>He also attended conferences outside Thailand in Vietnam and Spain in the year 2019.</div><div>Akareem currently has about 40 publications to his name with all journals published in reputable journals around the world.</div><div><br></div><div>The Epe born scholar is not only an academic genius but a networker who understands the essence of building human relations with colleagues, coursemates and friends. This quality won him the chairmanship of the National Association of Mathematical Science Students of Nigeria (UDUS Chapter) in 2016/2017. His brilliant contribution as the chairman of the year book printing committee was a panacea of the success recorded during our set's yearbook album production. </div><div><br></div><div>The 105 unforgettable are really proud of your achievement and we pray to celebrate your professorship soon. </div><div>Congratulations Dr. Ibrahim Hassan Abdulkareem </div><div><div class="separator" style="clear: both; text-align: center;">
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</div><i>Photo of Dr. Abdulkareem in fixed point research lab, KMUTT</i></div><div><br></div><div>You can contact Ibrahim Hassan Abdulkareem via:</div><div>Email:</div><div>ibrahimkarym@gmail.com abdulkarim.hassan@mail.kmutt.ac.th</div></div></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-63207000676754160792021-12-22T03:32:00.001-08:002021-12-22T03:32:27.798-08:00How to Solve Fractional Equation<div class="separator" style="clear: both; text-align: center;">
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</div><div>Evaluate $x$ in the equation $\frac{x^2-10}{x-3}+\frac{1}{x-3}=\frac{1}{5}$</div><div><br></div><div><b>Solution</b></div><div>First we need to find the Lowest Common Multiple (LCM) of the left hand side.</div><div><br></div><div>LCM=$x-3$</div><div><br></div><div>$\frac{x^2-10+1}{x-3}=\frac{1}{5}$</div><div>$\frac{x^2-9}{x-3}=\frac{1}{5}$</div><div>Recall that $(x-3)(x+3)=(x^2-9)$</div><div>$\frac{(x-3)(x+3)}{x-3}=\frac{1}{5}$</div><div>$x+3=\frac{1}{5}$</div><div>$x=-\frac{14}{5}$</div><div><br></div><div><br></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-82630941057385670132021-10-20T12:55:00.002-07:002021-10-20T13:17:40.969-07:00Echoes of Victory from KMUTT - An epistle of accolades on Dr. Jamilu Abubakar<div>Echoes of Victory from KMUTT - An epistle of accolades on Dr. Jamilu Abubakar</div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;">
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</div><br></div><div><br></div><div>Academia is a field of patience and resilience, only those who are willing to pay the price for hardwork receive the medal of PhD. The struggle can be daunting but at the end only the fierce emerge with victory. </div><div>Dr. Jamilu Abubakar has gone through the rigorous test and has emerged victorious, his name now concretized and his pen refilled with the golden ink.</div><div>Jamilu Abubakar is not an unpopular name at the department of mathematics Usmanu Danfodiyo University, Sokoto, he is a fine, gentle man with prestige who adores hardwork. He is a first class product of the same department and has just completed a PhD with the King Mongkut University of Technology Thonbury (KMUTT) Thailand, under the supervison of the erudite Prof. Poom Kumam of fixed point research laboratory.</div><div>Before his PhD defence, Dr. Jamilu completed a six month research visit at the Technical University of Cluj-Napoca, Romania.</div><div><div class="separator" style="clear: both; text-align: center;">
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</div><br></div><div>My first contact with the doctor was in 2014 when I had a missing grade in one of my courses, he was soft-spoken and guided me on what to do, he will later be the lecturer assigned to teach my set Real Analysis I, and I can still remember his recommendation of Classical Real Analysis textbook by M. bruckner, the book was helpful to many of us who had time to read it, I can't imagine my analysis skills without his recommendations. Dr. jamilu was also assigned to teach us functional analysis during our final year in 2017, the course was stressful but the doctor spared out time to ensure we understood its rudiments. His advise, recommendations and mentorship has seen many of his students prospering in their various fields of life. I can not forget his contribution towards the supervision of my final year project. He ensured I wrote a research paper out of the project, that was my major introduction to mathematical research.</div><div>Mathematics class of 2017 is indeed grateful to have passed through the mentorship of a man like Jamilu Abubakar. We pray for this new degree to be beneficial to him and those who will come across him, we pray his works have an everlasting impact on science and humanity in general and lastly we pray for longevity so that he shall enjoy the fruits of his hard work. Congratulations to the newest Doctor.</div><div><br></div><div><br></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-64817826644601954042021-10-18T05:25:00.006-07:002021-10-18T05:38:14.814-07:00How to proof an injective(one-to-one) function<div><div class="separator" style="clear: both; text-align: center;">
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</div><br /></div><div><br /></div>Let $f:A\rightarrow{A}$ with $f(x)=\frac{x+2}{x-3}$ for $A=\mathbb{R}/\{1,3\}$<div>Show that $f$ is a one-to-one function.</div><div><br /></div><div>Proof: we are given $f:A\rightarrow{A}$ with $f(x)=\frac{x+2}{x-3}$ and we are to proof that $f$ is a one-to-one function. Recall that a one-to-one function is defined thus: $\forall{x,y}\in{A}(f(x)=f(y)\Rightarrow{x=y})$.</div><div>Let $x,y\in{A}$ with $f(x)=f(y)$</div><div>$\Rightarrow\frac{x+2}{x-3}=\frac{y+2}{y-3}$</div><div>$\Rightarrow(x+2)(y-3)=(x-3)(y+2)$</div><div>$\Rightarrow{xy-3x+2y-6=xy+2x-3y-6}$</div><div>$5y=5x\Rightarrow{y=x}$</div><div>Hence, $y=x$ satisfies our definition of a one-to-one function.</div><div><br /></div><div><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com1tag:blogger.com,1999:blog-9101299148361109184.post-74180415571786282082021-10-11T02:31:00.006-07:002021-10-13T04:52:05.630-07:00How to proof Inequalities<div><br /></div><div class="separator" style="clear: both; text-align: center;">
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</div><div><br /></div><div><div>Let $a,b\in\mathbb{R}$. Prove th inequality $\frac{a}{b}+\frac{b}{a}\geq{2}$</div><div><br /></div><div><b>Proof</b></div><div><br /></div><div>We are given $\frac{a}{b}+\frac{b}{a}\geq{2}$, .</div><div>Let $(a-b)^2\geq{0}$ so that $a^2-2ab+b^2\geq{0}\Leftrightarrow{a^2+b^2}\geq{2ab}$</div><div>$\Leftrightarrow\frac{a^2+b^2}{ab}\geq{2}$</div><div>$\Leftrightarrow\frac{a}{b}+\frac{b}{a}\geq{2}$</div><div>Equality holds if and only if $a-b=0$ or $a=b$.</div></div><div><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com2tag:blogger.com,1999:blog-9101299148361109184.post-49382021376331274232021-08-31T10:11:00.008-07:002021-09-01T14:51:34.554-07:00How to find areas of similar triangles <div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9NyGUSTG9Sv64L5DpV9pPl_Kon554kMq-PVGPn0I3vxcILmJeR9zRgdZHO1o_-Sg-7ktpwPsxsiVNIKNwiC25Tl2EYK_3MYMF_mBV1ikkptAsDze5GVaS-HGT3oTn5SlKU-xTPLSWwkY-/s1600/New+Doc+68_2.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1640" data-original-width="1520" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9NyGUSTG9Sv64L5DpV9pPl_Kon554kMq-PVGPn0I3vxcILmJeR9zRgdZHO1o_-Sg-7ktpwPsxsiVNIKNwiC25Tl2EYK_3MYMF_mBV1ikkptAsDze5GVaS-HGT3oTn5SlKU-xTPLSWwkY-/s1600/New+Doc+68_2.jpg" width="296" /></a></div><br /></div>a) We are to find area of $\bigtriangleup{ABC}$<div>This is quite simple, all we need to do is to use the idea of similar triangles, to do this, simply extract the $\bigtriangleup{AMN}$ from $\bigtriangleup{ABC}$ so that we now have two similar triangles.</div><div>From this, it is evident that $AM=6cm$ and $AB=6+4=10cm$. $AM$ and $AB$ are corresponding sides therefore</div><div>$\frac {AM}{AB}=\frac{6}{10}=\frac{3}{5}$ is the scale factor. </div><div>Next we move to obtaining the ratio of the areas of the triangles known as area factor, </div><div>Area factor=$(scaler factor)^2=(\frac{3}{5})^2=\frac{9}{25}$</div><div>From the area factor we now obtain the area for $\bigtriangleup{ABC}$:</div><div>$\frac{Area of \bigtriangleup{AMN}}{Area of \bigtriangleup{ABC}}=\frac{9}{25}$</div><div>Area of $\bigtriangleup{AMN}=12cm^2$ is given from the figure, </div><div>$\frac{12}{Area of \bigtriangleup{ABC}}=\frac{9}{25}$</div><div>$Area of \bigtriangleup{ABC}=\frac{12\times{25}}{9}=\frac{300}{9}=33.3cm^2$</div><div><br /></div><div>b) Area of $MNCB$, $MNCB$ is not a triangle therefore we obtain its area from the area of the triangles in the shape. </div><div>Since the area of $\bigtriangleup{AMN}=12cm^2$ and the area of $\bigtriangleup{ABC}=33.3cm^2$ then we can obtain the area of $MNCB$ by simply subtracting area of the smaller triangle from the area of the bigger triangle. </div><div>=$33. 3-12=21.3cm^2$</div><div><br /></div><div><span style="color: red;">Reference:</span></div><div>New concept mathematics, JSS III, exercise 13.1, for Nigeria schools </div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-26636417093873413152021-07-26T03:43:00.004-07:002021-07-26T03:53:55.912-07:00How to proof for primes <div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_8lLI9hoH6W1L0-RP8pojPRXst0RSzHn9KLZUrhu-KaSTQjO_vXSYDnb1jEWlkKGnKk2eDR14aVMK03R0AAz2mtEVFLorjkxhxVqBnVKUD2WbYJ8O66jB-02eJ9a4x9CWB2pD5fDbZFTR/s1600/PhotoGridLite_1627295941019.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_8lLI9hoH6W1L0-RP8pojPRXst0RSzHn9KLZUrhu-KaSTQjO_vXSYDnb1jEWlkKGnKk2eDR14aVMK03R0AAz2mtEVFLorjkxhxVqBnVKUD2WbYJ8O66jB-02eJ9a4x9CWB2pD5fDbZFTR/s1600/PhotoGridLite_1627295941019.jpg" width="320" /></a></div><br /></div><h2 style="text-align: left;">Proof that for any integer $n>1$ the number $n^5+n^4+1$ is not a prime.</h2><div>Simply rewrite $n^5+n^4+1$ as $n^5+n^4+n^3-n^3-n^2-n+n^2+n+1$</div><div>Factor out common terms </div><div>$n^3(n^2+n+1)-n(n^2+n+1)+(n^2+n+1)$</div><div>$(n^3-n+1)(n^2+n+1)$</div><div>The product of two integers greater than $1$, hence $n^5+n^4+1$ is not a prime. </div><div><br /></div><div><br /></div><h2 style="text-align: left;">Reference </h2><div style="text-align: left;"><span style="color: red;">Andreescu, T., & Andrica, D. "<i>Number Theory Structures, Examples, and Problems</i>" page 25</span>.</div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com1tag:blogger.com,1999:blog-9101299148361109184.post-11482626298587740802021-07-24T06:45:00.006-07:002021-07-24T06:56:18.503-07:00Find the first term and common difference of the arithmetic sequence<div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibn807d4frNnWb-RSklyYjayuJ2zlZvC1_bwYQx4e4GSpZWRfW088oWdg0vvN0J_Q0OV7-k1fbywqbfMAgY-uUD6WdmR9GpWT7FR3q7qpiB0J3_e-2PNWb2-4IeBU1u5EUHqlL0RHIR6pb/s1600/PhotoGridLite_1627134173736.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibn807d4frNnWb-RSklyYjayuJ2zlZvC1_bwYQx4e4GSpZWRfW088oWdg0vvN0J_Q0OV7-k1fbywqbfMAgY-uUD6WdmR9GpWT7FR3q7qpiB0J3_e-2PNWb2-4IeBU1u5EUHqlL0RHIR6pb/s1600/PhotoGridLite_1627134173736.jpg" width="320" /></a></div><br /></div><h2 style="text-align: left;">The product of the third and the sixth terms of an arithmetic sequence is $406$. The ninth term of the sequence divided by the fourth term gives a quotient of $2$ and a remainder of $6$. Find the first term and the common difference of the arithmetic sequence.</h2>
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<div>Let $\{a_n\}$ be the arithmetic sequence so that we can construct an equation for each of the statements above.</div><div><br /></div><div>Let $a_3$ be the third term of the sequence and $a_6$ be the sixth term of the sequence.</div><div>Then their product is </div><div>$a_3.a_6=406$...β¦β¦β¦....β¦β¦β¦β¦ (1)</div><div>And</div><div>Let $a_9$ be the ninth term of the sequence, use the formula, Dividend=Divisor.Quotient+Remainder </div><div>$a_9=2.a_4+6$ β¦....β¦...β¦β¦β¦β¦....(2)</div><div>Rewrite (1) as</div><div>$(a_1+2d)(a_1+5d)=406$........β¦..β¦....β¦β¦..(3)</div><div>And rewrite (2) as </div><div>$a_1+8d=(a_1+3d)2+6$β¦....β¦β¦β¦β¦......... (4)</div><div>Simplify (4)</div><div>$a_1+8d=2a_1+6d+6$</div><div>$a_1=2d-6$β¦.β¦....β¦β¦....β¦β¦.(5)</div><div>Substitute (5) into (3) to get</div><div>$(4d-6)(7d-6)=406$</div><div>$14d^2-33d-185=0$</div><div>Solve the equation using quadratic formula </div><div>$d=5$ and $d=\frac{-37}{28}$</div><div>Two different values for a common difference will give us two different arithmetic sequences:</div>
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<div>For $d=5$ in (5)</div><div>$a_1=4$</div><div>And for $d=\frac{-37}{28}$ in (5)</div><div>$a_1=\frac{-79}{7}$</div><div><b>QED. </b></div><div><br /></div><h3 style="text-align: left;"><b>References</b> </h3><div><span style="color: red;">Ellina Grigoriva (2016), Methods of Solving Sequence and Series Problems, Springer international publishing AG. Page 15</span></div><div><br /></div><div><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-79360987106767988062021-07-19T14:20:00.006-07:002021-07-19T15:19:30.260-07:00Find the sum of all natural numbers between 1 and 1000 that are not divisible by 13<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWg45_421dB1cwsR1zb5kLo-723GAiBSUh1k2rs3YGZsZfvws419KW8w_RTMQiMHvCcW4awikGCc0k3bz2cByXyxcHQaGV7nPPdOVvNF5MQUIsRbnqdyQM_CBKMOLYJ9xOeYaH8vAvGHeV/s1600/PhotoGridLite_1626729512926.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWg45_421dB1cwsR1zb5kLo-723GAiBSUh1k2rs3YGZsZfvws419KW8w_RTMQiMHvCcW4awikGCc0k3bz2cByXyxcHQaGV7nPPdOVvNF5MQUIsRbnqdyQM_CBKMOLYJ9xOeYaH8vAvGHeV/s1600/PhotoGridLite_1626729512926.jpg" width="320" /></a></div><h2 style="clear: both; text-align: center;">Find the sum of all natural numbers between $1$ and $1000$ that are not divisible by $13$.</h2><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Let $S$ be the unknown sum and $M$ multiples of $13$ less than $1000$, then $S$ can be written in the form $S=S_{1000}-M$, where $S_{1000}$ is the sum of natural numbers between $1$ and $1000$ therefore $S_{1000}=1+2+3+...+999+1000$ with $a_1=1$ and $a_{1000}=1000$ </div><div class="separator" style="clear: both; text-align: center;">Let the $nth$ term be</div><div class="separator" style="clear: both; text-align: center;">$S_n=\frac{a_1+a_n}{2}.n$....................... (i)</div>
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<div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">$S_n=\frac{1+1000}{2}.1000=500500$</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">We now proceed to solving for multiples of $13$, let $13k$ represent all multiples of $13$ and let there sum be </div><div class="separator" style="clear: both; text-align: center;">$M=13(1)+13(2)+13(3)+...+13(k) =13.(1+2+3+...+k)$............................(ii)</div><div class="separator" style="clear: both; text-align: center;">then</div><div class="separator" style="clear: both; text-align: center;">$13k\leq{1000}$</div><div class="separator" style="clear: both; text-align: center;">$k\leq\frac{1000}{13}=76.923$</div><div class="separator" style="clear: both; text-align: center;">This means that the greatest natural number $k$ satisfying the inequality is $76$.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Recall that $M=13.(1+2+3+...+k)$</div><div class="separator" style="clear: both; text-align: center;">We are going to compute the $nth$ term of $M$ by simply Multiplying (1) and (2)</div><div class="separator" style="clear: both; text-align: center;">$M=13.\frac{a_1+a_k}{2}.n$</div><div class="separator" style="clear: both; text-align: center;">$M=13.\frac{1+76}{2}.76=13.\frac{77}{2}.77=38038$</div><div class="separator" style="clear: both; text-align: center;">Subtract $M$ from $S_n$ to obtain $S$</div><div class="separator" style="clear: both; text-align: center;">$S=500500-38038=462462$. The sum is therefore equal to $462462$.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><h2 style="clear: both; text-align: center;">
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<b>References</b> </h2><div class="separator" style="clear: both; text-align: center;"><span style="color: red;">Ellina Grigoriva (2016), Methods of Solving Sequence and Series Problems, Springer international publishing AG. Pages 12-13</span></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-44747974716900515862021-07-18T11:29:00.001-07:002021-07-18T11:31:08.599-07:00Lecture series by Dr Kamran Khan on Complex Integration<p> Lecture series by Dr Kamran Khan on Complex Integration</p>
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<p>Kamran Khan is an assistant professor at the department of mathematics, Aligarh Muslim University, Aligarh India. Kamran Khan's video will be posted from time to time for the benefits of students of complex analysis.</p><p><br /></p><p>You can subscribe to his channel on YouTube @ <a href="https://youtu.be/9WvguojSx2Q" target="_blank">Dr.Kamran Khan </a></p><p><br /></p>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-73956959191961779142021-07-16T10:25:00.004-07:002021-07-16T10:36:15.768-07:00How to solve hyperbolic trigonometric equations <div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFs4IY8DxSBVGy3SHjb8aKrK8sdo2-7WXMhxUnMonsI6uMTR8oIEymRFACtnkLoQ5m91u5voSLJ36nS2wl1Vs9Vg5gNiY8CCQkz9KHMq-cVI4iaKUHvoZyPnqAEbq6JJLA36Wt-pjxGBEJ/s1600/PhotoGridLite_1626456210220.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFs4IY8DxSBVGy3SHjb8aKrK8sdo2-7WXMhxUnMonsI6uMTR8oIEymRFACtnkLoQ5m91u5voSLJ36nS2wl1Vs9Vg5gNiY8CCQkz9KHMq-cVI4iaKUHvoZyPnqAEbq6JJLA36Wt-pjxGBEJ/s1600/PhotoGridLite_1626456210220.jpg" width="320" /></a></div><br /></div><div><br /></div><div><span style="color: red;">Solve the equation $2\cosh{2x}+10\sinh{2x}=5$</span>................. π</div><div><br /></div><div><span style="color: #2b00fe;">This is a hyperbolic trigonometric function equation, in order to solve the given equation, we must represent the hyper trig in exponential form, this is a regular JAMB and WAEC question on trig. </span></div><div><span style="color: #2b00fe;"><br /></span></div><div><span style="color: #2b00fe;">See the exponential forms of each of the identities below:</span>π</div><div>$\cosh{2x}=\frac{1}{2}(e^{2x}+e^{-2x})$ and </div><div>$\sinh{2x}=\frac{1}{2}(e^{2x}-e^{-2x})$.</div><div><br /></div><div><span style="color: #2b00fe;">substitute the exponential form into the equation </span>(π) </div>
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<div>$2.\frac{1}{2}(e^{2x}+e^{-2x})+10.\frac{1}{2}(e^{2x}-e^{-2x})=5$</div><div>$e^{2x}+e^{-2x}+5e^{2x}-5e^{-2x}=5$</div><div><br /></div><div>collect the like terms</div><div>$6e^{2x}-4e^{-2x}=5$</div><div>$6e^{2x}-4e^{-2x}-5=0$</div><div><span style="color: #2b00fe;">We are going to solve this equation by algebraic substitution because we want to represent the equation as a quadratic, it is easier to solve that way, but to do that, multiply through by $e^{2x}$</span></div><div><br /></div><div>$6e^{4x}-5e^{2x}-4=0$</div><div>$6e^{(2x)2}-5e^{2x}-4=0$</div><div><br /></div><div>Now let $e^{2x}=p$ so that </div><div>$6p^2-5p-4=0$</div><div><br /></div><div><span style="color: #2b00fe;">solving the equation quadratically yields</span> </div><div><br /></div><div>$p=-\frac{1}{2},\frac{4}{3}$</div><div>that also means $e^{2x}=-\frac{1}{2}$ and $e^{2x}=\frac{3}{4}$</div><div>looking at the roots, the only real solution is $e^{2x}>0$</div><div>Therefore,</div><div> $e^{2x}=\frac{4}{3}$</div><div><br /></div><div><span style="color: #2b00fe;">Introduce a $\ln$ to both sides to cancel the $e$ in the left hand side. </span></div><div><br /></div><div>$2x=\ln\frac{4}{3}$</div><div>And therefore </div><div>$x=\frac{1}{2}\ln\frac{4}{3}$.</div><div><br /></div><div><b>Which brings us to the end of the solution.</b></div><div><b>Thank you for reading, hope you've leaned something new, don't forget to subscribe to receive email notifications. </b></div><div><b>How do you find our use of colours for formatting texts?</b></div><div><b>Kindly drop a comment for observation and contribution. </b></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com1tag:blogger.com,1999:blog-9101299148361109184.post-50055903703694988962021-07-15T07:40:00.005-07:002021-07-15T07:50:54.844-07:00How to solve quadratic Diophantine equation problem<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX0SlNhx3LltaVUXruHXwL-kZ5CpriPGinGdHlUvN6-ZrWj-FKGw45UG3u4U6mJE-E_f7ZdU0ep0YgYxJLEBP3jfPdRPS8Rlkm2FSzNN2Yi73GLCgDrWaKwaIh1AnW1T-kv4mG1FLw2-XU/s1600/PhotoGridLite_1626359862257.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" class="imageResizeTarget" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX0SlNhx3LltaVUXruHXwL-kZ5CpriPGinGdHlUvN6-ZrWj-FKGw45UG3u4U6mJE-E_f7ZdU0ep0YgYxJLEBP3jfPdRPS8Rlkm2FSzNN2Yi73GLCgDrWaKwaIh1AnW1T-kv4mG1FLw2-XU/s1600/PhotoGridLite_1626359862257.jpg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both;">Find the integer solution $x^2+3x+9=9y^2$.............(1)</div><div class="separator" style="clear: both;"><br /></div>
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<div class="separator" style="clear: both;">This is a beautiful quadratic Diophantine's equation that requires an integer solution. So how do we go from here? πππ</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">I'm going to try as much as possible to make the solution easily comprehensible for those who might find it a bit confusing or difficult.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">Looking at the equation closely, you will realize the left hand side is just our normal everyday quadratic equation which can easily be factored. The right hand side is also a quadratic, but the problem is, the whole equation has two variables.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">We are going to solve this problem by treating the whole equation as a single quadratic equation, so move $9y^2$ to the left hand side so that you can equate the equation to zero.</div>
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<div class="separator" style="clear: both;">$x^2+3x+9-9y^2=0$</div><div class="separator" style="clear: both;">Use quadratic formula to find the factors,</div><div class="separator" style="clear: both;">Let $a=1,b=3,c=9-9y^2$</div><div class="separator" style="clear: both;">$x=\frac{-3\pm\sqrt{9-4(9-9y^2)}}{2}$</div><div class="separator" style="clear: both;">$x=\frac{-3\pm\sqrt{9-36+36y^2}}{2}$</div><div class="separator" style="clear: both;">$x=\frac{-3\pm\sqrt{36y^2-27}}{2}$</div><div class="separator" style="clear: both;">Factor our $9$</div><div class="separator" style="clear: both;">$x=\frac{-3\pm\sqrt{9(4y^2-3)}}{2}$</div><div class="separator" style="clear: both;">$x=\frac{-3\pm{3}\sqrt{(4y^2-3)}}{2}$......................(2)</div><div class="separator" style="clear: both;">Now that we have arrived at this, what do we do next? This is where it gets confusing and boring ππ€. </div><div class="separator" style="clear: both;">All we need to do is some algebraic manipulations, so your method might be different from mine, but what matters is the goal(integers).</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">What I'm going to do now is,</div><div class="separator" style="clear: both;">Let $\sqrt{4y^2-3}=k$ so that $4y^2-3=k^2$...............(3)</div><div class="separator" style="clear: both;">Therefore, $x=\frac{-3\pm{3k}}{2}$.</div><div class="separator" style="clear: both;">Rearrange (3) so it forms an equation with difference of two squares.</div><div class="separator" style="clear: both;">$4y^2-k^2=3$</div><div class="separator" style="clear: both;">$(2y-k)(2y+k)=3$</div><div class="separator" style="clear: both;">All we need to do now is find integers for which we will equate the factors to find $y$.</div>
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<div class="separator" style="clear: both;">We will equate each of the factors to $(1,3,-1,-3)$.</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">So equate $2y-k=1$ and $2y+k=3$</div><div class="separator" style="clear: both;">Sloving simultaneously for $y$ yields π</div><div class="separator" style="clear: both;">$y=1$</div><div class="separator" style="clear: both;">Also, equate $2y-k=-1$ and $2y+k=-3$ and solve simultaneously for $y$ to get π</div><div class="separator" style="clear: both;">$y=-1$</div><div class="separator" style="clear: both;">Now there are two values of $y=1,-1$ we will plug in these values of $y$ into equation to obtain the integer values of $x$.</div><div class="separator" style="clear: both;">πFor $y=1$</div><div class="separator" style="clear: both;">$x^2+3x+9=9(1)^2$</div><div class="separator" style="clear: both;">$x^2+3x+9-9=0$</div><div class="separator" style="clear: both;">$x^2+3x=0$</div><div class="separator" style="clear: both;">Solve quadratically to get π</div><div class="separator" style="clear: both;">$x=0,-3$</div><div class="separator" style="clear: both;"><br /></div><div class="separator" style="clear: both;">π For $y=-1$</div><div class="separator" style="clear: both;">$x^2+3x+9=9(-1)^2$</div><div class="separator" style="clear: both;">$x^2+3x+9-9=0$</div><div class="separator" style="clear: both;">$x^2+3x=0$</div><div class="separator" style="clear: both;">Solve quadratically to get π</div><div class="separator" style="clear: both;">$x=0,-3$</div><div class="separator" style="clear: both;">Hence, we conclude the integer solutions for the equation for both $x$ and $y$ are π</div><div class="separator" style="clear: both;">$x=0,-3$ and $y=1,-1$</div><div class="separator" style="clear: both;">And that's it. πππππππππ</div><div class="separator" style="clear: both;">Thank you for reading, do subscribe to our blog to receive emails on upcoming post and don't forget to leave a comment or observation. </div></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-44941460935881354212021-07-13T07:46:00.005-07:002021-07-13T07:59:00.663-07:00Find the value of x <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3CMcYx8UMjU9JrEdqgXGp2JA0UgPIqkJs9idogBxKM15WZoKF8Jg9ryIWtTl283JJOf9JfgDo9x4CMrAvzcWjOZRctzIfT2pqebBRffGTLr8j91um5UIeeXB0lORaTWNL1tRKlkUR61WT/s1600/PhotoGridLite_1626187049256.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3CMcYx8UMjU9JrEdqgXGp2JA0UgPIqkJs9idogBxKM15WZoKF8Jg9ryIWtTl283JJOf9JfgDo9x4CMrAvzcWjOZRctzIfT2pqebBRffGTLr8j91um5UIeeXB0lORaTWNL1tRKlkUR61WT/s1600/PhotoGridLite_1626187049256.jpg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Find the value of $x$</div><div class="separator" style="clear: both; text-align: center;">$\frac{x^{\sqrt{x}}} {\sqrt{x}}=\frac{{\sqrt{x}}^x}{\sqrt{x}}$</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Since both sides of the equation have a common denominator, multiply through by $\sqrt{x}$. </div><div class="separator" style="clear: both; text-align: center;">$x^\sqrt{x}=\sqrt{x}^x$</div><div class="separator" style="clear: both; text-align: center;">$x^{x^\frac{1}{2}}=(x^\frac{1}{2})^x$</div><div class="separator" style="clear: both; text-align: center;">$x^{x^\frac{1}{2}}={x^{\frac{1}{2}}}^x$</div><div class="separator" style="clear: both; text-align: center;">π€ Remember the axiom of indices with same bases on both sides of equation, applying that axiom yields, </div><div class="separator" style="clear: both; text-align: center;">π $x^\frac{1}{2}=\frac{1}{2}x$</div><div class="separator" style="clear: both; text-align: center;">Take the squares of both sides </div><div class="separator" style="clear: both; text-align: center;">$x=(\frac{1}{2})^2$</div><div class="separator" style="clear: both; text-align: center;">$x=\frac{1}{4}x^2$</div><div class="separator" style="clear: both; text-align: center;">This becomes a quadratic equations </div><div class="separator" style="clear: both; text-align: center;">$4x=x^2$</div><div class="separator" style="clear: both; text-align: center;">Rearrange the equation and solve quadratically using any method you like. </div>
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<div class="separator" style="clear: both; text-align: center;">$x^2-4x=0$</div><div class="separator" style="clear: both; text-align: center;">Therefore, </div><div class="separator" style="clear: both; text-align: center;">π $x=0,4$</div><div class="separator" style="clear: both; text-align: center;">And that brings us to the end of the solution. </div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">Thank you for visiting Mymathware, we hope to see your comments or observations regarding the solutions to this problem. </div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-51759782308148836732021-07-12T07:29:00.007-07:002021-07-13T02:42:43.842-07:00Find the integer solutions, 3Λ£+27ΚΈ+81αΆ»=10935 <div><b><div class="separator" style="clear: both; text-align: center;"><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjazqU0GHrGw8K-J6OeIUib0R2OGS-8pxZUqFB_OWiSScOQ_MoXih8lfFryztsGALWleYd9vS1KgpwX4-JCyeSSJNG0TKnVz-cjBBoY0RVjeI4ulUcMxdMB_YR0q-woA1A-QjVhskz7IMzB/s1600/PhotoGridLite_1626100977094.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2048" data-original-width="2048" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjazqU0GHrGw8K-J6OeIUib0R2OGS-8pxZUqFB_OWiSScOQ_MoXih8lfFryztsGALWleYd9vS1KgpwX4-JCyeSSJNG0TKnVz-cjBBoY0RVjeI4ulUcMxdMB_YR0q-woA1A-QjVhskz7IMzB/s1600/PhotoGridLite_1626100977094.jpg" width="320" /></a></div><br /></div></b></div><b>Find the integer solution for $3^x+27^y+81^z=10935$</b><div><br /></div><div>This is a rather thickish question so we have to be very careful in our simplifications.</div><div>The left hand side of the equation can be expressed in base $3$, so we must look for a way to express the right hand side in base $3$ also.</div><div>Let's take a walk.</div><div><br /></div><div>π $3^x+3^{3y}+3^{4z}=2187\times{5}$</div><div>$3^x+3^{3y}+3^{4z}=3^7\times{5}$</div><div>Factor out $3^x$ from the left hand side </div><div>$3^x(1+3^{3y-x}+3^{4z-x})=3^7(5)$</div><div>We can now equate the right hand side with the left hand side.</div>
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<div>π $3^x=3^7$</div><div>π$x=7$</div><div>And</div><div>$1+3^{3y-x}+3^{4z-x}=5$</div><div>Collect like terms</div><div>$3^{3y-x}+3^{4z-x}=4$</div><div>Both sides of the equation should be expressed in base 3 so that we can find $y$ and $z$.</div><div>$3^{3y-x}+3^{4z-x}=3+1$</div><div>$3^{3y-x}+3^{4z-x}=3^1+3^0$</div><div>There are two cases to finding $y$ and $z$</div><div>π CASE I</div><div>$3^{3y-x}=3^1$</div><div>$3y-x=1$</div><div>Substitute $x=7$</div><div>$y=\frac{8}{3}\notin\mathbb{Z}$</div><div>And</div><div>$3^{4z-x}=3^0$</div><div>$4z-x=0$</div><div>Substitute $x=7$</div><div>$z=\frac{7}{4}\notin\mathbb{Z}$</div><div><br /></div><div>πCASE II</div><div>$3^{3y-x}=3^0$</div><div>$3y-x=0$</div><div>Substitute $x=7$</div><div>$y=\frac{7}{3}\notin\mathbb{Z}$</div><div>And </div><div>$3^{4z-x}=3^1$</div><div>$4z-x=1$</div><div>Substitute $x=7$</div><div>$z=\frac{8}{4}=2\in\mathbb{Z}$</div>
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<div>Hence, there are no integer solution for $3^x+27^y+81^z=10935$ since only $x=7$ and $z=2$ are integers.</div><div><br /></div><div>You might use a different approach to solve the question, if you are getting a different answer or you feel my method is inappropriate, kindly drop a comment/observation in the comment section. Thank you. </div><div><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com3tag:blogger.com,1999:blog-9101299148361109184.post-65143175441030084132021-07-11T10:42:00.006-07:002021-07-11T11:25:44.690-07:00Solving logarithmic equations with different bases <br /><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEileHfmKONcfKK3HrqgKCDoIaWTCyExHm3hvLZ3Pqp6NKuCYGtQSKihsT0JCaeiA42PX3lTaLaiw6Vfdn9gtdEuuhnsYiyEzP2mfF2G1yRcpus1YB_p2JSPYdJt9rDXtTCMMBWhDN3WWu2k/s608/ghhv.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="207" data-original-width="608" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEileHfmKONcfKK3HrqgKCDoIaWTCyExHm3hvLZ3Pqp6NKuCYGtQSKihsT0JCaeiA42PX3lTaLaiw6Vfdn9gtdEuuhnsYiyEzP2mfF2G1yRcpus1YB_p2JSPYdJt9rDXtTCMMBWhDN3WWu2k/s320/ghhv.png" width="320" /></a></div><b>olve the logarithmic equations</b></div><div class="separator" style="clear: both; text-align: center;"><b>$\log_3(x+1)=\log_4(x+8)$</b></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">The bases of this equation is different so</div><div class="separator" style="clear: both; text-align: center;">Let $\log_3(x+1)=y$...............................(i)</div><div class="separator" style="clear: both; text-align: center;">And</div><div class="separator" style="clear: both; text-align: center;">Let $\log_4(x+8)=y$...............................(ii)</div><div class="separator" style="clear: both; text-align: center;">By the laws of logarithm </div><div class="separator" style="clear: both; text-align: center;">$3^y=x+1$...............................(iii)</div><div class="separator" style="clear: both; text-align: center;">$4^y=x+8$...............................(iv)</div><div class="separator" style="clear: both; text-align: center;">Solve eqn(iv) and (iii) simultaneously</div><div class="separator" style="clear: both; text-align: center;">$4^y-3^y=7$...............................(v)</div>
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<div class="separator" style="clear: both; text-align: center;">Find all the possible values for which equation (v) is true.</div><div class="separator" style="clear: both; text-align: center;">The equation is true if $y=2$</div><div class="separator" style="clear: both; text-align: center;">$4^2-3^2=16-9=7$</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><i>You can check your solution by back substituting $y=2$ into equation (iii) and (iv). </i></div><div class="separator" style="clear: both; text-align: center;"><i><br /></i></div><div class="separator" style="clear: both; text-align: center;"><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-4272830450029032932021-07-10T04:38:00.010-07:002021-07-11T11:05:05.320-07:00How to solve indicial equations with different bases<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgric9vOxB5D4o_wSRCd5CUys6LBjCq8XzS7TmCcKtIz8WkfGkvBESWq0haNS-N8oMYGhAzddKMOlzZVZfGu3iHFroo5S4SED8owiCiVdK9bqSVdlRTpC6e72OyrCTZiaYvUPmSkofriyt-/s1600/c+ifs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" class="imageResizeTarget imageResizeTarget" data-original-height="486" data-original-width="972" height="160" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgric9vOxB5D4o_wSRCd5CUys6LBjCq8XzS7TmCcKtIz8WkfGkvBESWq0haNS-N8oMYGhAzddKMOlzZVZfGu3iHFroo5S4SED8owiCiVdK9bqSVdlRTpC6e72OyrCTZiaYvUPmSkofriyt-/s1600/c+ifs.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Solve the indicial equation</div><div class="separator" style="clear: both; text-align: center;">$4^\frac{1}{x}-6^\frac{1}{x}=9^\frac{1}{x}$</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">To solve the equation, simply divide through by any of the term since the indices are all given in different bases. So we divide by $9^\frac{1}{x}$.</div><div class="separator" style="clear: both; text-align: center;">$\frac{4^\frac{1}{x}-6^\frac{1}{x}}{9^\frac{1}{x}}=1$</div><div class="separator" style="clear: both; text-align: center;">$(\frac{4}{9})^\frac{1}{x}-(\frac{6}{9})^\frac{1}{x}=1$</div><div class="separator" style="clear: both; text-align: center;">Express the bases in homogenous form.</div><div class="separator" style="clear: both; text-align: center;">$(\frac{2^2}{3^2})^\frac{1}{x}-(\frac{2}{3})^\frac{1}{x}=1$</div><div class="separator" style="clear: both; text-align: center;">$(\frac{2}{3})^\frac{2}{x}-(\frac{2}{3})^\frac{1}{x}=1$</div>
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<div class="separator" style="clear: both; text-align: center;">Solve the equation using substitution method. Let $p=(\frac{2}{3})^\frac{1}{x}$................................(*)</div><div class="separator" style="clear: both; text-align: center;">The equation is now a quadratic equation</div><div class="separator" style="clear: both; text-align: center;">$p^2-p=1$</div><div class="separator" style="clear: both; text-align: center;">Solve the equation using quadratic formula method and you will obtain two roots of the form</div><div class="separator" style="clear: both; text-align: center;">$p_{1,2}=\frac{1\pm\sqrt{5}}{2}$</div><div class="separator" style="clear: both; text-align: center;">Substitute back the value of $p_1=\frac{1+\sqrt{5}}{2}$ into equation (*) and find $x$. </div><div class="separator" style="clear: both; text-align: center;">$(\frac{2}{3})^\frac{1}{x}=\frac{1+\sqrt{5}}{2}$</div><div class="separator" style="clear: both; text-align: center;">Take the $\log$ of both sides so that we can make $x$ subject of formula. </div><div class="separator" style="clear: both; text-align: center;">$\frac{1}{x}\log(\frac{2}{3})=\log(\frac{1+\sqrt{5}}{2})$</div><div class="separator" style="clear: both; text-align: center;">$x=\frac{\log\frac{2}{3}}{\log(\frac{1+\sqrt{5}}{2})}=\frac{\log2-\log3}{\log(1+\sqrt{5})-\log2}$. </div><div class="separator" style="clear: both; text-align: center;">$\approx{-0.842591738...}$</div><div class="separator" style="clear: both; text-align: center;">There is no real value for $p_2$ because $p_2=\frac{1-\sqrt{5}}{2}<0$</div>
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<div class="separator" style="clear: both; text-align: center;"><br /></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-5647152778095375102021-06-13T06:51:00.002-07:002021-06-13T06:55:38.883-07:00Solving quadratic equations using the Poh-Shen Log method.<b>Solving quadratic equations using the Poh-Shen Log method.</b><div><b><br /></b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXQBo3DGFWsQOwUvg9ekwMkLvPGE-T8d0tTavhkePdz7iS0xeHKFTEyruVDaE6T6nqlNEpPrsL7-rOOoVzzn0_OzwcBBf0dNtgYlpnj8m4MfTxn_yncmUMeFW8I7JgC_U2qKAhz5omZ-Oe/s1600/vhjb.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" class="imageResizeTarget" data-original-height="486" data-original-width="972" height="160" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXQBo3DGFWsQOwUvg9ekwMkLvPGE-T8d0tTavhkePdz7iS0xeHKFTEyruVDaE6T6nqlNEpPrsL7-rOOoVzzn0_OzwcBBf0dNtgYlpnj8m4MfTxn_yncmUMeFW8I7JgC_U2qKAhz5omZ-Oe/s1600/vhjb.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">The Poh-Shen Loh method is another method for solving quadratic equations, the method was pioneered by Prof. Poh Shen-Loh, a professor of mathematics at the Carnegie Mellon University. Although the method was invented by ancient Babylonians, Poh Shen-Loh took it upon himself to refurbish and upgrade the method. The method although looks like an extension of the completing the square method and has been tested to work for all conditions of quadratics.<br /></div>
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</script><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><b><i>Solve $x^2-6x+8=0$ using the Poh Shen-Loh method.</i></b></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;">We are going to factor the equation into</div><div class="separator" style="clear: both; text-align: center;">$x^2-6x+8=(x-x_1)(x-x_2)$</div><div class="separator" style="clear: both; text-align: center;">$x^2-(x_1+x_2)x+x_1x_2$</div><div class="separator" style="clear: both; text-align: center;">Where:</div><div class="separator" style="clear: both; text-align: center;">Sum: $x_1+x_2=6$</div><div class="separator" style="clear: both; text-align: center;">Product: $x_1x_2=8$</div><div class="separator" style="clear: both; text-align: center;">Now we obtain the midpoint from the sum: $\frac{x_1+x_2}{2}=\frac{6}{2}=3$</div><div class="separator" style="clear: both; text-align: center;">Let $u$ be the distance between $x_1$ and $3$ and between $3$ and $x_2$.</div><div class="separator" style="clear: both; text-align: center;">Let $x_1=3-u$ and $x_2=3+u$</div><div class="separator" style="clear: both; text-align: center;">And $x_1x_2=8\Rightarrow(3-u)(3+u)=8$</div><div class="separator" style="clear: both; text-align: center;">$9-u^2=8\Rightarrow{u^2=9-1=1}$</div><div class="separator" style="clear: both; text-align: center;">Now that we have obtained the value of $u=1$, substitute $u$ into $x_1$ and $x_2$.</div><div class="separator" style="clear: both; text-align: center;">$x_1=3-u=3-1=2$ and $x_2=3+u=3+1=4$.</div><div class="separator" style="clear: both; text-align: center;">Hence our root is $x_{1,2}=2,4$.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><b><i>Now let's take another example with repeated roots. </i></b></div><div class="separator" style="clear: both; text-align: center;"><b><i>Solve $x^2+4x+4=0$ using the Poh Shen-Loh method. </i></b></div><div class="separator" style="clear: both; text-align: center;">We rewrite the equation into $x^2-(-4x)+4=0$</div><div class="separator" style="clear: both; text-align: center;">Because we need our equation to be in the form $x^2-(x_1+x_2)x+x_1x_2=0$.</div><div class="separator" style="clear: both; text-align: center;">Now, the sum=$x_1+x_2=-4$ and midpoint=$\frac{x_1+x_2}{2}=\frac{-4}{2}=-2$ and the product=$x_1x_2=4$.</div><div class="separator" style="clear: both; text-align: center;">Let $u$ be the distance between $x_1$ and $-2$ and between $-2$ and $x_2$ then $x_1=-2-u$ and $x_2=-2+u$.</div><div class="separator" style="clear: both; text-align: center;">$x_1x_2=4\Rightarrow(-2-u)(-2+u)=4$</div><div class="separator" style="clear: both; text-align: center;">$4-u^2=4$</div><div class="separator" style="clear: both; text-align: center;">$u^2=0$</div><div class="separator" style="clear: both; text-align: center;">$u=0$</div><div class="separator" style="clear: both; text-align: center;">Subsitute $u$ into $x_1$ and $x_2$.</div><div class="separator" style="clear: both; text-align: center;">$x_1=-2-u=-2-0=-2$</div><div class="separator" style="clear: both; text-align: center;">$x_2=-2+u=-2+0=-2$</div><div class="separator" style="clear: both; text-align: center;">Therefore, the roots of the equation is $x_{1,2}=-2,-2$</div><div class="separator" style="clear: both; text-align: center;"><br /></div></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com4tag:blogger.com,1999:blog-9101299148361109184.post-20388206634424098252021-06-10T15:51:00.002-07:002021-06-10T16:02:32.233-07:00Problem on differential equation<br><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk6jGv2n3cwn9saWcnFOwJVtoZ7-oVE4m4Ink6DNswIg7FrvNdvkGwSGfFoHdnIdsw-uVthrQUCURGUhazJtjkCxB0IT6AleHI-TSE3p6uyTlfgx02btGhXrQkXHvUYhdEXzA7-0A4lcAf/s1600/hkbvf.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk6jGv2n3cwn9saWcnFOwJVtoZ7-oVE4m4Ink6DNswIg7FrvNdvkGwSGfFoHdnIdsw-uVthrQUCURGUhazJtjkCxB0IT6AleHI-TSE3p6uyTlfgx02btGhXrQkXHvUYhdEXzA7-0A4lcAf/s1600/hkbvf.png" border="0" data-original-width="981" data-original-height="633" width="320" height="206"></a></div><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;">An object is thrown vertically upwards with an initial velocity, $u$ of $20m/s$. The motion of the object follows the differential equation $\frac{ds}{dt}=u-gt$, where $s$ is the height of the object in meters at time $t$ seconds and $g=9.8m/s^2$. Determine the height of the object after $3$ seconds if $s=0$ when $t=0$.</div><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;"><br></div><div class="separator" style="text-align: left; clear: both;">This is a first order differentiatial equation, and to determine the height of the object at the given conditions, then when have to obtain the general solution and particular solution then apply the boundary conditions.</div><div class="separator" style="text-align: left; clear: both;">$\frac{ds}{dt}=u-gt\Rightarrow\int{ds}=\int(u-gt)dt$</div><div class="separator" style="text-align: left; clear: both;">$s=ut-\frac{gt^2}{2}+c$ this is the general solution. Substitute the conditions at $s=0,t=0$</div><div class="separator" style="text-align: left; clear: both;">$0=u(0)-\frac{0^2}{2}+c$</div><div class="separator" style="text-align: left; clear: both;">$c=0$</div><div class="separator" style="text-align: left; clear: both;">The particular solution now is:</div><div class="separator" style="text-align: left; clear: both;">$s=ut-\frac{gt^2}{2}+0$</div><div class="separator" style="text-align: left; clear: both;">Substitute $u=20m/s$ and $g=9.8m/s^2$ and $t=3sec$</div><div class="separator" style="text-align: left; clear: both;">$s=20(3)-\frac{9.8(3)^2}{2}=60-44.1=15.9$</div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-63621861075184553722021-06-09T05:50:00.004-07:002021-06-09T07:14:12.978-07:00Find the minimal eigenvalue for the matrix AΒ²β°Β²ΒΉBelow is a beautiful proposition on matrix eigenvalues.<div><br><div class="separator" style="clear: both; text-align: center;"><table cellpadding="0" cellspacing="0" class="tr-caption-container"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvwhpBC8fdQVzdqJnZi2HE636pQBz8jWGPELDsgRXCk-itaaU7U6FrXj_LPjKFrUomkJ7hhi4NaYlb5WVZHE3OGAWrsB3v8P9GqzYeF0D_krYexMuqTCro2Y8u5hyphenhyphenIW0yuTYhCeud-1jNO/s1600/yojmh.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvwhpBC8fdQVzdqJnZi2HE636pQBz8jWGPELDsgRXCk-itaaU7U6FrXj_LPjKFrUomkJ7hhi4NaYlb5WVZHE3OGAWrsB3v8P9GqzYeF0D_krYexMuqTCro2Y8u5hyphenhyphenIW0yuTYhCeud-1jNO/s1600/yojmh.png" border="0" data-original-width="972" data-original-height="486" width="381" height="190" class=" " title="" alt=""></a></td></tr><tr><td class="tr-caption" style="text-align: center;"></td></tr></tbody></table></div><br></div><div><br></div><div>It says find the minimal eigenvalue of the matrix $A^{2021}$.</div><div>We find the eigenvalue using the determinant for $|A-I\lambda|$ where $I\lambda$ is an identity $\lambda$ matrix. </div><div><br><div class="separator" style="clear: both; text-align: center;"><table cellpadding="0" cellspacing="0" class="tr-caption-container"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBw3OnaurHafGkV-mcoL0BizmIlAtzeXeIX_J_hjz51rx7XnXqTMFaeM5kaLeQxLyAqgvc35arbXLxok4VIpB1biXPG3q4STfY0kJTF1REe6uBvW-67uQGj4A7BpCqR7iJvFseQz8IMMDt/s1600/tuhvv.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBw3OnaurHafGkV-mcoL0BizmIlAtzeXeIX_J_hjz51rx7XnXqTMFaeM5kaLeQxLyAqgvc35arbXLxok4VIpB1biXPG3q4STfY0kJTF1REe6uBvW-67uQGj4A7BpCqR7iJvFseQz8IMMDt/s1600/tuhvv.png" border="0" data-original-width="972" data-original-height="486" width="378" height="189" class=" imageResizeTarget" title="" alt=""></a></td></tr><tr><td class="tr-caption" style="text-align: center;"></td></tr></tbody></table></div></div><div><br></div><div>Next</div><div><br></div><div><br><div class="separator" style="clear: both; text-align: center;"><table cellpadding="0" cellspacing="0" class="tr-caption-container"><tbody><tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjO6ZmW15RIE6siLu0DqPaUWKFcRymIR0LzvTrPhV4fdieSmvEdsZNJQ5C-faRYsE-3io8Jm6P9oMmnuN8K24N_nDboZbWYSTrBXZfUmkyx_fY5S-vXABNa1TqH2RVPQiJCV72o3NVYYpb4/s1600/jkbv.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjO6ZmW15RIE6siLu0DqPaUWKFcRymIR0LzvTrPhV4fdieSmvEdsZNJQ5C-faRYsE-3io8Jm6P9oMmnuN8K24N_nDboZbWYSTrBXZfUmkyx_fY5S-vXABNa1TqH2RVPQiJCV72o3NVYYpb4/s1600/jkbv.png" border="0" data-original-width="972" data-original-height="486" width="417" height="208" class=" imageResizeTarget" title="" alt=""></a></td></tr><tr><td class="tr-caption" style="text-align: center;"></td></tr></tbody></table></div><br></div><div><br></div><div>=$(2-\lambda)[(-1-\lambda)(1-\lambda)]+0-0$</div><div>This yields a characteristic polynomial </div><div>$=(2-\lambda)(-1+\lambda^2)$</div><div>Hence, our eigenvalues are </div><div>$\lambda_1=\lambda_2=1,\lambda_3=-1$</div><div>$\min\{\lambda_i\}=-1$</div><div>And $A^{2021}=(-1)^{2021}=-1$</div><div><br></div><div><br></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-77570653449098446882021-06-09T04:58:00.002-07:002021-06-09T05:03:26.606-07:00Differentiate f(x)=(x+1)Λ£βΊΒΉ<div class="separator" style="clear: both; text-align: center;">The question below was obtained from one of the past questions for the Skoltech institute entrance examination for admission into computational science masters program. Skoltech is a reputable and prestigious institute in Moscow Russia that offers scholarships to international students. </div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_mNhEUf8-vNKbK8l-EhcAUN9zWJJ2fQ_Rq3ZsG0OqHALdVCYGGLZ_JjY6gdNm4FWy-g_Nrg7H8jVXSjnrMcBrtCXPy57NdFaB8euOo-Rwbi-TxGx_cNzQ18_uRGZ6XLM_ue8Okx30PBSD/s1600/gkbg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_mNhEUf8-vNKbK8l-EhcAUN9zWJJ2fQ_Rq3ZsG0OqHALdVCYGGLZ_JjY6gdNm4FWy-g_Nrg7H8jVXSjnrMcBrtCXPy57NdFaB8euOo-Rwbi-TxGx_cNzQ18_uRGZ6XLM_ue8Okx30PBSD/s1600/gkbg.png" border="0" data-original-width="649" data-original-height="124" width="320" height="61" class=" imageResizeTarget imageResizeTarget"></a></div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;">If $f(x)=(x+1)^{x+1}$, find $f'(x)$.</div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;">$f(x)=(x+1)^{x+1}$</div><div class="separator" style="clear: both; text-align: center;">Take the logarithm of both sides </div><div class="separator" style="clear: both; text-align: center;">$\ln{f(x)}=(x+1)\ln(x+1)$</div><div class="separator" style="clear: both; text-align: center;">Differentiate both sides </div><div class="separator" style="clear: both; text-align: center;">$\frac{1}{f(x)}f'(x)=\ln(x+1)+(x+1)\frac{1}{x+1}$</div><div class="separator" style="clear: both; text-align: center;">$\frac{1}{f(x)}f'(x)=\ln(x+1)+1$</div><div class="separator" style="clear: both; text-align: center;">$f'(x)=f(x)[\ln(x+1)+1]$</div><div class="separator" style="clear: both; text-align: center;">$=(x+1)^{x+1}[\ln(x+1)+1]$</div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;"><br></div><div class="separator" style="clear: both; text-align: center;"><br></div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0tag:blogger.com,1999:blog-9101299148361109184.post-752035586646386512021-06-04T10:51:00.004-07:002021-06-04T11:51:57.966-07:00Proof of Cassini-Samson Identity by induction <div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5KQI2iZC0vcyGAHR_LVkKACx2BdJTvXBTNuU7zj08Bwo8kgJcnTMYtD4FXSXwF2Tmq8IXvPddOXuQbrX9yyyR98Qh0w_xUiPQFEc5K2qM4L0o7yofVyUM-wSKzjVrb8aEx5eDsvpp-rFM/s1600/gjb.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="486" data-original-width="972" height="160" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5KQI2iZC0vcyGAHR_LVkKACx2BdJTvXBTNuU7zj08Bwo8kgJcnTMYtD4FXSXwF2Tmq8IXvPddOXuQbrX9yyyR98Qh0w_xUiPQFEc5K2qM4L0o7yofVyUM-wSKzjVrb8aEx5eDsvpp-rFM/s1600/gjb.png" width="320" /></a></div><br /></div><div>Prove of the Cassini-Samson Identity</div><div><br /></div><div>The Cassini-Samson Identity is an identity that involves the Fibonnaci numbers, it states that $F_{n-1}F_{n+1}-F_{n}^2=(-1)^n$ for $n\in\mathbb{N}$.</div><div>We thus prove the theorem by induction:</div><div>The base case $n=1$ is easily verified. </div><div>Next, assume $n=k$ and prove for $n=k+1$:</div><div>$F_{k}F_{k+2}-F_{k+1}^2$</div><div>$=F_{k}(F_{k}+F_{k+1})-F_{k+1}^2$</div><div>$=F_{k}^2+F_{k+1}(F_{k}-F_{k+1})$</div><div>$=F_{k}^2-F_{k+1}F_{k_1}=-(-1)^k=(-1)^{k+1}$. Which is thus proved. </div>Abubakar M. Shafiihttp://www.blogger.com/profile/06810660191820115724noreply@blogger.com0