{"id":241927,"date":"2025-07-04T04:27:11","date_gmt":"2025-07-04T04:27:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=241927"},"modified":"2025-07-04T04:27:13","modified_gmt":"2025-07-04T04:27:13","slug":"write-a-matlab-function-file-that-computes-the-sum-and-makes-a-semilog-x-plot-of-the-value-of-this-sum-as-a-function-of-k-for-k-12-1000","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/write-a-matlab-function-file-that-computes-the-sum-and-makes-a-semilog-x-plot-of-the-value-of-this-sum-as-a-function-of-k-for-k-12-1000\/","title":{"rendered":"Write a MATLAB function file that computes the sum and makes a semilog-x plot of the value of this sum as a function of k for k = 1,2,\u2026, 1000."},"content":{"rendered":"\n

k S = \\sum_{n=1} \\frac{1}{n(n+1)} Write a MATLAB function file that computes the sum and makes a semilog-x plot of the value of this sum as a function of k for k = 1,2,\u2026, 1000. Your numerical solution should suggest the value of this sum for k = \\infty. Derive the value of the infinite sum.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this problem, we need to evaluate the sum SkS_kSk\u200b given by:Sk=\u2211n=1k1n(n+1)S_k = \\sum_{n=1}^{k} \\frac{1}{n(n+1)}Sk\u200b=n=1\u2211k\u200bn(n+1)1\u200b<\/p>\n\n\n\n

Step 1: Simplifying the Expression<\/h3>\n\n\n\n

The given term 1n(n+1)\\frac{1}{n(n+1)}n(n+1)1\u200b can be decomposed using partial fractions:1n(n+1)=1n\u22121n+1\\frac{1}{n(n+1)} = \\frac{1}{n} – \\frac{1}{n+1}n(n+1)1\u200b=n1\u200b\u2212n+11\u200b<\/p>\n\n\n\n

Thus, the sum SkS_kSk\u200b becomes:Sk=\u2211n=1k(1n\u22121n+1)S_k = \\sum_{n=1}^{k} \\left( \\frac{1}{n} – \\frac{1}{n+1} \\right)Sk\u200b=n=1\u2211k\u200b(n1\u200b\u2212n+11\u200b)<\/p>\n\n\n\n

This is a telescoping series, meaning that most terms cancel out. When expanded, we get:Sk=(1\u221212)+(12\u221213)+(13\u221214)+\u22ef+(1k\u22121k+1)S_k = \\left( 1 – \\frac{1}{2} \\right) + \\left( \\frac{1}{2} – \\frac{1}{3} \\right) + \\left( \\frac{1}{3} – \\frac{1}{4} \\right) + \\dots + \\left( \\frac{1}{k} – \\frac{1}{k+1} \\right)Sk\u200b=(1\u221221\u200b)+(21\u200b\u221231\u200b)+(31\u200b\u221241\u200b)+\u22ef+(k1\u200b\u2212k+11\u200b)<\/p>\n\n\n\n

After cancellation, the result simplifies to:Sk=1\u22121k+1S_k = 1 – \\frac{1}{k+1}Sk\u200b=1\u2212k+11\u200b<\/p>\n\n\n\n

Step 2: MATLAB Function<\/h3>\n\n\n\n

You can now write a MATLAB function to compute this sum and plot it. Below is the MATLAB code to calculate the sum for k=1,2,\u2026,1000k = 1, 2, \\dots, 1000k=1,2,\u2026,1000 and create a semi-logarithmic plot:<\/p>\n\n\n\n

matlabCopyEditfunction semilogx_sum\n    % Define the number of terms\n    k_max = 1000;\n    \n    % Pre-allocate the array for storing sums\n    S = zeros(1, k_max);\n    \n    % Compute the sum for each k\n    for k = 1:k_max\n        S(k) = 1 - 1\/(k+1);\n    end\n    \n    % Plot the result in semilog-x scale\n    semilogx(1:k_max, S, 'LineWidth', 2);\n    title('Semilog-x Plot of the Sum S_k');\n    xlabel('k');\n    ylabel('S_k');\n    grid on;\nend\n<\/code><\/pre>\n\n\n\n
Step 3: Deriving the Infinite Sum<\/h3>\n\n\n\n
As kkk approaches infinity, the term 1k+1\\frac{1}{k+1}k+11\u200b becomes very small, so the sum SkS_kSk\u200b approaches:S\u221e=1\u22120=1S_{\\infty} = 1 – 0 = 1S\u221e\u200b=1\u22120=1<\/p>\n\n\n\n
Thus, the infinite sum is 1.<\/p>\n\n\n\n
Step 4: Plotting Interpretation<\/h3>\n\n\n\n
When you run the MATLAB function, you’ll notice that the plot levels off as kkk increases, approaching the value of 1, which corresponds to the sum as k\u2192\u221ek \\to \\inftyk\u2192\u221e. The semi-logarithmic plot helps highlight how the sum converges as kkk increases.<\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
k S = \\sum_{n=1} \\frac{1}{n(n+1)} Write a MATLAB function file that computes the sum and makes a semilog-x plot of the value of this sum as a function of k for k = 1,2,\u2026, 1000. Your numerical solution should suggest the value of this sum for k = \\infty. Derive the value of the infinite […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-241927","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241927","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=241927"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241927\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=241927"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=241927"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=241927"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}