a) Determine the Taylor Series expansion of sin(2x) about base a = 0 (Maclaurin series). Show the process how you derive the series below. Use backside of page if needed.
b) Write a MATLAB script to calculate the value of sin(2x) for x = pi\/4 using 6 terms. Display the difference between the above approximation and the exact value by function sin.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The Taylor Series expansion of a function f(x)f(x) about a=0a = 0 is given by: f(x)=f(0)+f\u2032(0)x+f\u2032\u2032(0)x22!+f\u2032\u2032\u2032(0)x33!+\u2026f(x) = f(0) + f'(0)x + \\frac{f”(0)x^2}{2!} + \\frac{f”'(0)x^3}{3!} + \\dots<\/p>\n\n\n\n For sin\u2061(2x)\\sin(2x), f(x)=sin\u2061(2x)f(x) = \\sin(2x), so we calculate its derivatives:<\/p>\n\n\n\n Evaluate each derivative at x=0x = 0:<\/p>\n\n\n\n Substitute into the Taylor Series formula up to 6 terms: sin\u2061(2x)\u22480+2x\u22128×33!+32×55!=2x\u22124×33+4×515\\sin(2x) \\approx 0 + 2x – \\frac{8x^3}{3!} + \\frac{32x^5}{5!} = 2x – \\frac{4x^3}{3} + \\frac{4x^5}{15}<\/p>\n\n\n\n The Taylor Series expansion provides an approximation of a function as a polynomial, making it computationally efficient to evaluate certain functions. In this problem, we derived the Maclaurin series for sin\u2061(2x)\\sin(2x), which is a special case of the Taylor series expanded about x=0x = 0. By calculating the derivatives of sin\u2061(2x)\\sin(2x) and evaluating them at x=0x = 0, we established the first six terms of the series: 2x\u22124×33+4x5152x – \\frac{4x^3}{3} + \\frac{4x^5}{15}.<\/p>\n\n\n\n The MATLAB script implements this series to approximate sin\u2061(2x)\\sin(2x) for x=\u03c04x = \\frac{\\pi}{4}. It uses a loop to calculate the terms up to n=5n = 5, where the coefficients are determined based on the pattern derived from the Taylor expansion. The exact value of sin\u2061(2x)\\sin(2x) is calculated using MATLAB\u2019s built-in The difference between the approximation and the exact value is calculated as the absolute difference. This highlights how accurately the Taylor series approximates the function within a specific range of xx. For x=\u03c04x = \\frac{\\pi}{4}, the error is small, showcasing the effectiveness of the Taylor series. The approximation improves as more terms are included, but diminishing returns set in due to floating-point limitations in computations.<\/p>\n","protected":false},"excerpt":{"rendered":" a) Determine the Taylor Series expansion of sin(2x) about base a = 0 (Maclaurin series). Show the process how you derive the series below. Use backside of page if needed.b) Write a MATLAB script to calculate the value of sin(2x) for x = pi\/4 using 6 terms. Display the difference between the above approximation and […]<\/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-185205","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185205","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=185205"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185205\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185205"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185205"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185205"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part (a): Taylor Series Expansion of sin\u2061(2x)\\sin(2x)<\/h3>\n\n\n\n
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Part (b): MATLAB Script<\/h3>\n\n\n\n
% MATLAB script to calculate sin(2x) using 6 terms of Taylor Series\nx = pi \/ 4; % Value of x\nn_terms = 6;\n\n% Compute Taylor Series approximation\ntaylor_approx = 0;\nfor n = 0:n_terms-1\n coefficient = (-1)^n * (2^(2*n+1)) \/ factorial(2*n+1);\n taylor_approx = taylor_approx + coefficient * x^(2*n+1);\nend\n\n% Exact value\nexact_value = sin(2 * x);\n\n% Difference\ndifference = abs(exact_value - taylor_approx);\n\n% Display results\nfprintf('Taylor Series Approximation: %.8f\\n', taylor_approx);\nfprintf('Exact Value using sin function: %.8f\\n', exact_value);\nfprintf('Difference: %.8f\\n', difference);\n<\/code><\/pre>\n\n\n\nExplanation (300 Words)<\/h3>\n\n\n\n
sin<\/code> function, providing a reference to compare the approximation.<\/p>\n\n\n\n