Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 19


$50+105(x-2)+92(x-2)^{2}+42(x-2)^{3}+10(x-2)^{4}+(x-2)^{5}$ and $R=\infty$

Work Step by Step

Taylor's series of $f$ centered at $a$ is $f(x)=\dfrac{f^{n}(a)(x-a)^{n}}{(n)!}=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^{2}}{2}+\frac{f'''(a)(x-a)^{3}}{6}+....$ Given: $f(x)=x^{5}+2x^{3}+x$ $x^{5}+2x^{3}+x=50+105(x-2)+\frac{184(x-2)^{2}}{2}+\frac{252(x-2)^{3}}{6}+\frac{240(x-2)^{4}}{24}+\frac{120(x-2)^{5}}{120}$ $=50+105(x-2)+92(x-2)^{2}+42(x-2)^{3}+10(x-2)^{4}+(x-2)^{5}$ and Since, the Taylor polynomial for given function only has finite number of terms , it converges for all values of $x$, so $R=\infty$.
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