Question

Use this list of Basic Taylor Series to find the Taylor Series for f(x) = cos(5x2) based at 0. Give your answer using summation notation, write out the first three non-zero terms, and give the interval on which the series converges. (If you need to enter [infinity], use the [infinity] button in CalcPad or type "infinity" in all lower-case.) The Taylor series for f(x)=cos(5x2) is: [infinity] Correct: Your answer is correct. k=0 The first three non-zero terms are: Correct: Your answer is correct. +

Answer 1

Answer:

$cos(5x^2) = \sum_{k=0}^{\infty} \frac{(-1)^k 5^{2k}x^{4x}}{(2k)!} = 1 - 25\frac{x^4}{2}+625\frac{x^8}{24}+....$

And it converges at  $(-\infty,\infty)$

Step-by-step explanation:

Remember that in general the taylor series of cosine is given by

$cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}$

and it converges at $(-\infty , \infty )$.

therefore

$cos(5x^2) \\= \sum_{k=0}^{\infty} \frac{(-1)^k (5x^2)^{2k}}{(2k)!}\\= \sum_{k=0}^{\infty} \frac{(-1)^k 5^{2k}x^{4x}}{(2k)!}\\\\= 1 - 25\frac{x^4}{2}+625\frac{x^8}{24}+....$

And it also converges at   $(-\infty,\infty)$