Question

We are asked to find the Maclaurin series for a function involving cos(x). Recall the Maclaurin series for cos(x). $$cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$$ The same equality would be true for any variable, and in particular for $$u = \frac{1}{13}x^2$$. Therefore, the Maclaurin series for $$cos\left(\frac{1}{13}x^2\right)$$ is $$\sum_{n=0}^{\infty} (-1)^n \frac{\left(\boxed{\phantom{\frac{1}{13}x^2}}\right)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{\boxed{\phantom{(2n)!}}(2n)!}$$

Name: we are asked to find the maclaurin series for a function involving cosx recall the maclaurin series for cosx cosx sumn0infty 1n fracx2n2n the same equality would be true for any variable an 18735
Uploaded: 2023-08-18T14:20:38-08:00
Duration: 2 min 58 s
Channel: Madhur L
Description: we are asked to find the maclaurin series for a function involving cosx recall the maclaurin series for cosx cosx sumn0infty 1n fracx2n2n the same equality would be true for any variable an 18735

          We are asked to find the Maclaurin series for a function involving cos(x). Recall the Maclaurin series for cos(x).
$$cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$$
The same equality would be true for any variable, and in particular for $$u = \frac{1}{13}x^2$$.
Therefore, the Maclaurin series for $$cos\left(\frac{1}{13}x^2\right)$$ is
$$\sum_{n=0}^{\infty} (-1)^n \frac{\left(\boxed{\phantom{\frac{1}{13}x^2}}\right)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{\boxed{\phantom{(2n)!}}(2n)!}$$

$we are asked to find the maclaurin series for a function involving cosx recall the maclaurin series for cosx cosx sumn0infty 1n fracx2n2n the same equality would be true for any variable an 18735$

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