Use the Taylor series for $f(x)=\cos (x)$ at x = 0 to find the Taylor series for $\cos (\sqrt{x})$ at x = 0.
A) $1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots$
B) $1-\frac{x}{2 !}+\frac{x^{2}}{4 !}-\frac{x^{3}}{6 !}+\cdots$
C) $\sqrt{x}-\frac{x}{2 !}+\frac{x^{3 / 2}}{4 !}-\frac{x^{2}}{6 !}+\cdots$
D) $\sqrt{x}-x+\frac{x^{3 / 2}}{2 !}-\frac{x^{2}}{3 !}+\cdots$

Question

Quizplus · Accepted Answer

The Taylor series for $\cos(x)$ at $x = 0$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$. Substituting $x$ with $\sqrt{x}$ gives $1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} + \cdots$, which matches option B.

Use the Taylor Series For f(x)=cos⁡(x)f(x)=\cos (x)f(x)=cos(x) At X = 0 to Find the Taylor Series For 0

Use the Taylor Series For $f(x)=\cos (x)$ At X = 0 to Find the Taylor Series For 0