Question

Find the first 4 terms of Maclaurin series for the function ( cos√x )

Answer 1

The first four terms of the Maclaurin series will be $- \frac{\sqrt x}{2}$,$+\frac{\sqrt{x^2}}{24}$,$- \frac{\sqrt{x^6}}{1200}$,$\frac{\sqrt x^8}{40320}$ respectively.

What exactly is a function?

A function is a statement, rule, or law that specifies the connection between two variables.

Functions are common in mathematics and are required for the formulation of physical connections.

The standard Maclaurin series for cos√x will be;

$\quad \sum _{n=0}^{\infty \:}\left(-1\right)^n\frac{\sqrt ({x})^{2n}}{\left(2n\right)!}$

Find the terms for the given equations;

n=1;

$\rm \rightarrow \left(-1\right)^1\frac{\sqrt ({x})^{2\times 1}}{\left(2\times 1\right)!} \\\\ \rm \rightarrow \left(-1\right)^1\frac{({x})^{1}}{2}} \\\\ \rightarrow -\frac{\sqrt x}{2}$

n=2

$\rm \rightarrow \left(-1\right)^2\frac{\sqrt ({x})^{2}}{\left(2\times 1\right)!} \\\\ \rm \rightarrow \left(-1\right)^2\frac{({x})^{2}}{2}} \\\\ \rightarrow +\frac{\sqrt{x^2}}{24}$

n=3

$\rightarrow \left(-1\right)^3\frac{\sqrt ({x})^{6}}{\left(1200)} }$

$\rightarrow - \frac{\sqrt{x^6}}{1200}$

n=4;

$\rightarrow \left(-1\right)^4\frac{\sqrt ({x})^{2\times 4}}{\left(2\times 4\right)!} \\\\ \rightarrow \frac{\sqrt x^8}{40320}$

Hence, the first four terms of the Maclaurin series will be $- \frac{\sqrt x}{2}$,$+\frac{\sqrt{x^2}}{24}$,$- \frac{\sqrt{x^6}}{1200}$,$\frac{\sqrt x^8}{40320}$ respectively.

To learn more about the function, refer to:

brainly.com/question/5245372

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