The first four terms of the Maclaurin series will be [tex]- \frac{\sqrt x}{2}[/tex],[tex]+\frac{\sqrt{x^2}}{24}[/tex],[tex]- \frac{\sqrt{x^6}}{1200}[/tex],[tex]\frac{\sqrt x^8}{40320}[/tex] 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;
[tex]\quad \sum _{n=0}^{\infty \:}\left(-1\right)^n\frac{\sqrt ({x})^{2n}}{\left(2n\right)!}[/tex]
Find the terms for the given equations;
n=1;
[tex]\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}[/tex]
n=2
[tex]\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}[/tex]
n=3
[tex]\rightarrow \left(-1\right)^3\frac{\sqrt ({x})^{6}}{\left(1200)} } \\\\[/tex]
[tex]\rightarrow - \frac{\sqrt{x^6}}{1200}[/tex]
n=4;
[tex]\rightarrow \left(-1\right)^4\frac{\sqrt ({x})^{2\times 4}}{\left(2\times 4\right)!} \\\\ \rightarrow \frac{\sqrt x^8}{40320}[/tex]
Hence, the first four terms of the Maclaurin series will be [tex]- \frac{\sqrt x}{2}[/tex],[tex]+\frac{\sqrt{x^2}}{24}[/tex],[tex]- \frac{\sqrt{x^6}}{1200}[/tex],[tex]\frac{\sqrt x^8}{40320}[/tex] respectively.
To learn more about the function, refer to:
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