prove that error between approximating function and first three terms of power series expansion is less than 1/1500

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Introduction: In this solution, we prove that error between approximating function and first three terms of power series expansion is less than 1/1500.

Question: Power series expansion of a function is

\[P\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n\left(x+2\right)^n}{3^n\left(n+1\right)}\]

prove that at point x =-2.5 error between approximating function and first three terms of power series expansion is less than 1/1500.

and test whether error is less than 1/1500. Determine first three terms and general term of derivative of power series

Solution :

Given power series of a function f(x)

\[P\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^n\left(x+2\right)^n}{3^n\left(n+1\right)}\]

writing it’s first four terms

\[=\frac{\left(x+2\right)}{1}-\frac{\left(x+2\right)^2}{6}+\frac{\left(x+2\right)^3}{27}-\frac{\left(x+2\right)^4}{108}+\ldots\ldots\]

Now to evaluate function at point x =-2.5

\[f\left(-2.5\right)=\frac{\left(-2.5+2\right)}{1}-\frac{\left(-2.5+2\right)^2}{6}+\frac{\left(-2.5+2\right)^3}{27}-\frac{\left(-2.5+2\right)^4}{108}\]

f(-2.5)=-0.54687

And evaluating approximating P(-2.5) using first three terms of the power series

\[P_3\left(-2.5\right)=\frac{\left(-2.5+2\right)}{1}-\frac{\left(-2.5+2\right)^2}{6}+\frac{\left(-2.5+2\right)^3}{27}\]

\[P_3\left(-2.5\right)=-0.546296296296\]

\[\left|f\left(-2.5\right)-P_3\left(-2.5\right)\right|=\left|-0.546875+0.546296296296\right| =\left|-0.0005787\right|\]

=0.0005787

Finding absolute value of the difference that is error

\[\left|f\left(-2.5\right)-P_3\left(-2.5\right)\right|=0.0005787<\frac{1}{1500}=0.00067\]

\[\left|f\left(-2.5\right)-P_3\left(-2.5\right)\right|<\frac{1}{1500}\]

Thus we proved that error between approximating function and first three terms of power series expansion is less than 1/1500.


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