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.