Question

Find the Taylor's polynomial approximation of f(x) = x Sinx upto n = 4 about the point a = 0 in the interval-1≤x≤1. Also estimate the maximum error.​

Answer 1

Answer:

Maybe the correct and helpful answer..

Answer 2

Tip:

• Taylor's theorem: $f(x)=P_n(x)+R_n(x)$
• $f(x)=f(x_0)+f'x_0(x-x_0)+\frac{f''x_0}{2!} (x-x_0)^2+\ldots+\frac{f^nx_0}{n!}(x-x_0)^n+\frac{f^{n+1}c}{(n+1)!}(x-x_0)^{n+1}$

Explanation:

• We have $f(x)=xSin[x]$
• We have to find Taylor's Polynomial approximation upto $n=4$ about the point $a=0$ in the interval $-1\leq x\leq 1$.
• We will find this by Taylor's theorem.

Steps:

Step 1 of 2:

$f(x)=P_n(x)+R_n(x)$ where $n=4$, so $f(x)=P_4(x)+R_4(x)$

$P_4(x)=f(x_0)+f'x_0(x-x_0)+\frac{f''x_0}{2!}(x-x_0)^2+\frac{f^3x_0}{3!}(x-x_0)^3+\frac{f^4x_0}{4!}(x-x_0)^4$

where, $x_0=0$ , $f(x)=xSin(x), f(x_0)=0$

$f'(x)=Sin(x)+xCos(x), f'(x_0)=0\\f''(x)=2Cos(x)-xSin(x), f''(x_0)=2\\f^3(x)=-3Sin(x)-xCos(x),f^3(x_0)=0\\f^4(x)=-4Cos(x)+xSin(x),f^4(x_0)=-4$

So, $P_4(x)=0+0+\frac{2}{2!}x^2+0+\frac{-4}{4!}x^4$

      $P_4(x)=x^2-\frac{x^4}{6}$

Step 2 of 2:

$R_n(x)=\frac{f^{n+1}(c)}{n+1}(x-x_0)^{n+1}$ , where $n=4$ and $c$ be any number between

so, $R_4(x)=\frac{f^5(c)}{5!}x^5$

$R_4(x)=\frac{5Sin(c)+cCos(c)}{120}x^5$ for some $c$ between $0$ and $x$.

If we take $x=1$, we get the approximation

$P_4(x)=1-\frac{1}{6}=0.83$

However, since $c>0$, then $R_4(1)\leq \frac{1}{120}$

$R_4(1)\leq 0.84\times 10^{-2}$

$|Sin(1)-0.83|<0.84\times 10^{-2}$

Final Answer:

The Taylor's polynomial approximation of $f(x)=xSin(x)$ upto $n=4$ about the point $a=0$ in the interval $-1\leq x\leq 1$ is $|Sin(1)-0.83|<0.84\times 10^{-2}$.