# Math Help - Taylor Series Problems

1. ## Taylor Series Problems

Hi all,

These have been giving me some trouble:

1. Use series to approximate the definite integral to within 3 decimal places

The integral from 0 to 1 : xcos(x^3) dx

2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

Thanks a lot for any help

-coolio

2. Originally Posted by coolio
Hi all,

These have been giving me some trouble:

1. Use series to approximate the definite integral to within 3 decimal places

The integral from 0 to 1 : xcos(x^3) dx

2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

Thanks a lot for any help

-coolio
1. $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$...therefore $\cos(x^3)=\sum+{n=0}^{\infty}\frac{(-1)^{n}(x^3)^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{ (-1)^{n}x^{6n}}{(2n)!}$

Now multiplying by x we get $x\cos(x^3)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+1}}{(2n)!}$

Integrating we get $\int{x\cos(x^3)}dx=\int\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+1}}{(2n)!}dx=\sum_{n=0}^{\infty}\frac{ (-1)^{n}x^{6n+2}}{(6n+2)(2n)!}$ and to get [tex] $\int_0^{1}x\cos(x^3)dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{6n+2}}{(6n+2)(2n)!}\bigg|_0^{1}=.440$

3. Originally Posted by coolio
Hi all,

These have been giving me some trouble:

1. Use series to approximate the definite integral to within 3 decimal places

The integral from 0 to 1 : xcos(x^3) dx

2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.

a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2

b. f(x)= e^(x^2), a=0, n=3, 0 < or equal to x < or equal to 0.1

Thanks a lot for any help

-coolio
For the other two use $f(x)\approx{f(c)+(x-c)f'(c)+frac{f''(c)(x-c)^2}{2!}+\frac{(x-c)^3f'''(c)}{3!}}$ $+...+\frac{(x-c)^nf^{(n)}}{n!}$

4. Originally Posted by Mathstud28
For the other two use $f(x)\approx{f(c)+(x-c)f'(c)+frac{f''(c)(x-c)^2}{2!}+\frac{(x-c)^3f'''(c)}{3!}}$ $+...+\frac{(x-c)^nf^{(n)}}{n!}$
$f(x)=\lim_{n\to\infty}\sum_{n=c}\frac{(x-c)^{n}f^{(n)}}{n!}$

Where $f^{(n)}$ denotes $\frac{dy^{n}}{d^{n}x}$

5. Originally Posted by coolio
1. Use series to approximate the definite integral to within 3 decimal places

The integral from 0 to 1 : xcos(x^3) dx
Note $x\cos x^3 = \sum_{n=0}^{\infty}(-1)^n\frac{x^{6n+1}}{(2n)!}$. This means, $\int_0^1 x\cos x^3 dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \int_0^1 x^{6n+1} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(6n+2)}$. Let $A_n$ be the $n$ term approximation. Then since this series is an alternating series it means $\left| \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!(6n+2)} - A_n\right| \leq \frac{1}{(2n)!(6n+2)}$. To approximate this to three decimal places means we want to make $\frac{1}{(2n)!(6n+2)} \leq \frac{1}{1000}$ which happens first at $n=2$.
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Mathstud do not make posts which do not answer the question, the question was not asking for an infinite series, it was asking for an approximation.

6. Originally Posted by coolio
2. Approximate f by a Taylor polynomial with degree n at the number a. Then use Taylor's inequality to estimate the accuracy of the approximation f(x) is approximately equal to Tn(x) when x lies in the given interval.
a. f(x)= sqrt(x), a=4, n=2, 4 < or equal to x < or equal to 4.2
Let $T_2$ be the 2nd degree Taylor approximating polynomial. If $x$ is a chosen point on $(4,4.2)$ then it means $\left|\sqrt{x} - T_2(x)\right| = \frac{|f'''(y)|}{3!}|x-4|^3 \leq \frac{1}{6}\cdot \frac{1}{5}\cdot |f'''(y)|$. Now put a bound on $f'''(y)$.