Maclauren series Polynominal

**zifnib** · November 23rd 2008, 07:49 PM

I need to Evaluate at X=.3 and .5 with precision of n=8 and I need to find the maximum error. I am very lost. Thank you.

$\int_{0}^{X} t e^-t^3$

**zifnib** · November 23rd 2008, 08:12 PM

i know that $e^x = \sum_{n=1}^\infty x^n/n!$ so then $t * \sum_{n=1}^\infty (-t^3)^n/n!$

Does that equal $\sum_{n=1}^\infty (-t^3)^{n+1}/n!$ or is it something else entirely? When I put it in my calculator I get something else entirely but someone else told me that it works??

**Mathstud28** · November 24th 2008, 03:27 AM

Originally Posted by zifnib

I need to Evaluate at X=.3 and .5 with precision of n=8 and I need to find the maximum error. I am very lost. Thank you.

$\int_{0}^{X} t e^-t^3$

Ok, first notice that

$\forall{t}\in\mathbb{R}~e^{t}=\sum_{n=0}^{\infty}\ frac{t^n}{n!}$

So

$\begin{aligned}\forall{-t^3}\in\mathbb{R}\implies{t}\in\mathbb{R}~te^{-t^3}&=t\sum_{n=0}^{\infty}\frac{\left(-t^3\right)^n}{n!}\\ &=t\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n}}{n!}\\ &=\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!} \end{aligned}$

Now notice that a power series is uniformly convergent, thus integrable on its interval of convergence so

$\begin{aligned}\int_0^{x}te^{-t^3}dy&=\int_0^x\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!}dt\\ &=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^nt^{3n+1}}{n!}dt\\ &=\sum_{n=0}^{\infty}\frac{(-1)^nx^{3n+2}}{(3n+2)n!}\end{aligned}$

Now input the appropriate values and note that if $S=\sum_{n=0}^{\infty}a_n$ and $S_N=\sum_{n=0}^{N}a_n$ that

$R_N=\left|S-S_N\right|\leqslant{a_{N+1}}$

Or in other words the error is less than or equal to the first neglected term. So lets say your error is $e$ , then you need to find $N$ such that $a_{N+1}\leqslant{e}$ and that will be the correct number of terms. But remember that since $N\in\mathbb{N}$ if you calcluate the correct $N$ to be lets say $\varphi$ you really want $\lceil{\varphi}\rceil$ .

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