# MacClaurin series discrepancy

• May 7th 2009, 01:17 PM
redsoxfan325
MacClaurin series discrepancy
This can happen with a bunch of different series, but let's just look at a really easy one as an example.

$f(x)=e^{x+1}$

There are two ways to calculate the Maclaurin polynomial:

1.) $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ so $e^{x+1} = \sum_{n=0}^{\infty}\frac{(x+1)^n}{n!}$

2.) $f^{(n)}(x) = e$ for all $n$ so the polynomial is $\sum_{n=0}^{\infty}\frac{ex^n}{n!}=e\sum_{n=0}^{\i nfty}\frac{x^n}{n!}$

Clearly, these are the same if you're summing from $0$ to $\infty$, but what if you are just trying to approximate the function? Which is a better estimate? (I would assume it's the second one.)

Or, the true reason behind my asking this: My friend's professor asked on an exam what the coefficient on the $x^2$ term was, yet depending on how you calculate this, you could get different answers. In the first method you get $\frac{1}{2}$; in the second, you get $\frac{e}{2}$.

I'm assuming that if $a_n^2$ is the coefficient on $x^2$ (in the first method) when taking the sum from $0$ to $n$, these coefficients will form a sequence such that $\lim_{n\to\infty}a_n^2 = \frac{e}{2}$

Does anyone have anything to add to this or is that reasoning all pretty much fine?
• May 7th 2009, 01:55 PM
Calculus26
Good ol' Mathcad e/2 = 1.359

just first seven terms of the coefficients of x^2 for http://www.mathhelpforum.com/math-he...3f7a5086-1.gif

form the sequence

.5,1,1.25,1.333,1.354,1.358,1.359

for what its worth
• May 7th 2009, 02:06 PM
Calculus26
If you consider the binomial theorem

For(x+1)^n then the coefficient is nCr = n!/[(n-2)!*2]

In the series the coefficient is then 1/[(n-2)!*2] for each n

If you add these coeeficients you get

1/2 sum{n=2toinfinity}[1/(n-2)!] = e/2