approximate with Taylor remainder

• May 17th 2008, 07:27 AM
karma
approximate with Taylor remainder
Hello everyone! (Wink) I am trying to prove that the following inequality is right for all n equal to or larger than 9. I tried using the triangle inequality and then mathematical induction but none of it seems to work. please help!

l e^-1 - (sigma sign from k=0 to n) (1)^-k / k!l

• May 17th 2008, 07:47 AM
TheEmptySet
Quote:

Originally Posted by karma
Hello everyone! (Wink) I am trying to prove that the following inequality is right for all n equal to or larger than 9. I tried using the triangle inequality and then mathematical induction but none of it seems to work. please help!

l e^-1 - (sigma sign from k=0 to n) (1)^-k / k!l

First question what are we trying to show that it is less than(where is the inequality)?

$\bigg| e^{-1}-\sum_{k=0}^{n}\frac{(-1)^{-k}}{k!} \bigg|$

Second what class is this for do you know Taylor's Remainder theorem?.
• May 17th 2008, 07:51 AM
karma
Sorry!! I forgot to write that.(Doh) The question is that it is smaller than or equal to 10^(3-n). I actually do not know what the taylor remainder theorem is, but yes we are doing taylor series.
• May 17th 2008, 07:58 AM
TheEmptySet
Quote:

Originally Posted by karma
yes we are doing taylor series.

Does this series look familiar to you?

$f(x)=e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

consider
$f(-1)=e^{-1}=?$

Do you know the Alternating series Test error bound?

If so you can use it to finish this problem.
• May 17th 2008, 08:11 AM
karma
I know that lerrorl = lxl^(n+1) / (n+1)! for a Pn(x), is that what you mean by alternating series error bound? If that is so, I dont really see how you can use that.

But if you replace f(-1) (which I hadnt seen before) than you get an infinite sum starting from k=n, right? What do you do with the absolute value then?
• May 17th 2008, 08:40 AM
TheEmptySet
Quote:

Originally Posted by karma
I know that lerrorl = lxl^(n+1) / (n+1)! for a Pn(x), is that what you mean by alternating series error bound? If that is so, I dont really see how you can use that.

But if you replace f(-1) (which I hadnt seen before) than you get an infinite sum starting from k=n, right? What do you do with the absolute value then?

The error in an alternating series is always less than the absolute value of the next term

$\bigg| e^{-1}-\sum_{k=0}^{n}\frac{(-1)^k}{k!}\bigg| < |a_{n+1}|$

So lets find out how big the next term can be using n=9

$\bigg| \frac{(-1)^{n+1}}{(n+1)!}\bigg|=\frac{1}{10!}=\frac{1}{362 8800} \approx 2.756 \times 10^{-7} < 10^{-6}=10^{3-9}=10^{3-n}$

I hope this clears it up.

Good luck
• May 17th 2008, 09:37 AM
karma
Right! That really helps. Thank you very much!=)