approximate with Taylor remainder

**karma** · May 17th 2008, 06:27 AM

Hello everyone!

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

thanks in advance

**TheEmptySet** · May 17th 2008, 06:47 AM

Originally Posted by karma

Hello everyone!

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

thanks in advance

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?.

**karma** · May 17th 2008, 06:51 AM

Sorry!! I forgot to write that.

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.

**TheEmptySet** · May 17th 2008, 06:58 AM

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.

**karma** · May 17th 2008, 07:11 AM

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?

**TheEmptySet** · May 17th 2008, 07:40 AM

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

**karma** · May 17th 2008, 08:37 AM

Right! That really helps. Thank you very much!=)

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