# Thread: approximate with Taylor remainder

1. ## approximate with Taylor remainder

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

2. 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

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?.

3. 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.

4. 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.

5. 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?

6. 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

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