# Error in Taylor polynomial of e^x

1. Apr 25, 2008

### flash

Find the Taylor polynomial of degree 9 of

$$f(x) = e^x$$

about x=0 and hence approximate the value of e. Estimate the error in the approximation.

I have written the taylor polynomial and evaluated for x=1 to give an approximation of e.
Its just the error that is confusing me. I have:

$$R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1} = \frac{e^z}{10!}$$

and I need to find an upper bound for this to give the maximum error of the approximation.
So far I have

$$0 < z < 1, e^0 < e^z < e^1$$
and then

$$\frac{1}{10!} < \frac{e^z}{10!} < \frac{e}{10!}$$

but then the upper bound has an e in it. In a similar example in the book they have just put in 3 instead of e, I guess making the assumption that e < 3. But I'm thinking if i do this I should somehow show that e < 3.

Thanks for any help :)

2. Apr 25, 2008

### HallsofIvy

Have you first approximated e using your Taylor polynomial? That should suggest that e< 3.

3. Apr 25, 2008

### flash

I have, the approximation less than 3 but that doesn't necessarily mean that the actual value (taylor approximation + error) is < 3, does it?

4. Apr 25, 2008

### rootX

You do not need to show e < 3. Everyone know e is 2.71828 18284 59045 23536... :D

To be honest, I never knew approximation would give lower value (surprised!), but it should be obvious .. see the red line http://upload.wikimedia.org/wikipedia/commons/5/50/Exp_derivative_at_0.svg

You are trying to get upper error bound; mine prof said it doesn't matter what you pick (50,100,..) it just have to be above the max function value!

so e^1 = 1+1+1/2+1/6+.. (you know little more than 2)
Why should take risk of using something like 2.5..
simply pick 3!

Last edited: Apr 25, 2008
5. Apr 25, 2008

### flash

Thanks for the replies. I had another idea, I could show (by taking the lower sum of small interval slices) that the area under the graph of 1/x between 1 and 3 has to be greater than 1, therefore 3 > e.