Error in Taylor polynomial of e^x

In summary, In trying to approximate e, the taylor polynomial gives an error that is less than 3, but does not necessarily mean that the actual value (taylor approximation + error) is < 3.
  • #1
flash
68
0
Find the Taylor polynomial of degree 9 of

[tex]
f(x) = e^x
[/tex]

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:

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

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

[tex]
0 < z < 1, e^0 < e^z < e^1
[/tex]
and then

[tex]
\frac{1}{10!} < \frac{e^z}{10!} < \frac{e}{10!}
[/tex]

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 :)
 
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  • #2
Have you first approximated e using your Taylor polynomial? That should suggest that e< 3.
 
  • #3
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
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:
  • #5
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.
 

1. What is a Taylor polynomial?

A Taylor polynomial is a polynomial approximation of a function that can be used to estimate the value of the function at a certain point. It is created by taking the derivatives of the function at a specific point and using those values to construct a polynomial.

2. What is the error in a Taylor polynomial?

The error in a Taylor polynomial is the difference between the actual value of the function and the estimated value given by the polynomial. The error is due to the fact that a polynomial can only approximate a function, and is not an exact representation of the function.

3. How is the error in a Taylor polynomial calculated?

The error in a Taylor polynomial can be calculated using the remainder term (also known as the Lagrange remainder). This term takes into account the value of the next term in the Taylor polynomial, and can be used to determine the maximum possible error for a given degree of the polynomial.

4. How does the error in a Taylor polynomial change with increasing degree of the polynomial?

As the degree of the Taylor polynomial increases, the error typically decreases. This is because a higher degree polynomial is able to better approximate the function, resulting in a smaller difference between the actual value and the estimated value.

5. How can the error in a Taylor polynomial be used to improve the accuracy of the approximation?

The error in a Taylor polynomial can be used to determine the number of terms needed in the polynomial to achieve a desired level of accuracy. By increasing the number of terms, the polynomial can better approximate the function and minimize the error. Additionally, using a smaller interval around the point of approximation can also improve the accuracy of the Taylor polynomial.

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