- #1
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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 :)
[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 :)