Taylor polynomial question

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hencethus
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Taylor polynomial question

Postby hencethus » Wed Nov 17, 2010 10:31 pm UTC

OK, this was a homework problem. I've already solved it, but I have a question that i hope someone finds interesting.

Here's the problem:
Taylor's theorem states that if [imath]T_n(x)[/imath] is the [imath]n[/imath]th degree Taylor polynomial of [imath]f(x) = cos(x)[/imath] centered at [imath]a = 0[/imath], then for all [imath]\beta[/imath] there exists an [imath]\alpha[/imath], [imath]0 < \alpha < \beta[/imath], such that [math]f(\beta) = T_n(\beta) + {f^{n+1}(\alpha)\over(n + 1)!}(\beta)^{n+1}[/math] How large must [imath]n[/imath] be so that the [imath]n[/imath]th degree Taylor polynomial of [imath]f(x) = cos(x)[/imath] centered at [imath]a = 0[/imath] approximates [imath]cos(x)[/imath] to within [imath]\epsilon = .0001[/imath] for all [imath]x \in [0,10][/imath]?


Here's how I solved it:
We need [imath]|f(10) - T_n(10)| \le .0001[/imath] or [imath]|{f^{n+1}(\alpha)\over(n + 1)!}(10)^{n+1}| \le .0001[/imath], so
[math]|f^{n+1}(\alpha)| \le 1 \iff {|f^{n+1}(\alpha)|\over(n + 1)!}(10)^{n+1} \le {(10)^{n+1}\over(n + 1)!}[/math]
[math]{(10)^{n+1}\over(n + 1)!} \le {1\over10000} \iff 10^{n+5} \le (n+1)! \iff n \ge 34[/math]
[math]\therefore n \ge 34 \Rightarrow |f(10) - T_n(10)| \le .0001[/math]

It turns out that that's the answer my professor was looking for, but before I knew that I wanted to check my answer, so I wrote a python script to evaluate the Taylor polynomial:

Code: Select all

from math import factorial
print('Cosine Taylor polynomial evaluator.')
degree=input('What degree?')
x=input('Evaluate at x=')
degree = degree/2
sum=0
for i in range(0,degree+1):
   sum += (float(((-1)**i))/factorial(2*i)) * (x**(2*i))
print(sum)


According to my script, if [imath]n \ge 32[/imath] then [imath]|f(10) - T_n(10)| \le .0001[/imath]

Code: Select all

What degree?31
Evaluate at x=10
-0.83942020518

Cosine Taylor polynomial evaluator.
What degree?32
Evaluate at x=10
-0.839040166105

Cosine Taylor polynomial evaluator.
What degree?33
Evaluate at x=10
-0.839040166105

Cosine Taylor polynomial evaluator.
What degree?34
Evaluate at x=10
-0.83907403768


According to my calculator:
[imath]cos(10) = -0.839071529[/imath]

So here's my question: Where's the discrepancy come from? Is it that my script is dealing with crazy numbers like 1/32! and python isn't precise enough? Or is it because there's some wiggle room in the Taylor remainder when I used [imath]f^{n+1}(\alpha)\le1[/imath] ?

gorcee
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Re: Taylor polynomial question

Postby gorcee » Wed Nov 17, 2010 11:28 pm UTC

Specifically, which discrepancy do you mean? It might be more helpful to output your absolute error. It looks to me like your values at n = 34 are within the 0.0001 error tolerance. Isn't this what you expect?

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hencethus
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Re: Taylor polynomial question

Postby hencethus » Thu Nov 18, 2010 12:18 am UTC

Yeah, I'd expect n = 34 to work. The question is, does n = 32 also work, and my method of solving didn't get me the lowest possible value for n? Or is there something wrong with my script such that n = 32 doesn't really work?

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skeptical scientist
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Re: Taylor polynomial question

Postby skeptical scientist » Thu Nov 18, 2010 2:09 am UTC

hencethus wrote:Yeah, I'd expect n = 34 to work. The question is, does n = 32 also work, and my method of solving didn't get me the lowest possible value for n?

That's right. By Taylor's theorem, the error from the nth degree polynomial at x is equal to \(\frac{f^{(n+1)}(\alpha)}{(n + 1)!}x^{n+1}\) for some \(0<\alpha<x\). Using the fact that we only care about |x|≤10 in this problem, and that all derivatives of cos are bounded by 1, this means that the difference between the actual value and the nth degree Taylor polynomial is no more than 10n+1/(n+1)!. When n≥33,* this quantity is less than .0001, which means that the 33rd degree Taylor approximation is sufficient. But the actual error may be smaller than the 10n+1/(n+1)! bound we obtain from Taylor's theorem, so it may be that a lower degree Taylor approximation would also work.

*I get 33 rather than 34 here; you might have been forgetting that the exponent is n+1 rather than n.
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gorcee
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Re: Taylor polynomial question

Postby gorcee » Thu Nov 18, 2010 3:11 am UTC

What SS said. The error bound is an upper bound. You could get under that bound earlier, but if you do, it's generally considered to just be fortuitous.

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hencethus
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Re: Taylor polynomial question

Postby hencethus » Thu Nov 18, 2010 4:06 am UTC

Ah, thank you. That clears that up.
Also, I don't know where I got 34 from. It's definitely 33. Interestingly, the 33rd degree is the same as the 32nd degree, since every other term has sin(0) in the numerator. So you can infer that if the 33rd degree is sufficient, then so is the 32nd. And n = 32 actually is the lowest value of n for which the error is less than .0001


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