Question on Taylor expansion of first order

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  • #1
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Hello,

according to my textbook, the Taylor expansion of first order of a scalar function f(t) having continuous 2nd order derivative is supposed be: [tex]f(t) = f(0) + f'(0)t + \frac{1}{2}f''(t^*)t^2[/tex] for some [itex]t^*[/itex] such that [itex]0\leq t^* \leq 1[/itex]

Quite frankly, I have never seen such a formulation and I don't understand how one could derive the identity [tex]\frac{1}{2}f''(t^*)t^2 = \frac{1}{2}f''(0)t^2 + \frac{1}{3!}f'''(0)t^3 + \frac{1}{4!}f''''(0)t^4 + \ldots [/tex]

Can anyone help me with this?
 

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  • #2
Erland
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This is not how it is done. First one proves Taylor's therem. Then one observes that for some infinitely differentiable functions (but not all), the Taylor series converges to f(x) on some interval. In this case, the identity you asked about holds, but I can't se that this is of any interest.
 
  • #3
arildno
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That is incorrect, Erland!

We may start with the fundamental theorem of calculus,
[tex]f(t)=f(0)+\int_{0}^{t}f'(\tau)d\tau[/tex]
We now make the clever use of integration by parts:
[tex]f(t)=f(0)+(f'(\tau)(\tau-t))|_{\tau=0}^{\tau=t}-\int_{0}^{t}f''(\tau)(\tau-t)d\tau[/tex]
We now make use of the intermediate value theorem, which says that for functions a(y), b(y), there exists an 0<=X*<=x, so that
[tex]\int_{0}^{x}a(y)b(y)dy=a(X*)\int_{0}^{x}b(y)dy[/tex]

Applying this to the above, we may write (and evaluate the first expression with f' as well)
[tex]f(t)=f(0)+f'(t)t-f''(t*)\int_{0}^{t}(\tau-t)d\tau[/tex]
Or, performing a trivial integration,
[tex]f(t)=f(0)+f'(t)t+\frac{1}{2}f''(t*)t^{2}, 0\leq{t*}\leq{t}[/tex]

This formula is valid for all C_2 functions, and is very important in order to estimate the effective accuracy of a linearization procedure for a function f(t)

See for example:
https://en.wikipedia.org/wiki/Taylor's_theorem#Explicit_formulae_for_the_remainder
 
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  • #4
Erland
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That is incorrect, Erland!
What is incorrect? I wrote "First one proves Taylor's theorem", and that is what you did, but only for n=2. The identity I was referring to (and perhaps that was unclear) is the second of OP:s identities, which seems to be of no interest at all.
 
  • #5
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Hello,

according to my textbook, the Taylor expansion of first order of a scalar function f(t) having continuous 2nd order derivative is supposed be: [tex]f(t) = f(0) + f'(0)t + \frac{1}{2}f''(t^*)t^2[/tex] for some [itex]t^*[/itex] such that [itex]0\leq t^* \leq 1[/itex]

Quite frankly, I have never seen such a formulation and I don't understand how one could derive the identity [tex]\frac{1}{2}f''(t^*)t^2 = \frac{1}{2}f''(0)t^2 + \frac{1}{3!}f'''(0)t^3 + \frac{1}{4!}f''''(0)t^4 + \ldots [/tex]

Can anyone help me with this?
(I seem to have switched from t to x -- hope that is not a problem).

The formulation with the ##f''(x^*)x^2## is standard. It measure the difference between f(x) and its linear approximation g(x) = ##f(0) +f'(0)x## in the vicinity of x = 0. As such it is considered an "error term" and is useful if you are hoping to prove something about f(x) by first proving it for g(x) and then claiming the error term doesn't matter.

It can be derived as Erildno shows below, or using the mean value theorem for derivatives. However, what is the question? Is it to expand ##f''(x^*)x^2## into the expression you mention? Because you can't based on what you were told. You know only that f has two derivatives. You can't expand anything in terms of the higher derivatives, which might not exist.
 
  • #6
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Thank you all.
My main question was to understand how to obtain that remainder in the right-most term.
 
  • #7
Erland
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Thank you all.
My main question was to understand how to obtain that remainder in the right-most term.
There are at least two ways. One is arildno's. The other is based upon the generalized mean value theorem (also called Cauchy mean value theorem). I can return to this if you want.
 
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  • #8
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You know, interpreting the OP's question is not always trivial. Often there is ambiguity, and I certainly think there was ambiguity here. I myself am not good at guessing, and you'll observe I asked OP in my post what he was really asking.

Since I've answered the wrong question many times, I feel friendly towards anyone who tries to help.
 
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  • #9
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Hi guys!

I re-read my original post and I must confess that I feel very sorry for having formulated my question in a confusing way, which then triggered a discussion between arildno and Erland.

In this regard I have to say that *all* the responses that I have gotten have been very useful, and my original question (the one I actually meant to ask) has been fully answered; actually I even got more information than what I was hoping for.

Thanks again.
 
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