Taylor's Formula: Proof & Usage?

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Hi I've got a question on the following proof of Taylor's Formula, Is it correct?

Using the Fundamental Theorem of Calculus:

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f(b) \ - \ f(a) [/latex]

We'll rewrite this in integration by part form:

[latex] u \ = \ f'(t) \ , \ du \ = \ f''(t) dt \ ,\ dv \ = \ dt \ , \ v \ = \ - (b - t)[/latex]

(The "v" term includes the minus due to a chain rule differentiation removing it anyway! Very sneaky thing that confused me for a few minutes but I like it! ).

[latex] \int_{a}^{b} u dv \ = \ u v |^b_a \ - \ \int_{a}^{b} v du [/latex]

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ - f'(t)(b \ - \ t) |^b_a \ - \ \int_{a}^{b} - f''(t)(b \ - \ a)dt [/latex]

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ \int_{a}^{b} f''(t)(b \ - \ a)dt [/latex]

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ - \ f''(t)\frac{(b \ - \ a)^2}{2} |^b_a \ - \ \int_{a}^{b} - f'''(t)\frac{(b \ - \ a)^2}{2}dt [/latex]

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt [/latex]

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt [/latex]

Okay, before I go on I'd like to qualify that the left hand side can be
rewritten as

[latex] \int_{a}^{b} f ' (t)\,dt \ = \ f(b) \ - \ f(a) [/latex]

so that the above becomes

[latex] \ f(b) \ - \ f(a) \ = \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt [/latex]

or better yet:

[latex] \ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + \ \int_{a}^{b} f'''(t)\frac{(b \ - \ a)^2}{2}dt [/latex]

Okay, well you'd keep going with the right hand term until you've
calculated the (n - 1) term & then you want the "n"th term

[latex] \ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ + \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ f^n (t) dt [/latex]

[latex] \ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ - \ \ f^n (t) \frac{(b \ - \ t)}{(n )!}^{(n)} |^b_a \ - \ \ f^{(n + 1)}(t) dt \frac{(b \ - \ t)}{(n )!}^{(n)} [/latex]

[latex] \ f(b) \ = \ f(a) \ + \ \ f'(a)(b \ - \ a) \ + \ f''(a)\frac{(b \ - \ a)^2}{2} \ + ... \ + \ f^n (a) \frac{(b \ - \ a)}{(n )!}^{(n)} \ + \ \int_{a}^{b} \ f^{(n + 1)}(t) \frac{(b \ - \ t)}{(n )!}^{(n)} dt [/latex]

Alright, I think all of that is right. My old calculus book gave some
horrendous proof that I don't think anyone would bother memorizing
but this one just falls out of combining I.B.P. with the FTC.

EDIT: The book I got this from is Serge Lang's "A First Course In Calculus"
unbelievably clear book - for the most part

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As for the proof that

[latex] \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ f^n (t) dt \ = \ \frac{f^n(c)}{n!}(b \ - \ a)^n[/latex]

[latex] f^n(u) \ = \ m \ \le \ M \ = \ f^n(v) \ [/latex]

These are constants and we'll squeeze the integral in between them


[latex] m \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ dt \ \le \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ f^n (t) dt \ \le \ M \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ dt[/latex]

[latex] m \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} dt \ \le \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} \ f^n (t) dt \ \le \ M \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} dt[/latex]

[latex] \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} f^n (u) dt \ \le \ R_n \ \le \ \int_{a}^{b} \frac{(b \ - \ t)}{(n \ - \ 1)!}^{(n - 1)} f^n (v) dt[/latex]

[latex] \frac{(b \ - \ a)}{n!}^{n} f^n (u) \ \le \ R_n \ \le \ \frac{(b \ - \ a)}{n!}^{n} f^n (v) [/latex]

So, by the Intermediate Value Theorem ∃ c ∈ (a,b) : [latex] \frac{f^n(c)}{n!}(b \ - \ a)^n[/latex]

I'm wondering, is there a way to find this "c" numerically because
neither of the two books I've looked in give you a way to find this?

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I hope you see how easy & intuitive all of this is, well as for the rest of the
way the chapter goes it's quite confusing.

There's a theorem that says

If ∃ [latex] M_n [/latex] : [latex] | f^n(x) | \ \le \ M_n[/latex] then

[latex] | \frac{f^n(c)}{n!}(b \ - \ a)^n | \le \ \frac{M_n|b - a|^n}{n!}[/latex]

I don't even know what this means, I think it means that if there is a
number such that the "n"th derivative is less than this number then
it's less than the right hand side.

Furthermore, I don't know how to use this in practice
The following bits of the chapter all rely on this specific concept & I
can't advance until I get it.

Example, compute sin(0.1) to three decimals.

sin(0.1) ≈ sin(0) + cos(0)•(0.1) - sin(0)•(0.1)²/2 - cos(0)•(0.1)³/6 + sin(0)•(0.1)⁴/24 + cos(0)•(0.1)⁵/120 - sin(0)•(0.1)^6/6! - ...

sin(0.1) ≈ 0 + 0.1 - 0 - (0.1)³/6 + 0 + (0.1)⁵/120 - 0 - ...

sin(0.1) ≈ 0 + 0.1 - 0 - (0.1)³/6 + 0 + (0.1)⁵/120 - 0 - ...

sin(0.1) ≈ 0 + 0.1 - 0 - 0.001/6 + 0 + 0.00001/120 - 0 - ...

sin(0.1) ≈ 0.1 - 0.001/6 + 0.00001/120 - ...

That would give you accuracy to 3 decimal places but I don't see
how the "error" function thing even fits in here or is needed?
If you know, how would I find "c" in this particular example?

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Alright, I think I've got it but there's still a lot of questions...

If you look at the proof in the second post I have a [latex]M_n[/latex] term.
That represents the maximum value on the interval & I think the
formula is saying that you take the maximum possible value of the
"n"th derivative on the interval you're evaluating and you can safely
assume that to be an upper bound.

Because it's a trigonometric function we have a bound of 1 so we
can safely set [latex]M_n \ = \ 1[/latex]

Now, if I am to use the error term in the specific example I gave
I need to find a value of "n" so that I come out with 3 decimal
places of error. 3 seems like a good term! This is kind of a trial-&-error
aspect of the whole process right?

Anyway, if I do this then I'm fine for my example but I have a further
question, how are you supposed to deal with this in general?

If you have a strange polynomial function with lots of curves in
the interval how do you find as suitable [latex]M_n[/latex] ?
Do you just use the normal calculus process of finding max & min
on this seperate term?

Is it not just a hell of a lot easier to determine the "c" term & use that?