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Soo we all know Taylor series expansion formula for expansion around expansion point $A(a,f(a))$:

$$f(x) \approx \underbrace{f(a)}_{1st~term} + \underbrace{\frac{f'(a)\, (x-a)}{1!}}_{2nd~term}+ \underbrace{\frac{f''(a)\, (x-a)^2}{2!}}_{3rd ~term}+\underbrace{\frac{f'''(a)\, (x-a)^3}{3!}}_{4th~term}+...$$

We ve all used it but rare people know where does it come from. This is soo rare that i haven't even found it on the web! I decided i want to know:

  • how did Brook Taylor construct this formula,
  • how did Brook obtain 1st, 2nd, 3rd and 4th terms? Was it something like on this webpage?
  • In this video author starts explaining "Taylor series expansion" with "power series expansion" which goes like this: $$f(x) \approx \underbrace{c_0}_{1st~term} + \underbrace{c_1(x-a)}_{2nd~term} + \underbrace{c_2(x-a)^2}_{3rd~term} + \underbrace{c_3(x-a)^3}_{4th~term}+...$$ If you absolutely need to use this "power series expansion" to explain "Taylor series expansion" please focus more on explaining the "power series expansion" as i never quite understood this. Please focus on explaining:

    • how did we construct this formula,
    • how do we know that 1st, 2nd, 3rd and 4th term in are the way they are,
    • what gives us right to approximate function $f(x)$ with power series $c_0 + c_1(x-a) + c_2(x-a)^2+...$. In other words how do we know we can do this? I think that i will need a minimal geometric explaination here.
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This is not the place to teach you Taylor series. The questions you ask are answered in any rigorous analysis textbook. I presume you studied it from a calculus textbook, these tend to just throw formulas at you without explaining much. Take an analysis textbook and all will be revealed. –  Ittay Weiss Feb 8 '13 at 0:49
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Taylor series is a representation of the function as an infinite sum of terms. Not all functions have Taylor series which converge, and even if the Taylor series converges, the limit may not be equal to the function. Functions which agree with their Taylor series are known as analytic functions. So, no, we can't always approximate $f(x)$ with power series, but when we can, it turns out to be quite useful. –  Ivan Loh Feb 8 '13 at 0:58
    
I am sorry you closed this thread as it is a VERY basic question which would help many students. It is WELL defined as i provided multiple list items which i need to be explained. If you don't know how to anwser, please don't state it is incomplete. You mathematics always just use formulas out of the box not even knowing its origins? GOSH! I am angry! –  71GA Feb 8 '13 at 9:41
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closed as not a real question by Ittay Weiss, Micah, Brett Frankel, Ron Gordon, tomasz Feb 8 '13 at 4:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

For $n$ given points on the plane (with different $x$-coordinates) there exists a polynomial of degree $n-1$ that connects the points. If $n\to\infty$, one might guess that all nice functions can be written in a (possibly infinite) polynomial form.

Assume that a function has a power series, then $$f(x)=a_0+a_1x+a_2x^2+\dots$$ for some numbers $a_n$, plugging in $x=0$ we get $f(0)=a_0$. Calculating $f'$, $$f'(x)=a_1+2a_2x+3a_3x^2+\dots$$ So $f'(0)=a_1$. And so on. The factorials come from the differentiation of decreasing $x^n$ members.

For example, the function which is itself's derivative, $f:x\mapsto e^x$ satisfies $$a_0=a_1,\ a_1=2a_2,\ a_2=3a_3,\ \dots$$ And, if we already know that $e^0=1$, then it gives $a_0=1$ and all the coefficients will be determined (yielding $e^x=1+x+\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^3}{3!}+\dots$)

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