Proof of a function's representation with Taylor Series
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\title{Proof of Function Representation with Taylor Series}
\author{Adrian D'Costa}
\begin{document}
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We know that a power series is:
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\begin{align*} f(x) =& a_0 + a_1(x-a) + a_2(x-a)^2 + a_3(x-a)^3 + a_4(x-a)^4 \\&+ a_5(x-a)^5 + ... + a_n(x-a)^{n}... \text{[where }\left|x-a\right| < R\text{]}\end{align*}
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\section{Section (1)}
Now:
$f(a) = a_0$
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$\therefore a_0 = \frac{f^{(0)}(a)}{0!} \text{ where } f^{(0)}(a) = f(a) \text{ and } 0! = 1$
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\section{Section (2)}
Taking the first derivative of f(x):
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$f^{(1)}(x) = a_1 + 2a_2(x-a) + 3a_3(x-a)^2 + 4a_4(x-a)^3 + 5a_5(x-a)^4 + ...$
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$f^{(1)}(a) = a_1.....\text{(i)}$
$a_{1} = \frac{f^{(1)}(a)}{1!}.....\text{[rearranging (i)]}$
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\section{Section (3)}
Same way taking the second derivative of $f(x)$:
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$f^{(2)}(x) = 2a_2 + 6a_3(x-a) + 12a_4(x-a)^2 + 20a_5(x-a)^3 + ...$
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$f^{(2)}(a) = 2a_2.....\text{(ii)}$
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$ a_2 = \frac{f^{(2)}(a)}{2!}....\text{[rearranging (ii)]}$
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\section{Section (4)}
Taking third derivative:
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$f^{(3)}(x) = 6a_3 + 24a_4(x-a) + 60a_5(x-a)^2 + ...$
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$f^{(3)}(a) = 6a_3.....\text{(iii)}$
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$a_3 = \frac{f^{(3)}(a)}{3!}....\text{[rearranging (iii)]}$
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\section{Section (5)}
Taking fourth derivative:
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$f^{(4)}(x) = 24a_4 + 120a_5(x-a) + ...$
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$f^{(4)}(a) = 24a_4.....\text{(iv)}$
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$a_4 = \frac{f^{(4)}(a)}{4!}....\text{[rearranging (iv)]}$
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\section{Section (6)}
Taking fifth derivative:
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$f^{(5)}(x) = 120a_5 + ...$
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$f^{(5)}(a) = 120a_5.....\text{(iv)}$
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$a_5 = \frac{f^{(5)}(a)}{5!}....\text{[rearranaging (v)]}$
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By plugging in values of $a_{0}, a_{1}, a_{2}, a_{3}, a_{4} \text{ and } a_5$ into $f(x)$ we get:
\begin{align*}f(x) =&\frac{f^{(0)}(a)(x-a)^0}{0!} + \frac{f^{(1)}(a)(x-a)^1}{1!} + \frac{f^{(2)}(a)(x-a)^2}{2!} + \frac{f^{(3)}(a)(x-a)^3}{3!} \\&+ \frac{f^{(4)}(a)(x-a)^4}{4!} + \frac{f^{(5)}(a)(x-a)^5}{5!} + ...\end{align*}
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So the pattern is:
$f(x) = \mathlarger{\mathlarger{\sum}}\limits_{n = 0}^n\frac{f^{(n)}(a)(x-a)^n}{n!}$
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And that's it. That's the Taylor series.
$\text{[Q.E.D]}$
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