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Taylor series

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Definition ^[?]

Proof

See Taylor's theorem

Series

General formula

An intuitive explanation of the Taylor series is that, in order to approximate the value of $f (x)$ , as a first approximation we use the value at another point $a$ , i.e. $f (a)$ . If $x$ and $a$ are close together and $f$ varies only slowly, this can be a good approximation. Then we refine the approximation step by step. The derivative of $f$ is used to calculate approximately how much $f$ would be expected to change between $a$ and $x$ , and this amount is added as a correction. But we assume we only know the derivative of $f$ at $a$ , and the derivative may change between the two numbers, so another correction is needed, involving the second derivative which is a measure of how much the first derivative changes. So it continues, adding corrections to corrections, and in the limit, if it converges then it converges to the actual value of $f (x)$ even if $x$ and $a$ are far apart.

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots+\frac{f^{(r)}(a)}{r!}(x-a)^r+\cdots$

$=\sum_{n=0}^{\infin} \frac{f^{(n)}(a)}{n!} (x-a)^{n}$

where $\scriptstyle f'$ is the first derivative of the function $\scriptstyle f$ , and $\scriptstyle f''$ is the second derivative, and so on. Most practically used smooth functions satisfy this formula but not all of them. The smooth function $f (x) = exp( - 1 / x 2)$ have all derivatives equal to zero but the function itself is not equal to zero. In order to prove Taylor formula for particular functions, mathematicians reduce the problem to proving Taylor formula for basic functions using a variety of theorems. Taylor formula for basic functions can be proved directly by estimating their derivatives.

Exponential & Logarithmic functions

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^r}{r!}+\cdots \qquad \forall x$

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+(-1)^{r+1}\frac{x^r}{r}+\cdots \qquad (-1 < x \le 1)$

Trigonometric functions

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^{r}\frac{x^{2r+1}}{(2r+1)!}+\cdots \qquad \forall x$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+(-1)^{r}\frac{x^{2r}}{(2r)!}+\cdots \qquad \forall x$

$\tan x=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\cdots+\frac{B_{2r} (-4)^r (1-4^r)}{(2r)!} x^{2r-1}+\cdots \qquad |x|<\frac{\pi}{2}$
where B_k=k^th Bernoulli number.