PX153 - F1 - taylor series

taylor's theorem gives a way of expressing functions as a power series
it uses the derivatives of a function to approximate its value around a point

\frac{d f}{d x} |_{x = a} ≃ \frac{f (a + Δ x) - f (a)}{Δ x}

for small $Δ x$ :

f (a + Δ x) ≃ f (a) + \frac{d f}{d x} |_{x = a} Δ x

make a more accurate approximation by adding higher order derivatives. let's assume that we can approximate the function as a power series, a polynomial in $(x - a)$ , that will be valid for some interval around $x = a$ , ie:

f (x = a) ≃ A_{0} + A_{1 (x - a)} + A_{2} (x - a)^{2} + A_{3} (x - a)^{3} + . . .

f (x = a) ≃ A_{0} + \sum_{n = 1}^{N} A_{n} (x - a)^{n} $ $ w h e r e $ A_{0} $, $ A_{1} $, $ A_{2} . . . A_{n} $ a r e c o n s t a n t s w e m u s t f i n d $ $ \frac{d f}{d x} = A_{1} + 2 A_{2} (x - a) + 3 A_{3} (x - a^{2}) + . . .

\frac{d^{2} f}{d x^{2}} = 2 A_{2} + 3.2 . A_{3} (x - a) + 4.3 . A_{4} (x - a)^{2} + . . .

evaluate these at $x = a$ , allows us to find $A_{0}$ , $A_{1}$ , $A_{2}$ , etc

f (x) ≃ f (a) + f^{'} |_{x = a} (x - a) + \frac{1}{2!} f^{″} |_{x = a} (x - a)^{2} + . . . + \frac{1}{n!} f^{n^{'}} |_{x = a} (x - a)^{n}

f (x) ≃ f (a) + \sum_{n = 1}^{\infty} \frac{1}{n!} f^{n^{'}} |_{x = a} (x - a)^{n}

$\frac{1}{n!} f^{n^{'}} |_{x = a} = A_{n}$ is just a number
this is the taylor expansion of $f (x)$
if $a = 0$ , this is the maclaurin expansion
eg: find the firs 3 non-zero terms of the taylor expansion of $f (x) = \sin x$ about $x = 0$ (ie: maclaurin expansion)
$$f(x) = \sin x \simeq \sin 0 + \sum\limits_{n=1}^{\infty}\frac{1}{n!} f^{n'}|_{x=a}(x)^{n}$$
$f (0) = \sin 0 = 0$
$f^{'} (0) = \cos (0) = 1$
$f^{″} (0) = - \sin (0) = 0$
$f^{‴} (0) = - \cos (0) = - 1$
$f^{‴} (0) = - \cos (0) = - 1$
$f^{4^{'}} (0) = \sin (0) = 0$
$f^{5^{'}} (0) = \cos (0) = 1$
$⟹ \sin x = x - \frac{1}{3!} x^{3} + \frac{1}{5!} x^{3} + . . .$
- $\sin x$ is an odd function, so, we only have odd powers of $x$
- note: $x$ is in $r a d i a n s$
eg: estimate the value of $\cos 62 °$ by expanding $\cos x$ around $60 °$
- switch to radians: $60 ° = \frac{π}{3}$ , $62 ° - 60 ° = 2 ° = \frac{π}{90}$ $\cos (\frac{π}{3} + \frac{π}{90}) ≃ \cos (\frac{π}{3}) + \sum_{n = 1}^{\infty} \frac{1}{n!} f^{n^{'}} |_{\frac{π}{3}} (\frac{π}{90})^{n}$ $\cos (\frac{π}{3}) = \frac{1}{2}$
  $f^{'} (\frac{π}{3}) = \frac{- \sqrt{3}}{2}$
  $f^{″} (\frac{π}{3}) = \frac{- 1}{2}$
  $f^{‴} (\frac{π}{3}) = \frac{\sqrt{3}}{2}$
$\cos 62 ° ≃ \frac{1}{2} - \frac{\sqrt{3}}{2} (\frac{π}{90}) - \frac{1}{2!} \frac{1}{2} (\frac{π}{90})^{2} + . . .$ $\cos 62 ° ≃ 0.5 - 0.302 - 00003 + . . . ≃ 0.4695$
(accurate to 4 sf)
a useful trick is the relationship the derivative/integral of the taylor series is equal to the derivative/integral of the function
if the expansion is around $0$ , you can find the expansion of a power of the independent variable (eg: $x^{3}$ ), by substituting it into the taylor expansion of the function of $x$
eg: if you want to find the expansion of $\sin x^{3}$ about $x = 0$ , find the expansion for $\sin x$ and replace $x$ with $x^{3}$

⟹ \sin x^{3} = x^{3} + \frac{x^{9}}{3!} + \frac{x^{15}}{5!} + . . .

non analytical function

every differential is zero