Fformiwla Euler

Portread 1753 gan Emanuel Handmann o Leonhard Euler. Mae'n bosibl fod ganddo broblem ar ei lygad dde (strabismus o bosib)..

Mae fformiwla Euler yn nodi fod:

$e^{i\theta }=\cos (\theta )+\sin (\theta )i$

ble mae $i$ yn rhif dychmygol sydd yn sgwario i roi $-1$.

Daw'r enw "fformiwla Euler" ar ôl y mathemategydd Leonhard Euler.

Prawf

Mae hyn yn deillio o ehangiadau Cyfres Taylor sy'n nodi fod:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+...+\frac{x^p}{p!}$

$\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\frac{{\theta }^6}{6!}+...+\frac{(-1{)}^p{\theta }^{2p}}{(2p)!}+...$

$\sin \theta =\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\frac{{\theta }^7}{7!}+...+\frac{(-1{)}^p{\theta }^{2p+1}}{(2p+1)!}+...$

Wedyn o gyfnewid $x=i\theta$ yn ehangiad Cyfres Taylor ar gyfer $e^x$ rydym yn cael:

$e^{i\theta }=1+(i\theta )+\frac{(i\theta {)}^2}{2!}+\frac{(i\theta {)}^3}{3!}+\frac{(i\theta {)}^4}{4!}+\frac{(i\theta {)}^5}{5!}+\frac{(i\theta {)}^6}{6!}+\frac{(i\theta {)}^7}{7!}+...$

$=1+(i\theta )+\frac{i^2{\theta }^2}{2!}+\frac{i{i}^2{\theta }^3}{3!}+\frac{i^2{i}^2{\theta }^4}{4!}+\frac{i{i}^2{i}^2{\theta }^5}{5!}+\frac{i^2{i}^2{i}^2{\theta }^6}{6!}+\frac{i{i}^2{i}^2{i}^2{\theta }^7}{7!}+...$

$=1+(i\theta )-\frac{{\theta }^2}{2!}-i\frac{{\theta }^3}{3!}+\frac{{\theta }^4}{4!}+i\frac{{\theta }^5}{5!}-\frac{{\theta }^6}{6!}-i\frac{{\theta }^7}{7!}+...$

$=\{1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\frac{{\theta }^6}{6!}+...\}+i\{\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-i\frac{{\theta }^7}{7!}+...\}$

$=\cos (\theta )+\sin (\theta )i$