Fformwla Euler

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Neidio i: llywio, chwilio
Portread 1753 gan Emanuel Handmann o Leonhard Euler. Mae'n bosibl fod ganddo broblem ar ei lygad dde (strabismus o bosib)..[1]

Daw enw fformwla Euler ar ôl Leonhard Euler.

Mae fformwla Euler yn nodi fod:

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

ble mae i yn rif dychmygol sydd yn sgwario i roi -1.

Prawf[golygu]

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} yr ydym 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{ii^{2}\theta^{3}}{3!} + \frac{i^{2}i^{2}\theta^{4}}{4!} + \frac{ii^{2}i^{2}\theta^{5}}{5!} + \frac{i^{2}i^{2}i^{2}\theta^{6}}{6!} + \frac{ii^{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} \,

Cyfeiriadau[golygu]

  1. Calinger, Ronald (1996). "Leonhard Euler: The First St. Petersburg Years (1727–1741)". Historia Mathematica 23 (2): 154–155. doi:10.1006/hmat.1996.0015. 
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