I need the proof for (e^z1)(e^z2)=e^(z1+z2)
using the power series expansion for e^z=1+z+(z^2)/2! + (z^3)/3! +....
I've tried it a couple of times but cant seem to get it.
thanks for any help!!
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I need the proof for (e^z1)(e^z2)=e^(z1+z2)
using the power series expansion for e^z=1+z+(z^2)/2! + (z^3)/3! +....
I've tried it a couple of times but cant seem to get it.
thanks for any help!!
$\displaystyle e^{z1}e^{z2} = e^{z1+z2}$ is a result of one of the rules of exponents. It requires no proof, at least not if I'm teaching.
yeah i've proved it using exponent laws no problem but our prof said to do it the other way and wants it proved using the power series.
Not really. The exponent rule $\displaystyle a^ba^c=a^{b+c}$ works for $\displaystyle a,b,c>0$. But you cannot use this rule here because $\displaystyle z_1,z_2$ are complex numbers. Thus, would does it mean to raise an exponent?Quote:
Originally Posted by icemanfan
Use the Cauchy product formula,Quote:
Originally Posted by gidget
$\displaystyle e^ae^b = \left( \sum_{n=0}^{\infty} \frac{a^n}{n!} \right) \left( \sum_{n=0}^{\infty}\frac{b^n}{n!} \right) = \sum_{n=0}^{\infty} c_n$
Where $\displaystyle c_n = \sum_{k=0}^n \frac{a^k}{k!}\cdot \frac{b^{n-k}}{(n-k)!} = \frac{1}{n!} \sum_{k=0}^n {n\choose k}a^kb^{n-k} = \frac{(a+b)^n}{n!}$
Therefore,
$\displaystyle e^ae^b = \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{a+b}$.
Hello, gidget!
Quote:
I need the proof for: .$\displaystyle e^a\!\cdot\!e^b\;=\;e^{a+b}$
using the power series expansion: .$\displaystyle e^z\:=\:1+z+\frac{z^2}{2!} + \frac{z^3}{3!} +\cdots$
We have: .$\displaystyle \begin{array}{cccccccc}
e^a &=& 1 &+\; a & +\; \dfrac{a^2}{2!} & +\; \dfrac{a^3}{3!} & +\; \dfrac{a^4}{4!} & + \hdots \\ \\[-3mm]
e^b & = & 1 & +\; b & +\; \dfrac{b^2}{2} &+\; \dfrac{b^3}{3!} &+\; \dfrac{b^4}{4!} &+ \hdots \\ \\[-3mm] \hline \end{array}$
Multiply: . $\displaystyle \begin{array}{cccccccc}
e^a\!\cdot\!e^b &=& 1 & +\;a & +\;\dfrac{a^2}{2!} & +\;\dfrac{a^3}{3!} & +\;\dfrac{a^4}{4!} & + \hdots \\ \\[-3mm]
& & & +\;b & + \;ab & +\; \dfrac{a^2b}{2!} & +\;\dfrac{a^3b}{3!} & + \hdots \\ \\[-3mm]
& & & & + \;\dfrac{b^2}{2!} & +\;\dfrac{ab^2}{2!} & +\;\dfrac{a^2b^2}{2!2!} & +\hdots \\ \\[-3mm]
& & & & & + \;\dfrac{b^3}{3!} & +\;\dfrac{ab^3}{3!} & + \hdots \\ \\[-2mm] \end{array}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \begin{array}{cc}\dfrac{b^4}{4!} & + \hdots\end{array}$
Add down the columns . . .
$\displaystyle e^a\!\cdot\!e^b \;=\;1 + (a + b) + \frac{a^2+2ab + b^2}{2!} + \frac{a^3+3a^2b+3ab^2 + b^3}{3!}$.$\displaystyle +\:\frac{a^4+4a^3b + 6a^2b^2 + 4ab^3 + b^4}{4!} + \hdots$
$\displaystyle e^a\!\cdot\!e^b\;=\;1 + (ab) + \frac{(a+b)^2}{2!} + \frac{(a+b)^3}{3!} + \frac{(a+b)^4}{4!} + \hdots \;=\;e^{a+b}$
