is-possible-to-proof-that-f-x-e-cx-obey-f-x-y-f-x-f-y-using-e-x-n-0-x-n-n-

Question Number 4962 by 123456 last updated on 27/Mar/16

is possible to proof that

f (x) = e^{c x}

obey

f (x + y) = f (x) f (y)

using

e^{x} = \underset{n = 0}{\sum^{+ \infty}} \frac{x^{n}}{n!}

?

Commented by 123456 last updated on 27/Mar/16

e^{x + y} = \underset{n = 0}{\sum^{+ \infty}} \frac{{(x + y)}^{n}}{n!}

= \underset{n = 0}{\sum^{+ \infty}} \frac{1}{n!} \underset{m = 0}{\sum^{n}} (\begin{matrix} n \\ m \end{matrix}) x^{m} y^{n - m}

= \underset{n = 0}{\sum^{+ \infty}} \frac{1}{n!} \underset{m = 0}{\sum^{n}} \frac{n!}{m! (n - m)!} x^{m} y^{n - m}

= \underset{n = 0}{\sum^{+ \infty}} \underset{m = 0}{\sum^{n}} \frac{1}{m! (n - m)!} x^{m} y^{n - m}

Commented by prakash jain last updated on 08/Jan/17

e^{x + y} = \underset{i = 0}{\sum^{\infty}} \frac{{(x + y)}^{i}}{i!}

{(x + y)}^{m} = \underset{i = 0}{\sum^{m}}^{m} C_{i} x^{i} y^{m - i}

c o e f f i c i e n t o f x^{k} i n e^{x + y}

\underset{m = k}{\sum^{\infty}} \frac{1}{m!} \cdot \frac{m!}{k! (m - k)!} y^{m - k} = \frac{1}{k!} \underset{m = k}{\sum^{\infty}} \frac{y^{m - k}}{(m - k)!}

= \frac{1}{k!} \underset{m = 0}{\sum^{\infty}} \frac{y^{m}}{m!} = \frac{e^{y}}{k!}

so e^{x + y} = \underset{k = 0}{\sum^{\infty}} \frac{e^{y}}{k!} x^{k} = e^{y} e^{x}