Prove that $\exp\left(\frac{p}{q} x\right) = \exp(x)^\frac{p}{q}$, considering $\displaystyle\exp(x) =\sum_{n=0}^\infty\frac{x^n}{n!}$.

I must use only the definition of exp(x) being a power series to show that $\exp(\frac{p}{q} \cdot x)$. I already have that $\exp(n\cdot x) = \exp(x)^n$ for $n \in \mathbb{N}$, which I used induction on $n$ to prove.

I wanted to consider the case p=1, then I tried to make an induction on $q$ but it wasn't useful. How could I prove it?

  • $\begingroup$ From $\exp(nx)=\exp(x)^n$, you can prove that $\exp\left(\frac{x}{q}\right)=\exp(x)^{\frac1q}$. $\endgroup$ – Batominovski Jan 22 '17 at 16:26

If you already know that $\exp(nx)=\exp(x)^n$, then you can set $y=x/q$, so $$ \exp(py)=\exp(y)^p $$ On the other hand, $$ \exp(x)=\exp(qy)=\exp(y)^q $$

so $$\exp(y)=\exp(x)^{1/q}$$

This works for $p,q>0$. The case $p<0$ follows easily from $\exp(-x)=\exp(x)^{-1}$

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