# How to prove that $\exp(x_1+x_2)=\exp(x_1)\exp(x_2)$?

In my analysis class, we covered the properties of the exponential function (as of now we use $\exp$ instead of $e$). One of the properties of $e$ is that $\exp(x_1+x_2)=\exp(x_1)\exp(x_2)$. In high school this was just assumed knowledge but now we have to prove that this statement is indeed true, which is proving to be quite difficult.

I assume that one would need to work out the derivatives of both side and see if they are equal. Then if $f(0)=1$ we can assume that both sides of the equation are true.

• How do you define $\text{exp}(x)$? – Olivier Oloa Jan 5 '18 at 12:59
• Show that $\displaystyle x \mapsto e^{x+y}-e^{x}e^{y}$ is constant. – Atmos Jan 5 '18 at 13:02
• How to prove it depends strongly on the used definition of $\exp$. If that's defined via the power series, a straightforward multiplication of power series is the way to go. If it's defined as the solution to the initial value problem $y' = y,\, y(0) = 1$, differentiating is the way to go. – Daniel Fischer Jan 5 '18 at 13:06
• @DanielFischer it's the second one. So the second part of the equations's derivative would be found using the product rule and then it should equal the derivative of the left side. – Ski Mask Jan 5 '18 at 13:09
• @croraf It's just the Cauchy product. And one uses $\frac{1}{k!n!} = \frac{1}{(k+n)!}\binom{k+n}{n}$ plus the binomial theorem. If one defines it as the limit of $(1+x/n)^n$, one needs to check that $$\biggl(1 + \frac{xy}{n^2\bigl(1 + \frac{x+y}{n}\bigr)}\biggr)^n \to 1\,$$ which isn't too hard either. – Daniel Fischer Jan 5 '18 at 13:27

From your comments it appears that you are using the definition of $\exp(x)$ as the unique solution to $f'(x) =f(x), f(0)=1$. It can be proved that such a solution must be non-zero for all $x$ (see second part of this answer). Thus $\exp(x) \neq 0$ for all $x$. Let $a$ be any arbitrary real number and consider the function $g$ defined by $g(x) =\exp(x+a) /\exp(x)$. We have $$g'(x) =\frac{\exp (x) \exp(x+a) - \exp(x+a) \exp(x) } {(\exp(x)) ^2}=0$$ and thus $g$ is a constant. It follows that $g(x) =g(0)=\exp(a)$ and thus $$\exp(x+a) =\exp(x) \exp(a)$$ Now replace $x$ by $x_1$ and $a$ by $x_2$.

The above technique does not work if one chooses $$g(x) =\exp(x+a) - \exp(x) \exp(a)$$ because one can't see that $g'(x) =0$ in very obvious manner. But this can be done with some effort. Using above definition of $g$ one gets $g'(x) =g(x),g(0)=0$. Ideally when one studies the definition of exponential function as a solution of differential equation $f'(x) =f(x), f(0)=1$ then the first step is to show that if the solution exists then it must be unique. And that uniqueness is shown by the following:

Theorem: If function $g:\mathbb{R} \to\mathbb{R}$ satisfies $g'(x) =g(x), g(0)=0$ then $g(x) =0$ for all values of $x$.

On the contrary assume that there is some $a$ such that $g(a) \neq 0$ and consider $$h(x) =g(a+x) g(a-x)$$ then $$h'(x) =g(a+x) g(a-x) - g(a+x) g(a-x) =0$$ so that $h$ is constant and $h(x) =h(0)=g(a)g(a)>0$ or in other words $g(a+x) g(a-x) >0$. Putting $x=a$ and noting that $g(0)=0$ we see that $0>0$ and this contradiction shows that $g(x) =0$ for all $x$ and we are done. The same proof can be used to show that $\exp(x) \neq 0$. We just need to consider $h(x) =\exp(x) \exp(-x)$.

if you are familiar with the Cauchy-product $CP$ you can proof this like follows:

$$exp(z) \cdot exp(w)\sum_{n=0}^{\infty} \frac{z^n}{n!}\sum_{n=0}^{\infty} \frac{w^n}{n!} \overset{CP}{=}\\ \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} \right)=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^{\infty} \frac{n!}{k!(n-k)!}z^kw^{n-k} \right)= \\ \sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^{\infty}\binom{n}{k} z^k w^{n-k} \right) = \sum_{n=0}^{\infty} \frac{1}{n!}(z+w)^n = exp(z+w)$$

This is one of the basic properties of an exponential function. Assuming that you know the basic connection between it and natural logarithm, this is that $e^x$ is the inverse of $\ln{x}$. With that being said, we can write the following

$$\ln{e^{x+y}} = x + y = \ln{e^x} + \ln{e^y} = \ln{e^x \cdot e^y}$$

Ultimately, since natural logarithm and exponential function are fundamentally related (trivially speaking one-to-one), we conclude that

$$e^{x+y} = e^x \cdot e^y$$

Be aware that such proof does not use derivatives. I hope that helps.

• And how do you prove $\ln a+\ln b=\ln a\cdot b$, that you use for the last step? At the level the OP describes it probably safe to assume that one should not use a property of the natural logarithm that is basically equivalent to what is asked, but that the definition should be used. – Henrik - stop hurting Monica Jan 5 '18 at 13:21
• I can only now see the problem in my answer, thank you for pointing it out. I suggest this post. – Pero Alex Jan 5 '18 at 13:55