How to simplify a sum of exponential equation?

Question

Suppose I have three constants $a, b, c\in R$. I have a formulation as $f=e^{ab}+e^{ac}$. Can I have some result like $f'=e^{a(b+c)}$. I know $f'$ does not hold. But I just want to combine the two terms of the right-hand side in $f$ together. Could somebody give me some hints? Thanks a lot.

Answer 1

$$e^{ab}+e^{ac}=e^{ac}\left(\frac{e^{ab}}{e^{ac}}+1\right)=e^{ac}\left(e^{ab-ac}+1\right)=e^{ac}\left(e^{a(b-c)}+1\right)$$

Answer 2

For the complex extension of $\cos(x)$, we have $$\cos(x)=\frac{e^{ix}+e^{-ix}}2$$Which is how we evaluate $\cos(i)$.

From here, we note your expression:$$e^{ab}+e^{ac}$$If we allow $b=-c$, then we have $$e^{ab}+e^{-ab}=2\cos(\frac{ab}i)$$If not, then we can't really do much.

We can also factor some things out:$$e^{ab}+e^{ac}=[e^a]^b+[e^a]^c$$

We notice that if we have $x=e^a$, then we get$$x^b+x^c$$First, let's assume $b>c$. Then we factor $x^c$ out to get $$x^c[x^{b-c}+1]$$

From here, I will note that by setting this equal to zero, we can find factors.$$x^c[x^{b-c}+1]=0$$

$x^c$ cannot equal $0$, therefore, we set the inside equal to zero.$$x^{b-c}+1=0$$$$x^{b-c}=-1$$$$x=(-1)^{\frac1{b-c}}$$

Find all $b-c$-th roots of $-1$.

You may ask why we aren't just assuming normal form, and that is because we are trying to factor our polynomial, where complex parts are permitted.

We will call these roots $x_1,x_2,x_3,\cdots x_{b-c}$, noting there will be exactly $b-c$ different roots of $-1$. You can find a similar case here. Then, when dealing with polynomials and finding roots, we can always change our polynomial into the form:$$(x-root)(x-root)(x-root)\cdots$$For an example:$$x^2+1\to(x+i)(x-i)\to x=(-1)^{\frac12}=i,-i$$

Substitute this back in to get:

$$x^c\Pi_{n=0}^{b-c}(x-x_n)$$

Where $/Pi$ is the brother of $\sum$, it multiplies its terms instead of adding them.

And that's it, factored all the way.

Oh yes, substitute $x=e^a$ back in to get $$e^{ac}\Pi_{n=0}^{b-c}(e^a-x_n)$$

We note that I have assumed $b>c$. If it is the other way around, use the commutative property of addition to switch the positions of $b$ and $c$ and solve just the same way.

Also note that this only works if $b-c$ is a positive whole number. Reason is simple, we can't do the following$$\Pi_{n=0}^{2.5}$$Or, similarly:$$\sum_{n=0}^{2.5}$$At least we can't use them without more advanced skills, skills I don't have, only reference to: here.