Compare exp(a(x+y)) to a(exp(x) + exp(y))

Question

I would like to compare $\exp(a(x+y))$ to $a(\exp(x) + \exp(y))$ for $a>0$. How do I approach this?

This is false as stated. For the related true statement, please tell us what you are assuming. — Dunham, Jun 25 '19 at 13:55
I'm working on this with a constant multiplied out front: https://math.stackexchange.com/questions/285227/prove-expxy-expx-expy-for-expx-sum-n-0-infty-frac-xn — MadProgrammer, Jun 25 '19 at 13:59
Please look again at that post. The right hand side is a multiplication, not addition. — mjw, Jun 25 '19 at 14:01
Even if you replace plus with multiplication its still not correct. — kingW3, Jun 25 '19 at 14:02
Do you want $\exp(a(x+y))=\exp(ax)\exp(ay)$ or $a\exp(x+y)=a\exp(x)\exp(y)$ — Dunham, Jun 25 '19 at 14:02
You are right. I have updated the question. I am really interested in comparing aexp(x+y) and aexp(x)exp(y). My assumption was that they are equal. — MadProgrammer, Jun 25 '19 at 14:07
These are two different two-variable functions, neither is always greater than the other. Are you interested in asymptotics or some particular point of the domian? — Dunham, Jun 25 '19 at 14:14
Well at 0, exp(a(x+y)) is 1, and a(exp(x)+exp(y)) is a. What happens as they go to infinity? Which gets there faster? Are they ever equal? Suppose a = 1/2? Suppose a = 2? — MadProgrammer, Jun 25 '19 at 14:26
Remember that $\exp a(x+y) = \mathrm e^{a(x+y)}$. Can you expand that using exponent rules? — gen-ℤ ready to perish, Jun 25 '19 at 14:27

score 1 · Answer 1 · 2019-06-25T14:26:02.073

1

Set $y=0$ and your identity becomes

$$\exp(ax)=a\exp(x),$$ which is notoriously false.

Still not convinced ? With $x=1$,

$$\exp(a)=a\,e\ ???$$

A correct statement is

$$(\exp(x+y))^a=\exp(a(x+y))=\exp(ax+ay)=\\\exp(ax)\exp(ay)=(\exp(x))^a(\exp(y))^a=(\exp(x)\exp(y))^a.$$

edited Jun 25 '19 at 14:26

answered Jun 25 '19 at 14:05

“notoriously” seems a bit extra – gen-ℤ ready to perish Jun 25 '19 at 14:26
@ChaseRyanTaylor: read obviously then. – Jun 25 '19 at 14:28
I don’t know what you’re referring to – gen-ℤ ready to perish Jun 25 '19 at 14:29

score 0 · Answer 2 · answered Jun 25 '19 at 14:10

0

The expression is not valid. For example if you let $a=0$ you get $1=0$

What you may ask is $$e^{a( x+y)}=e^{ax}e^{ay}$$

answered Jun 25 '19 at 14:10

score 0 · Answer 3 · answered Jun 25 '19 at 14:12

0

Careful with typos!

$$e^{a(x+y)}=e^{ax} e^{a y}$$

i.e.,

$$\exp(a(x+y))=\exp(ax)\exp(a y)$$

answered Jun 25 '19 at 14:12

Narasimham

score 0 · Accepted Answer · answered Jun 25 '19 at 20:45

It is probably easiest to restrict to lines in the plane and consider what happens. Here are a few cases:

Case 1: $y = 0$ (By symmetry, similar to $x=0$)

We are comparing $\exp(ax)$ to $a\exp(x)$.

Taking the natural log of each, we see that they are equal when $x = \log(a)/(a-1)$.

If $a>1$, $\exp(ax)$ grows faster as $x\rightarrow \infty$, and decays to 0 faster as $x\rightarrow -\infty$

If $a<1$, $a\exp(x)$ grows faster as $x\rightarrow \infty$, and decays to 0 faster as $x\rightarrow -\infty$

Case 2: $x = y$

We are comparing $\exp(2ax)$ to $2a\exp(x)$.

The analysis is very similar, with $a$ replaced by $2a$.

Note

Without too much work, you can also do asymptotics for $y = mx$ (arbitrary line of slope $m$ in the plane).

Thanks for working with me while I clarified the question. Sorry to confuse everyone else. — MadProgrammer, Jun 26 '19 at 15:10

4 Answers4