I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e^x$and it becomes $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ and the same for $e^X$. This image shows what I'm thinking, and what happens when I multiply them on each other. But I want to go more deeper and show the full equation of the numbers shown, for example, showing $\frac{1}{6}x^3$ and others as well. If I did, there would be a total of 16 numbers shown that are arranged in order of degrees. So I try to prove it like this picture. But the problem shows when I try to prove it more. I tried proving it more, but $\frac{1}{6}(x+X)^3$ won't work. Than what should I do to make it work? Or what formula should I use?

1The title was wrong and I have edited it. I hope you agree with the new title. – Kavi Rama Murthy Jul 18 '20 at 04:52

@KaviRamaMurthy Could you tell me the change? I can't access the change history. – Heech.p Jul 18 '20 at 04:54

1You had typed $e^{x}+X$ instead of $e^{x+X}$. – Kavi Rama Murthy Jul 18 '20 at 04:55

Use $$ \left( {\sum\limits_{n = 0}^\infty {a_n x^n } } \right)\left( {\sum\limits_{n = 0}^\infty {b_n x^n } } \right) = \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^n {a_k b_{n  k} } } \right)x^n } $$ and the binomial theorem. – Gary Jul 18 '20 at 04:56

which term of $\frac{1}{6}(x+X)^3$ you could not find? – amitava Jul 18 '20 at 04:58

@amitava I'm confused how they add up, as $(x+X)^3$ is x³+3x²y+3xy²+y³ – Heech.p Jul 18 '20 at 05:00

Yes, and you need to multiply 1/6 to each term, then which term you did not get in the expansion of $e^{x+X}$ – amitava Jul 18 '20 at 05:03

For all who looked at my question, please look at the end as well, I've updated and added another question I'm not sure about – Heech.p Jul 18 '20 at 05:04

9Using both uppercase and lowercase versions of the same variable in a calculation on a blackboard is pretty diabolical. I reckon proving $e^xe^y=e^{x+y}$ is easier than proving $e^xe^X=e^{x+X}$. – Angina Seng Jul 18 '20 at 05:04

1@Heech.p $(1)(\frac{1}{6}X^3) + (x)(\frac{1}{2}X^2) +(\frac{1}{2}x^2)(X) + (\frac{1}{6}X^3)(1) = \frac{1}{6}(x + X)^3$ – Learning Mathematics Jul 18 '20 at 05:05

1The next four terms will but $\frac 16 X^3 + \frac 12 X^2 x + \frac 12 Xx^2 + \frac 16x^3 = \frac 16(x+X)^3$. It is assumed you will recognize the pattern and take it for granted without needing any proof that the sum is in increasingly larger "chunks". The $n$th chunk will be $n+1$ terms: $\frac 1{n!}X^n + \frac 1{(n1)!}X^{n1}x + ...... + \frac 1{n!}x^n = \frac 1{n!}(x+X)^n$. If you need to prove it I'd suggest a proof by induction. – fleablood Jul 18 '20 at 06:12

@fleablood I would give this answer an accept and an upvote but it's a comment...:) – Heech.p Jul 18 '20 at 06:17

1Using pictures instead of text doesn't give math.stackexchange.com a chance to find related question. Using the variables x and X makes the formulas very confusing.So 1 – miracle173 Jul 18 '20 at 06:18

@miracle173 I understand the second problem, but the first problem I think is not a problem, it depends if someone wants to use pictures or not – Heech.p Jul 18 '20 at 06:20

Do you have to prove it using series expansion? Will you accept an alternative proof? – Koro Jul 18 '20 at 06:53

@Koro the exact answer, that's still not here, that I wanted, was an answer that would show proof, not using formulas and etc, but taking the exact numbers I showed and using them. – Heech.p Jul 18 '20 at 07:08

See also: [Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series](https://math.stackexchange.com/q/414061) (and other [questions linked there](https://math.stackexchange.com/questions/linked/414061)). – Martin Sleziak Jul 24 '20 at 10:46
5 Answers
Perhaps you're confused about the arrangement. We should get $e^xe^x=e^{2x}$, which can be written $$e^{2x} = 1 +(2x) + \tfrac{(2x)^2}{2!} + \tfrac{(2x)^3}{3!}+\cdots$$ $$=1 + 2x + \tfrac{4x^2}{2!} + \tfrac{8x^3}{3!} + \cdots$$
So let's multiply series: $$e^x\cdot e^x = \left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)$$ $$= 1\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + x\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + \tfrac{x^2}{2!}{\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)}+\cdots$$ $$= (1) + (1\cdot x + x\cdot 1) + (1\cdot \tfrac{x^2}{2!} + x\cdot x + \tfrac{x^2}{2!}\cdot 1) + \cdots $$ $$= 1 + (2)x + (\tfrac1{2!} + 1 + \tfrac1{2!})x^2 + \cdots$$ $$=1 + 2x + \tfrac4{2!}x^2 + \cdots$$ You collect the products of a fixed degree $n$ as $$1\cdot \tfrac{x^n}{n!} + x\cdot\tfrac{x^{n1}}{(n1)!} + \tfrac{x^2}{2!}\cdot\tfrac{x^{n2}}{(n2)!} + \cdots + \tfrac{x^{n2}}{(n2)!}\cdot \tfrac{x^2}{2!} +\tfrac{x^{n1}}{(n1)!}\cdot x + \tfrac{x^{n}}{n!}\cdot 1$$ $$= (\tfrac1{n!} + \tfrac{n}{n!} + \tfrac{n(n1)}{n!} + \tfrac{n(n1)(n2)}{n!} + \cdots)x^n$$ Your job is to show that this is $\tfrac{2^n}{n!}x^n$ (expand $(1+1)^n)$ using the binomial formula).
Retry:
I think you are doing it right, you just didn't collect all terms correctly.
You are really just using the distributive law, like $$(a + b + c+\cdots)(\textrm{terms}) = a\cdot(\textrm{terms}) + b\cdot(\textrm{terms}) + c\cdot(\textrm{terms})+\cdots$$ For each of the products on the RHS, you need to look for results of the same degree (we're really looking at the exponential series here, of course).
Constant terms only occur as $1\cdot 1$, in the first product on the RHS.
Degree 1 terms occur as $1\cdot x$ or $x\cdot 1$ (in the first and second products on the RHS).
Degree 2 terms occur as $1\cdot x^2$, $x\cdot x$, or $x^2\cdot 1$ (in the first, second, and third products on the RHS).
Degree 3 terms occur as $1\cdot x^3$, $x\cdot x^2$, $x^2\cdot x$, or $x^3\cdot 1$ (in the 1st, 2nd, 3rd, and 4th products on the RHS).
And so on.
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This is probably the clearest, but I'm not sure how you got to the end equation which correct :). But all the answers, including your's doesn't help much, because I asked why it would work, so it would be better if you showed how it worked. – Heech.p Jul 18 '20 at 05:53


If you would answer using examples or just showing how it would work using the numbers I gave, I would give an upvote and an accept, but this edit is still a bit confusing to me – Heech.p Jul 18 '20 at 06:22

The expression $x+X$ is meaningless as $x$ and $X$ are quite different things (one is scalar, the other one a matrix), hence it is not $$ e^{x+X}=1+{x+X}+{(x+X)^2\over 2!}+\cdots $$ But the following identity turns out to be true:$$e^{xI+X}=1+{xI+X}+{(xI+X)^2\over 2!}+\cdots$$where $I$ is the identity matrix of the same order of $X$. This is indeed true because of the following implication:$$AB=BA\implies e^{A}e^{B}=e^{A+B}$$
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I asked why it something worked, and you said something different, just didn't help much as it came to an answer I knew, but wanted to prove – Heech.p Jul 18 '20 at 05:54


@MostafaAyaz I'm happy you caught your mistake, but I hope you understood that I wanted proof using the numbers I gave, as for your answer, it proved something I already know, but I wanted to prove – Heech.p Jul 18 '20 at 05:57
\begin{align} & \left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) \\[10pt] = & \Big(\bullet\Big) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left( \bullet \cdot \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left(\left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \frac{b^k}{k!} \right) \\[8pt] = {} & \sum_{k=0}^\infty \sum_{j=0}^\infty \left( \frac{a^j}{j!} \cdot \frac{b^k}{k!} \right). \end{align} This sum includes all the cases where $j+k=0,$ and all cases where $j+k=1$, and all cases where $j+k=2,$ and so on. What are the cases where $j+k=4,$ for example? They are these: $$ \frac{a^0b^4}{0!4!} + \frac{a^1 b^3}{1!3!} + \frac{a^2b^2}{2!2!} + \frac{a^3 b^1}{3!1!} + \frac{a^4b^0}{4!0!} $$ That is the same as \begin{align} & \frac 1 {4!} \left( \frac{4!}{0!4!} a^0b^4 + \frac{4!}{1!3!} a^1 b^3 + \frac{4!}{2!2!} a^2b^2 + \frac{4!}{3!1!} a^3 b^1 + \frac{4!}{4!0!} a^4b^0 \right) \\[8pt] = {} & \frac 1 {4!} (a+b)^4. \end{align} And similarly for other numbers than $4.$ Thus the sum is $$ \sum_{\ell=0}^\infty \frac 1 {\ell!} (a+b)^\ell = e^{a+b}. $$
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Lets look at a couple functions. $Y=e^X$ and $y=e^x$.
Then $\ln Y=X$ and $\ln y = x$. $$X+x=\ln (Y)+\ln (y) = \ln (Yy) \Longrightarrow e^{X+x}=Yy=e^Xe^x$$
Ok, so there is a little informal proof using some properties of the natural log.
$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ (Note: I am not using z as a complex variable, only a stand in for x because it is used so much in the question.)
So, $$e^x=1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+...$$
$$e^X=1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}+...$$
$$\begin{align*} e^xe^X &= \bigg(1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+... \bigg)\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg) \\&= 1\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg) + x\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg)+ \frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg)...\\&= \bigg(1+X+\frac{X^2}{2} + \frac{X^3}{6}...\bigg) + \bigg(x+xX+x\frac{X^2}{2}+x\frac{X^3}{6}...\bigg)+\bigg(\frac{x^2}{2} + \frac{x^2}{2}X + \frac{x^2}{2}\frac{X^2}{2}+\frac{x^2}{2}\frac{X^3}{6}...\bigg)+...\\&= 1+(x+X)+\bigg(\frac{x^2}{2}+xX+\frac{X^2}{2}\bigg)+\bigg(\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}\bigg)+...\\ &= 1+(x+X)^1 +\frac{(x+X)^2}{2} + \frac{(x+X)^3}{6} + ...\\&= e^{x+X}\end{align*}$$
Another way: $$\begin{align*} e^{x+X}=\sum_{j=0}^{\infty} \frac{1}{j!}(x+X)^j&= \sum_{j=0, 0\le k\le j}^{\infty} \frac{1}{j!} {j\choose k}x^{jk}X^k \\ &= \frac{1}{0!}\bigg[{0\choose 0}(x^{00}X^0)\bigg] + \frac{1}{1!}\bigg[{1\choose 0}(x^{10}X^0)+{1\choose 1}(x^{11}X^1)\bigg] +\frac{1}{2!}\bigg[{2\choose 0}(x^{20}X^0)+{2\choose 1}(x^{21}X^1) + {2\choose 2}(x^{22}X^2)\bigg] +...\\&= 1+x+X + \frac{x^2}{2} +xX + \frac{X^2}{2} +\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+\bigg(x +xX + x\frac{X^2}{2}+...\bigg) + \bigg(\frac{x^2}{2}+\frac{x^2 X}{2} +\frac{x^2X^2}{2\cdot 2}+...\bigg)+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+x\bigg(1+X+\frac{X^2}{2}+...\bigg)+\frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+...\bigg)+...\end{align*}$$
You can finish it from there.
The exact reason this works is hidden in the way the Taylor series defines a function. I will direct you to this answer, which in my opinion, is a really nice and different way of understanding the intuition behind the Taylor series. The top answer on that page directs to another link of the proof of the Taylor series as well.
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You can try the following without resorting to series expansion of $e^x$:
We'll use the following famous limit :
$ e^x= \lim_{n \to \infty} (1+n^{1})^{nx} $
Hence, $e^x\cdot e^X=\lim_{n \to \infty} (1+n^{1})^{nx}\lim_{n \to \infty} (1+n^{1})^{nX}=\lim_{n\to \infty} (1+n^{1})^{n(x+X)}=e^{x+X} $
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