exponential-infinite-product

Exponential Infinite Products

Summary: §155–§157: applying the §151 cyclotomic formula to $(1+z/j{)}^j=e^z$ with $j$ infinite gives infinite-product expansions for $e^x-1$, $(e^x-e^{-x})/2$, and $(e^x+e^{-x})/2$.

Sources: chapter9

Last updated: 2026-04-29

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$e^x-1$ (§155–§156)

From Chapter 7, $e^x=(1+x/j{)}^j$ with $j$ infinitely large, so

$e^x-1=(1+x/j{)}^j-1.$

Compare with the §151 form for $a^n-z^n$ at $a=1+x/j$, $n=j$, $z=1$. Each trinomial factor has the form

$(1+x/j{)}^2-2(1+x/j)\cos \frac{2k\pi }{j}+1.$

For $k=0$ this is the square $(x/j{)}^2$; take its square root, giving the linear factor $x/j$, i.e. simply $x$ once we pull out the constant.

For $k\ne 0$: since $j$ is infinite the arc $2k\pi /j$ is infinitesimal, so by §134 $\cos (2k\pi /j)=1-2k^2{\pi }^2/j^2$. Substituting, expanding $(1+x/j{)}^2$, and dropping terms with $j^3$ or higher in the denominator, each factor is

$\frac{x^2}{j^2}+\frac{4k^2}{j^2}{\pi }^2+\frac{4k^2}{j^3}{\pi }^{2}x=\frac{4k^2{\pi }^2}{j^2}\left(1+\frac{x}{j}+\frac{x^2}{4k^2{\pi }^2}\right).$

Discarding the multiplicative constants (they get absorbed into the leading factor of $e^x-1$, which is $x$), Euler obtains

$e^x-1=x\left(1+\frac{x}{1\cdot 2}+\frac{x^2}{1\cdot 2\cdot 3}+\cdots \right)=x\left(1+\frac{x}{j}+\frac{x^2}{4{\pi }^2}\right)\left(1+\frac{x}{j}+\frac{x^2}{16{\pi }^2}\right)\left(1+\frac{x}{j}+\frac{x^2}{36{\pi }^2}\right)\cdots$

(source: chapter9, §155). The infinitesimal $x/j$ in each factor is necessary: there are $\frac{1}{2}j$ factors and discarding it would lose a finite total $x/2$. Euler will eliminate this nuisance by combining $e^x-1$ with $1-e^{-x}$.

$(e^x-e^{-x})/2$ — the sine–hyperbolic product (§156)

Compute $e^x-e^{-x}=(1+x/j{)}^j-(1-x/j{)}^j=2(x+x^3/3!+x^5/5!+\cdots )$. Now compare with §151 for $a=1+x/j$, $z=1-x/j$, $n=j$. Each factor has the form

$a^2-2az\cos \frac{2k\pi }{j}+z^2=\frac{4x^2}{j^2}+\frac{4k^2{\pi }^2}{j^2}-\frac{4k^2{\pi }^2{x}^2}{j^4}.$

Drop the term with $j^4$ in the denominator; the resulting factor is proportional to $1+x^2/(k^2{\pi }^2)$. The $k=0$ factor, after square-rooting, contributes $x$ (not $2x$, because the square root of the constant leading term gets folded into the overall normalization). So

$\boxed{\frac{e^x-e^{-x}}{2}=x\prod _{k=1}^{\infty }\left(1+\frac{x^2}{k^2{\pi }^2}\right)=x\left(1+\frac{x^2}{{\pi }^2}\right)\left(1+\frac{x^2}{4{\pi }^2}\right)\left(1+\frac{x^2}{9{\pi }^2}\right)\cdots }$

(source: chapter9, §156). The series side is $x+x^3/3!+x^5/5!+\cdots$, that is $\sinh x$.

$(e^x+e^{-x})/2$ — the cosine–hyperbolic product (§157)

$e^x+e^{-x}=(1+x/j{)}^j+(1-x/j{)}^j=2(1+x^2/2!+x^4/4!+\cdots )$. Use the §150 form for $a^n+z^n$ with the same $a,z$. Each factor has arc $(2k+1)\pi /n$:

$a^2-2az\cos \frac{(2k+1)\pi }{j}+z^2=\frac{4x^2}{j^2}+\frac{(2k+1{)}^2{\pi }^2}{j^2}-\cdots .$

Cleaning up gives factor $1+4x^2/((2k+1{)}^2{\pi }^2)$. There is no exceptional $k=0$ term to square-root; instead $2k+1$ runs over all positive odd integers:

$\boxed{\frac{e^x+e^{-x}}{2}=\prod _{k=0}^{\infty }\left(1+\frac{4x^2}{(2k+1{)}^2{\pi }^2}\right)=\left(1+\frac{4x^2}{{\pi }^2}\right)\left(1+\frac{4x^2}{9{\pi }^2}\right)\left(1+\frac{4x^2}{25{\pi }^2}\right)\cdots }$

(source: chapter9, §157). The series side is $1+x^2/2!+x^4/4!+\cdots =\cosh x$.

What §158 does next

Substitute $x=iz$ into the boxed formulas: $(e^{iz}-e^{-iz})/(2i)=\sin z$ and $(e^{iz}+e^{-iz})/2=\cos z$ from eulers-formula. The factor $1+x^2/(k^2{\pi }^2)$ becomes $1-z^2/(k^2{\pi }^2)$, and the products turn into the infinite products for $\sin z$ and $\cos z$. See sine-infinite-product and cosine-infinite-product.

Why this is striking

These formulas are the first appearance in mathematics of an infinite product representation of an analytic function. They are not power series — they encode the zeros of the function. $\sin z$ vanishes at $z=k\pi$, and the product makes this transparent: each factor $(1-z^2/k^2{\pi }^2)$ vanishes at $z=\pm k\pi$. Likewise $\cosh x$ has no real zeros (every factor $1+4x^2/((2k+1{)}^2{\pi }^2)$ is positive), reflecting that $\cosh x\ge 1$ for real $x$.

Euler exploits these products in Chapter 10 to evaluate $\sum 1/k^2={\pi }^2/6$ and the Basel-family sums. The bridge is Newton’s identities applied to the “infinite polynomial” $\sin z/z=\prod (1-z^2/k^2{\pi }^2)$.

§159–§164 — generalizations

Sections 159–164 repeat the recipe with the more general $e^x-2\cos g+e^{-x}$ in place of $e^x\pm e^{-x}$, producing infinite products for $(\cos v+\cos g)/(1+\cos g)$, $(\cos v-\cos g)/(1-\cos g)$, $(\sin g+\sin v)/\sin g$, and a fourth variant. They are mostly catalog material — the master technique is what matters.

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• sine-infinite-product
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• eulers-formula
• newtons-identities
• basel-problem
• zeta-at-even-integers
• chapter-9-on-trinomial-factors
• chapter-10-on-the-use-of-the-discovered-factors-to-sum-infinite-series