Prove that product from n=1 to infinity cosh x/(2^n)=frac
Prove that $\displaystyle\prod_{n=1}^{\infty}\cosh\frac{x}{2^n}=\frac{\sinh x}x$.
Explanation of Method
The hyperbolic double-angle identity $\sinh(2\theta)=2\sinh\theta\cosh\theta$, rearranged as $\cosh\theta=\dfrac{\sinh2\theta}{2\sinh\theta}$, turns each factor $\cosh(x/2^n)$ into a ratio of consecutive $\sinh$ values. Chaining these ratios telescopes the entire product down to $\sinh x$ over $2^N\sinh(x/2^N)$, and the limit follows from $\lim_{y\to0}\sinh(y)/y=1$.
- Rewrite each $\cosh(x/2^n)$ using the hyperbolic double-angle identity.
- Telescope the resulting chain of $\sinh$-ratios.
- Take the limit $N\to\infty$ using $\sinh(y)/y\to1$ as $y\to0$.
Step 1 · Double-angle rewrite
Since $\sinh(2\theta)=2\sinh\theta\cosh\theta$, we have $\cosh\theta=\dfrac{\sinh2\theta}{2\sinh\theta}$. With $\theta=x/2^n$: $$ \cosh\frac{x}{2^n}=\frac{\sinh\dfrac{x}{2^{n-1}}}{2\sinh\dfrac{x}{2^n}}. $$
Step 2 · Telescoping the partial product
$$ P_N=\prod_{n=1}^N\cosh\frac{x}{2^n}=\frac1{2^N}\prod_{n=1}^N\frac{\sinh\dfrac{x}{2^{n-1}}}{\sinh\dfrac{x}{2^n}}=\frac1{2^N}\cdot\frac{\sinh x}{\sinh\dfrac{x}{2^N}}=\frac{\sinh x}{2^N\sinh\dfrac x{2^N}}. $$ (Verified by induction: the $N\to N+1$ step multiplies by $\cosh(x/2^{N+1})=\dfrac{\sinh(x/2^N)}{2\sinh(x/2^{N+1})}$, which correctly advances the formula.)
Step 3 · Take the limit
Writing $y=x/2^N\to0$ as $N\to\infty$, and using $2^N=x/y$: $$ 2^N\sinh\frac x{2^N}=x\cdot\frac{\sinh y}y\;\longrightarrow\;x\cdot1=x. $$ $$ \boxed{\lim_{N\to\infty}P_N=\frac{\sinh x}x} $$ (Numerical check at $x=1.3$: $P_{20}\approx1.306448$ vs. $\sinh(1.3)/1.3\approx1.306448$ — matching to 6 decimal places.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{\sinh x}{2^N\sinh(x/2^N)}$ |
| Limit | $\dfrac{\sinh x}x$ |
| Domain | all real $x$ (with $x=0$ read as the limiting value $1$) |
Additional Notes
- This is the exact hyperbolic counterpart of the trigonometric identity $\prod_{n=1}^\infty\cos(\varphi/2^n)=\sin(\varphi)/\varphi$ — the same telescoping argument works verbatim after replacing $\sin,\cos$ by $\sinh,\cosh$, since both pairs satisfy an identical double-angle relation.
- Unlike the trigonometric version, which is only meaningfully bounded for real $\varphi$, this hyperbolic product converges for every real (and in fact every complex) $x$, since $\cosh$ has no zeros on the real axis to cause the partial products to misbehave.
- Structurally this belongs with Problems 363 and 366: all three build an infinite product indexed by powers of $2$ that telescopes via a doubling identity down to a strikingly simple closed form.