Prove that cosh x=product from k=1 to infinity (1+
Prove that $\displaystyle\cosh x=\prod_{k=1}^{\infty}\left(1+\frac{4x^2}{(2k-1)^2\pi^2}\right)$.
Explanation of Method
Since $\cosh x=\cos(ix)$, substituting $x\to ix$ into Problem 393's product for $\cos x$ and using $i^2=-1$ flips every minus sign to a plus sign, giving the result directly. A second, independent route builds the product from the (purely imaginary) zeros of $\cosh x$ via Weierstrass factorization.
- Substitute $x\to ix$ into $\cos x=\prod\left(1-\dfrac{4x^2}{(2k-1)^2\pi^2}\right)$.
- Simplify using $\cosh x=\cos(ix)$ and $i^2=-1$.
- Cross-check via the zeros of $\cosh x$ directly.
Step 1 · Substitute $x\to ix$ in the cosine product
$$ \cos(ix) = \prod_{k=1}^\infty\left[1-\frac{4(ix)^2}{(2k-1)^2\pi^2}\right] = \prod_{k=1}^\infty\left[1-\frac{-4x^2}{(2k-1)^2\pi^2}\right] = \prod_{k=1}^\infty\left[1+\frac{4x^2}{(2k-1)^2\pi^2}\right]. $$
Step 2 · Identify $\cosh x=\cos(ix)$
$$ \boxed{\cosh x = \prod_{k=1}^{\infty}\left(1+\frac{4x^2}{(2k-1)^2\pi^2}\right)} $$ (Numerical check at $x=0.7$: partial product to $10^5$ terms gives $\approx1.255168$, matching $\cosh(0.7)\approx1.255169$.)
Step 3 · Cross-check via zeros
Solving $\cosh x=0$: $e^{2x}=-1\implies x=\pm i\tfrac{(2k-1)\pi}2$. For a purely imaginary zero at $x=\pm i\alpha_k$, the Weierstrass factor is $1-x^2/(i\alpha_k)^2=1-x^2/(-\alpha_k^2)=1+x^2/\alpha_k^2$. With $\alpha_k=(2k-1)\pi/2$: $$ \cosh x=\prod_{k=1}^\infty\left(1+\frac{x^2}{\alpha_k^2}\right)=\prod_{k=1}^\infty\left(1+\frac{4x^2}{(2k-1)^2\pi^2}\right), $$ matching Step 2 exactly.
Summary Table
| Function | Infinite product |
|---|---|
| $\cos x$ | $\prod_{k\ge1}\left(1-\dfrac{4x^2}{(2k-1)^2\pi^2}\right)$ |
| $\cosh x$ | $\prod_{k\ge1}\left(1+\dfrac{4x^2}{(2k-1)^2\pi^2}\right)$ |
Additional Notes
- This completes the family of four classical Weierstrass products in this batch: $\sin x$ and $\cos x$ (Problem 393) each acquire a hyperbolic twin ($\sinh x$ in Problem 395, $\cosh x$ here) via the single substitution $x\to ix$, which uniformly changes every $-$ sign inside the products to a $+$.
- All factors are positive for real $x$, consistent with the classical bound $\cosh x\ge1$ for all real $x$ — the product can never vanish or go negative on the real line, matching $\cosh$'s zero set being purely imaginary.
- Taking the logarithmic derivative of this product (as in Problems 397–398 for $\cot$ and $\csc$) yields the partial-fraction expansion of $\tanh x$, with poles at the same imaginary points $x=\pm i(2k-1)\pi/2$ found in Step 3.