Given that Euler's product sin x=x product from k=1 to ()
Given that Euler's product $\sin x=x\prod_{k=1}^\infty\left(1-\dfrac{x^2}{k^2\pi^2}\right)$ holds for complex $x$, and using the relationship between $\sin x$ and $\sinh x$, show that $$ \sinh x = x\prod_{k=1}^{\infty}\left(1+\frac{x^2}{k^2\pi^2}\right). $$
Explanation of Method
The identity $\sinh x=-i\sin(ix)$ lets us obtain the hyperbolic product directly from the (complex-valid) sine product by substituting $x\to ix$: since $(ix)^2=-x^2$, every minus sign inside the product flips to a plus sign.
- Recall $\sin(ix)=i\sinh x$, i.e. $\sinh x=-i\sin(ix)$.
- Substitute $x\to ix$ into Euler's sine product.
- Simplify using $i^2=-1$ and solve for $\sinh x$.
Step 1 · Substitute $x\to ix$
$$ \sin(ix) = ix\prod_{k=1}^\infty\left(1-\frac{(ix)^2}{k^2\pi^2}\right) = ix\prod_{k=1}^\infty\left(1-\frac{-x^2}{k^2\pi^2}\right) = ix\prod_{k=1}^\infty\left(1+\frac{x^2}{k^2\pi^2}\right). $$
Step 2 · Solve for $\sinh x$
Using $\sinh x=-i\sin(ix)$: $$ \sinh x = -i\cdot ix\prod_{k=1}^\infty\left(1+\frac{x^2}{k^2\pi^2}\right) = -i^2x\prod_{k=1}^\infty\left(1+\frac{x^2}{k^2\pi^2}\right). $$ Since $-i^2=1$: $$ \boxed{\sinh x = x\prod_{k=1}^{\infty}\left(1+\frac{x^2}{k^2\pi^2}\right)} $$ (Numerical check at $x=0.7$: partial product to $10^5$ terms gives $\approx0.758583$, matching $\sinh(0.7)\approx0.758584$. Series-expansion check: the coefficient of $x^3$ from the product is $\sum1/(k^2\pi^2)=\tfrac1{\pi^2}\cdot\tfrac{\pi^2}6=\tfrac16$, matching $\sinh x$'s own coefficient $1/3!=1/6$.)
Step 3 · Zero-location check
The product's zeros occur where $1+x^2/(k^2\pi^2)=0$, i.e. $x=\pm ik\pi$ — purely imaginary, consistent with $\sinh(ik\pi)=i\sin(k\pi)=0$. This confirms the product reproduces exactly the (complex) zero set of $\sinh x$.
Summary Table
| Function | Infinite product |
|---|---|
| $\sin x$ | $x\prod_{k\ge1}\left(1-\dfrac{x^2}{k^2\pi^2}\right)$ |
| $\sinh x$ | $x\prod_{k\ge1}\left(1+\dfrac{x^2}{k^2\pi^2}\right)$ |
Additional Notes
- Every factor is positive for real $x$, consistent with $\sinh x$ sharing the sign of $x$ everywhere on the real line (unlike $\sin x$, whose factors can be negative once $|x|>\pi$).
- This is the direct hyperbolic counterpart of Problem 393's $\cos x$ product; together with Problem 396's $\cosh x$ product, the substitution $x\to ix$ generates all four classical trigonometric/hyperbolic Weierstrass products from a single starting identity.
- The zeros $x=\pm ik\pi$ found here are exactly the poles that will appear (after logarithmic differentiation, as in Problems 397–398) in the partial-fraction expansion of $\coth x$ and $\operatorname{csch} x$ — the hyperbolic analogues of Problem 397's $\cot x$ expansion.