Since ez e z is never zero for any z ∈C z ∈ C, no solution exists for e−z = 0 e − z = 0:

2015年10月8日 · Free Online Scientific Notation Calculator. Solve advanced problems in Physics, Mathematics and Engineering. Math Expression Renderer, Plots, Unit Converter, Equation …

2015年9月7日 · The hyperbolic functions are quite different from the circular ones. For one thing, they are not periodic. For your equation, the double-"angle" formula can be used: sinh x cosh x …

So if cosh x = 0 cosh x = 0, then ex +e−x = 0 e x + e − x = 0. But this can't happen because both ex e x and e−x e − x are positive, no matter what real value x x has. And the sum of two …

But if we restrict the domain of cosh cosh suitably, then there is an inverse. The usual definition of cosh−1 x cosh − 1 x is that it is the non-negative number whose cosh cosh is x x.

2017年2月4日 · For your first point, this depends on what else you know about cosh cosh. It is true that f(z)2 ≥ 1 f (z) 2 ≥ 1 does not imply f(z) ∈ [1, ∞) f (z) ∈ [1, ∞). But you may know that …

2015年10月1日 · I know that for large positive numbers cosh (x) and sinh (x) would almost be equal to ex/2 e x / 2 as e−x/2 e − x / 2 would become negligible given the magnitude of x in …

2016年10月7日 · If f(x) = 0 f (x) = 0 then x x satisfies cos x = − 1 cosh x cos x = − 1 cosh x. For large x x, cosh x cosh x is large, so x x must be such that cos x cos x is small.

Prove the following theorem: The zeros of sinh z sinh z and cosh z cosh z in the complex plane all lie on the imaginary axis. To be specific

This gives solutions x = 0, x = ln(25 7 ± 24 7) x = 0, x = ln (25 7 ± 24 7) However, when solving for cosh instead initially (and working in terms of sinh), the solutions are x = 0, x = ln(25 7 + 24 7) …