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Original question: Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x)=\dfrac{a^x-a^{-x}}{2}$, where $a>0$ and $a\ne 1$, and $\alpha$ be a real number such that $f(\alpha)=1$. Find $f(2\alpha)$.$^1$

A few years ago, I was a high school student and solved it. Now I am reading the book again, because I has begun teaching my cousin since last week. When I revisited the question, suddenly I wanted to find $f(2\alpha)$, $f(3\alpha)$, $f(4\alpha),\;\dots$ \begin{align} f(2\alpha)&=\frac{a^{2\alpha}-a^{-2\alpha}}{2}\\ &=\frac{(a^{\alpha}-a^{-\alpha})(a^{\alpha}+a^{-\alpha})}{2}\\ &=f(\alpha)\sqrt{a^{2\alpha}+2+a^{-2\alpha}}\\ &=f(\alpha)\sqrt{(a^{\alpha}-a^{-\alpha})^2+4}\\ &=f(\alpha)\sqrt{4(f(\alpha))^2+4}\\ f(3\alpha)&=f(\alpha)(4(f(\alpha))^2+3)\;(\text{calculations skipped})\\ f(4\alpha)&=f(\alpha)(4(f(\alpha))^2+2)\sqrt{4(f(\alpha))^2+4} \end{align}

My question: Can we express $f(n\alpha)$ in terms of $f(\alpha)$? (Here $f(\alpha)$ isn't necessarily $1$.)

Attempt: It is known that $x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+y^{n-1})$ for $n\in \mathbb{N}$, so $$f(n\alpha)=\frac{a^{n\alpha}-a^{-n\alpha}}{2}=\frac{(a^{\alpha}-a^{-\alpha})(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})}{2}.$$ However, $(a^{(n-1)\alpha}+a^{(n-3)\alpha}+\cdots+a^{(3-n)\alpha}+a^{(1-n)\alpha})$ term is annoying me. I think it will be $2(f((n-1)\alpha)+f((n-3)\alpha)+\cdots+?)$, but I have no idea how to do next.

Partial solution is also appreciated.

$^1$ It was translated from Korean to English by me. Reference: Sunwook Hwang and 12 other authors (2010).『수학Ⅰ 익힘책』. Seoul: (주)좋은책신사고. page 62.

choco_addicted
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3 Answers3

4

We can rewrite the function as

$$f(x)=\sinh(x\ln a)$$

then $f(nx)=\sinh(nx\ln a)$

let, $x\ln a=t$ then, $f(t)=\sinh(t)$ and $f(nt)=\sinh(nt)$ also, $\cosh(t)=\sqrt{1+f(t)^2}$

Finally use the fact that $$\sinh(nt)=\dfrac{(\cosh(t)+\sinh(t))^n-(\cosh(t)-\sinh(t))^n}{2}$$

Someone
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3

From $f(\alpha)=\frac{a^\alpha - a^{-\alpha}}{2}$, we can induce the quadratic equation for $a^{\alpha}$: $$ a^{2\alpha}-2f(\alpha)a^{\alpha}-1=0. $$ By the quadratic formula, we get $$ a^{\alpha}=f(\alpha)\pm \sqrt{(f(\alpha))^2+1} $$ and using $a^{\alpha}>0$ we eliminate one possibility. Thus $$ a^{n\alpha}=(a^{\alpha})^n =\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n $$ and \begin{align} a^{-n\alpha}&=\frac{1}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n}\\ &=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}\\ &=\frac{\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n}{(f(\alpha)^2+1-f(\alpha)^2)^n}\\ &=\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n. \end{align} Therefore $$ f(n\alpha)=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^n -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^n\right) $$ Checking the formula for $n=2$: \begin{align} f(2\alpha)&=\frac{1}{2}\left(\left(\sqrt{f(\alpha)^2+1} + f(\alpha)\right)^2 -\left(\sqrt{f(\alpha)^2+1} - f(\alpha)\right)^2\right)\\ &=\frac{1}{2}(f(\alpha)^2+1+2f(\alpha)\sqrt{f(\alpha)^2+1} + f(\alpha)^2 -f(\alpha)^2-1+2f(\alpha)\sqrt{f(\alpha)^2+1}-f(\alpha)^2)\\ &=\frac{1}{2}\cdot 4f(\alpha)\sqrt{f(\alpha)^2+1}\\ &=2f(\alpha)\sqrt{f(\alpha)^2+1} \end{align}

choco_addicted
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2

Let $a^x = y$, then $y$ can be solved in terms of the RHS. Let $f(\alpha) = A$, from:

$$ y - \frac{1}{y} = 2A $$ with the quadratic root formula, we have $$y = g(A) = A + \sqrt{1 + A^2}$$ After this it is a matter of substitution and simplification: $$ f(n \alpha) = \frac{1}{2} (g(A)^n - g(A)^{-n}) $$

Taozi
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