extract x: $((x+y)^n - x^n)^{1/n}$ expressed in the form $ba^{c}$ where all that matters is b contains x

Question

I've done a bit of calculus/diff.eq./lin.alg. in the past. But that was a long time ago and to be honest I was always better at the actual applying than the pure math.

Anyway, I have a function which I am trying to get into a certain form. I want to do this by isolating/extracting one variable.

I've been trying on paper, Mathematica, Wolfram|Alpha and it's just not making sense to me anymore.

Here's the function:

$$((x+y)^n - x^n)^{1/n}$$

And I'm trying to extract the x out of this so I get something of the form:

$$xa^\text{exponent}$$

Whatever the exponent is or 'a' is or if x has any +something/-something/divided by/exponents, or even if 'a' still contains x's I want x times something to an exponent.

What I have come up with myself so far is driving me mad: I get two forms for 'a' depending on if I do

$$(xa)^{1/n}$$

or

$$x(a^{1/n})$$

As I said, I am really rusty and tried a lot of things which I think are oh-so wrong, but this is what I got trying to extracate x:

$$x^n ((1+((1/x)y)^n -1)^{1/n}$$

OR

$$x^n ((((1/x) + ((1/(x^n))y))^n)-1)^{1/n}$$

I'm not too proud to say I even got:

$$x((1+(1/x)y)^n - x(1/n))^{1/n}$$

I'm doing this all for a hobby programming project of mine (lead android programmer) and I just can't get it to work out ... plus I would like to be sure the answer is correct!

If it helps, $x$,$y$ and $n$ are all positive integers. $y<x$ and $y\geq 1$, $x\geq3$ and $n\geq2$.

Answer 1

Your given equation is equivalent to $$\sqrt[n]{\binom{n}{1} x^{n-1} y + \binom{n}{2} x^{n-2}y^2 + \cdots + \binom{n}{n} y^n}$$ which you could rewrite into $$x \left[ \binom{n}{1} \frac{y}{x} + \binom{n}{2} \frac{y^2}{x^2} + \cdots + \binom{n}{n} \frac{y^n}{x^n} \right]^{1/n}$$ or possibly more usefully as $$ x \left[ \left( \frac{y}{x} + 1\right)^n - 1 \right]^{1/n} $$ but I doubt there's any better simplification.

Answer 2

How about $$((x+y)^n-x^n)^{1/n}=\left(x^n\frac{(x+y)^n}{x^n}-x^n\right)^{1/n}=(x^n)^{1/n}\left(\left(1+\frac{y}{x}\right)^n-1\right)^{1/n}=x\left(\left(1+\frac{y}{x}\right)^n-1\right)^{1/n},$$ and since $a^x=e^{x\log a}$ (since $x\log a = \log a^x$ and $e^{\log x}=x$) then this is equivalent to $$x\exp\left(\frac{1}{n}\log\left(\left(1+\frac{y}{x}\right)^n-1\right)\right)\equiv ba^c,$$ so $b=x$, $a=e$, and $c$ is the exponent of $a=e$. Note that $\exp(z)\equiv e^z$.

Answer 3

Consideration of the issue function $$f(x,y,n) = x\left(\left(1+\dfrac yx\right)^n-1\right)^{\frac1n}\tag1$$ in the form of $$f(x,y,n) = x a^{\text{exponent}}$$ is equivalent to looking for the presentation $$\left(1+\dfrac{y}x\right)^n-1 = a^b.\tag2$$ Exact approach is $b=1$ and the form $(1),$ wherein $(2)$ allows factorization $$\left(1+\dfrac{y}x\right)^{2k+1}-1 = \left(\dfrac{y}x+1-1\right) {\left(\left(\dfrac{y}x+1\right)^2-2\left(\dfrac{y}x+1\right)\cos\dfrac{2\pi}{2k+1}+1\right)}\times {\left(\left(\dfrac{y}x+1\right)^2-2\left(\dfrac{y}x+1\right)\cos\dfrac{4\pi}{2k+1}+1\right)}\dots \times\left(\left(\dfrac{y}x+1\right)^2-2\left(\dfrac{y}x+1\right)\cos\dfrac{2k\pi}{2k+1}+1\right),$$ or $$\left(1+\dfrac{y}x\right)^{2k+1}-1 = \dfrac{y}x {\left(\left(\dfrac{y}x\right)^2+4\left(\dfrac{y}x+1\right)\sin^2\dfrac{\pi}{2k+1}\right)}\times {\left(\left(\dfrac{y}x\right)^2+4\left(\dfrac{y}x+1\right)\sin^2\dfrac{2\pi}{2k+1}\right)}\dots \left(\left(\dfrac{y}x\right)^2+4\left(\dfrac{y}x+1\right)\sin^2\dfrac{k\pi}{2k+1}\right),\tag3$$ in the case of odd $n=2k+1,$ and $$\left(1+\dfrac{y}x\right)^{2k}-1 = \left(\left(\dfrac{y}x+1\right)^k-1\right)\left(\left(\dfrac{y}x+1\right)^k+1\right)\tag4$$ in the case of even $n=2k.$

And there are approximations $$\left(1+\dfrac{y}x\right)^n\approx e^{(ny)/x},\quad f(x,y,n)\approx x\left( e^{\frac {ny}x}-1\right)^{\frac1n},\quad \text{ if }\ \dfrac {ny}x \approx 1,\ n >> 1,\tag{5}$$ $$\left(1+\dfrac{y}x\right)^n-1 \approx \dfrac {ny}x,\quad f(x,y,n)\approx x\left(\dfrac{ny}x\right)^{\frac1n},\quad \text{ if }\ \dfrac {ny}x << 1.\tag{6}$$ On the other hand, the polynomial $(2)$ reduces to the degree $(n-1),$ so the consideration in the form of $$f(x,y,n) = x^{\frac{n-1}n} a^{\frac1n}\tag7$$ can be useful too. In this case $$a = x\left(\left(1+\dfrac{y}x\right)^n-1\right),\tag8$$ $$a = y\left(n +\binom{n}{2}\dfrac yx +\dots + \binom{n}{n-2}\left(\dfrac yx\right)^{n-3} + n\left(\dfrac yx\right)^{n-2} + \left(\dfrac yx\right)^{n-1}\right).\tag9$$