Any recommendations on how to solve this using a power series $x^{x + 1} = (x + 1)^x$? [closed]

Any recommendations on how to solve this using a power series:

$$x^{x+1}=(x+1)^x$$

Figured out you could not do it algebraically so decided to think outside the box.

• Well, do you think there is closed form or do you just want an approximation? Commented Feb 17, 2017 at 15:39
• I know the solution from graphing is approx 2.29 so I believe there should be a closed form? Not sure so thought it might be easier to use a log property, but not sure how or which series to use and steps I should use to combine them. Commented Feb 17, 2017 at 15:42
• Well, I don't know but computation doesn't reveal a possible closed form. Commented Feb 17, 2017 at 15:44
• How close to a close form do you think I could approximate? Just out of curiosity? Commented Feb 17, 2017 at 15:46
• @John It doesn't make sense to approximate a closed form. It either is closed form, or it isn't. Commented Feb 17, 2017 at 15:55

Note that

$$x^{x+1}=x^x\cdot x=(x+1)^x$$

Divide both sides by $x^x$ to get

$$x=\left(1+\frac1x\right)^x$$

Now from here I'll tell you I don't believe there is a closed form, but we can do some quick fixed-point iteration:

$$x_{n+1}=\left(1+\frac1{x_n}\right)^{x_n}$$

With $x_0=2.3$, we get

$x_1=2.2940772541106$

$x_2=2.2932879508444$

$x_3=2.2931825401271$

$x_4=2.2931684586440$

$x_5=2.2931665774723$

Which is the first few digits of the solution.

• So is it possible to tell what this solution is besides guessing based on initial value? Was wondering if it would be possible to use a Taylor series approximation on a series that converged to determine what the solution is? Since Algebraically this is all that it seems to be simplified down to ? Commented Feb 19, 2017 at 19:06
• Since we know as $x\to\infty$, the right side monotonically approaches $e$ from below, and as $x\to0$, the right side approaches one, so we may, without guessing, deduce that we can have $1<x_0<e$. Choosing the average of the bounds gives $x_0=1.9$ Commented Feb 19, 2017 at 19:50

It's actually not that difficult to get the power series. Using the fact that any number $a^x$ is equal to $e^{x\ln a}$ we can express the power series as $\sum\limits_{n=0}^{\infty}\frac{(x\ln a)^n}{n!}$ from the power series for $e^x$. If a is another variable say $x$ then we just replace it with x. So here we have a power series for both the expressions given