${(e-1)}^{(e+1)}\stackrel{?}{\lessgtr}{(e+1)}^{(e-1)}$

Question

Without the use of calculators find which one is greater:$${(e-1)}^{(e+1)}\stackrel{?}{\lessgtr}{(e+1)}^{(e-1)}$$

More generally, find out which is greater (without the use of calculators): $${(e-x)}^{(e+x)}\stackrel{?}{\lessgtr}{(e+x)}^{(e-x)}$$ for $0<x<(e-1)$

Since $(e-1) \text{ and } (e+1)$ are both greater than zero, any irrational power is real.
It is well known that $x^{1/x}$ obtains a maximum value at $x=e$.

The general rule of thumb for these $x^{y} \text{ vs } y^{x}$ comparisons is that the number closer to e as a base is greater. That very fact helps us prove that $\phi^{\pi} < \pi^\phi$ since $\pi$ is closer to $e$ than $\phi$ is.

In my case however, both the bases are equally distant to $e$ and so I don't see a way to figure this out.
Foreseeing the complexity, the idea of using a Taylor expansion for the logarithm of both sides was dropped. (I don't think that would work)

Answer 1

We have that for $x=(0,e-1)$

$$f(x)=\frac{\log (e-x)}{(e-x)}<\frac{\log (e+x)}{(e+x)}=f(-x)=g(x)$$

indeed

• both functions are strictly decreasing with $f(0)=g(0)$ and $f(e-1)<g(e-1)$

and we also have $f'(x) < g'(x)$, indeed

• both derivatives are strictly decreasing with $f'(0)=g'(0)$ and $f'(e-1)<g'(e-1)$

and we also have $f''(x) < g''(x)$, indeed

• $f''(0)=g''(0)$ and $f''(e-1)<g''(e-1)$

with $f'''(x)<0$ and $g'''(x)>0$.

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Derivatives for f and g

• $f'(x)=\frac{\log (e-x)-1}{(e-x)^2},\;\;\;g'(x)=-f'(-x)=\frac{1-\log (e+x)}{(e+x)^2}$

• $f''(x)=\frac{2\log (e-x)-3}{(e-x)^3},\;\;\;g''(x)=f''(-x)=\frac{2\log (e+x)-3}{(e+x)^3}$

• $f'''(x)=\frac{6\log (e-x)-11}{(e-x)^4},\;\;\;g'''(x)=-f'''(-x)=\frac{11-6\log (e+x)}{(e+x)^4}$

Answer 2

I will use this result from If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$.:

If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$.

Therefore, if $1<x<e < y < e^2/x$ (or $xy < e^2$) then $x^{1/x} < y^{1/y}$ or $x^y < y^x$.

Let $x=e-1, y=e+1$. Then $xy=e^2-1 < e^2$ so $x^y<y^x$.

More generally, if $x=e-c, y=e+c$ where $0 < c < e-1$ then $xy=e^2-c^2<e^2$ so $x^y < y^x$.

Answer 3

In this answer we allow ourselves a "calculator" that can perform exact additions, subtractions, and multiplications with arbitrarily large whole numbers, which we use to save time versus the corresponding hand calculations. Divisions and more advanced operations that generally do not give exact whole-number results, and inputs other than whole numbers, can be used to set up the calculation but not perform it.

Our starting point is the continued fraction for $e$ due to Euler:

$e=[2,1,2,1,1,4,1,1,6,...]$

From the third and fourth convergents of this continued fraction we have the bounds

$8/3<e<11/4.$

We now define $x=(e+1)^{e-1},y=(e-1)^{e+1}$ and use the subscripts $+$ and $-$ to define calculated upper and lower bounds for these quantities. We will therefore get a definitive result if $x_->y_+$ or if $y_->x_+$.

So

$x_-=(11/3)^{5/3}$

$x_+=(15/4)^{7/4}$

$y_-=(5/3)^{11/3}$

$y_+=(7/4)^{15/4}$

By our rules we must eliminate the fractions in the exponents and in the bases. We first raise our quuantities to the 12th power:

$x_-^{12}=(11/3)^{20}$

$x_+^{12}=(15/4)^{21}$

$y_-^{12}=(5/3)^{44}$

$y_+^{12}=(7/4)^{45}$

Since the relevant comparisons are specifically between one lower bound and the opposing upper bound, we plan to consider two cases with slightly different least common deominators. This will cut down somewhat on the size of the numbers compared with using a common denominator for all four quantities.

Comparing first $x_+$ with $y_-$, we have

$3^{44}×4^{21}x_+^{12}=15^{21}×3^{44}=49119...[46]$

$3^{44}×4^{21}y_-^{12}=5^{44}×4^{21}=25000...[44]$

The brackets indicate the total number of digits in each number. In the second number above all the digits not shown are fairly obviously zero, but anyway we failed to obtain the definitive result $y_->x_+$. So we cannot render $y>x$.

We try our second case, comparing $x_-$ with $y_+$. Note the different common denominator, which allows a simplification of the calculation:

$3^{20}×4^{45}x_-^{12}=11^{20}×4^{45}=(11^4×4^9)^5=\color{blue}{3838050304}^5$

$3^{20}×4^{45}y_+^{12}=7^{45}×3^{20}=(7^9×3^4)^5=\color{blue}{3268642167}^5$

OK, I should have done the second case first to avoid those 44- and 46-digit numbers. Little did I realize the implications of $8-3$ happening to divide $11+4$!

In any event we have our answer:

$3837050304>3268642167\implies x_->y_+\implies x>y\implies \color{blue}{(e+1)^{e-1}>(e-1)^{e+1}}.$

Answer 4

You may see a list of Maclaurin series here. \begin{array}{|c|c|c|} \hline \text{Function}&\text{Maclaurin series}\\\hline \ln(1+x)&\sum_{k=1}^\infty\frac{(-1)^{k+1}}kx^k,\,|x|<1\\\hline \ln(1-x)&-\sum_{k=1}^\infty\frac1kx^k,\,|x|<1\\\hline \end{array} Let $P_n(x)=\sum_{k=1}^n\frac{(-1)^{k+1}}kx^k$ be the $n$-th degree Maclaurin polynomial of $\ln(1+x)$. Then, for $0<x<1$, we have $$P_{2n}(x)<\ln(1+x)<P_{2n+1}(x)$$ On the other hand, if $Q_n(x)=\sum_{k=1}^n\frac{(-1)^{k+1}}kx^k$ is the $n$-th degree Maclaurin polynomial of $\ln(1-x)$, then, for $0<x<1$, we have $$\ln(1-x)<Q_{n}(x)$$ for all $n$. Why?

We wanted to show that $$(e+1)\ln(e-1)<(e-1)\ln(e+1)$$ which can be written as $$(e+1)(1+\ln(1-\frac1e))<(e-1)(1+\ln(1+\frac1e)).$$ Now, it is enough to show that $$(e+1)(1+Q_2(\frac1e))<(e-1)(1+P_2(\frac1e))$$ that is $$(e+1)(1-\frac1e-\frac1{2e^2})<(e-1)(1+\frac1e-\frac1{2e^2}).$$ After simplifications luckily we get $$-\frac1{2e^2}<\frac1{2e^2}$$ which is true. If it wasn't true, we had to try with $4$-th degree polynomials.

Answer 5

By convexity of $(x+1)^{x-1}$ and $(x-1)^{x+1}$, equality of both functions when $x=3$ and because they can be equal just in one point and because obviously for $x$ at neighborhood of $1$ we have $(x+1)^{x-1}\gt(x-1)^{x+1}$ and because $e\lt 3$ we conclude that $$(e+1)^{e-1}\gt(e-1)^{e+1}$$