Using Induction to Show a Function is Decreasing

[Dschumanji]

I have recently been studying the following function for fun (math nerd):

f(x) = x^(1/x) where x is an integer greater than 0

I want to prove that the function is decreasing for x > 2. It seems that the straight forward approach to this proof is to use induction. It can easily be shown that 3^(1/3) > 4^(1/4), which becomes my base case. I then assume that the following inequality holds:

(k-1)^(1/(k-1)) > k^(1/k)

Using the above inequality I must show that:

k^(1/k) > (k+1)^(1/(1+k))

I have tried everything in my mathematical toolkit to rewrite the first inequality into something that looks like the second but to no avail. My last attempt involved taking the logarithm to the base k-1 on the left side and the logarithm to the base k on the right side. It seems to work but I run into some technicalities that I can't seem to get around when I try to raise each side to a different base. Someone please help!

 

[Petr Mugver]

Try using the fact that log(k + 1) - log(k) < 1, which is true because in your conditions
1 + 1/k < 1/e

 

[Dschumanji]

Try using the fact that log(k + 1) - log(k) < 1, which is true because in your conditions
1 + 1/k < 1/e

I'm not seeing how using log(k+1) - log(k) < 1 should help me out. I am able to go from

(k-1)^(1/(k-1)) > k^(1/k)

to

log(k+1) - (k/(k-1))log(k-1) < 1

Also where in my conditions do I imply 1 + 1/k < 1/e?

 

[Nebuchadnezza]

The fallacy i see is that the function is not strictly decreasing when x>2.

If we find derive the function and look at its extremum, we see that.

The maximum of the function is $$( \, e \, , \, \sqrt[\large{e}]{e} )$$

Therfore the function is increasing between x=2 and x=e , and decreases after this.

 

[Maxter]

I get a local max at e also.

I always like turning these into exponential functions with base e:

x^(1/x)=e^((ln x)/x)

But that's coming at it from a different angle than the OP.

 

[Dschumanji]

The fallacy i see is that the function is not strictly decreasing when x>2.

If we find derive the function and look at its extremum, we see that.

The maximum of the function is $$( \, e \, , \, \sqrt[\large{e}]{e} )$$

Therfore the function is increasing between x=2 and x=e , and decreases after this.

Yes, the maximum of the function occurs at x = e when the function is defined over the real numbers. The function is only defined for integers greater than zero.

 

[Maxter]

Just reread the title, this is by induction - sorry, my comment is a red herring!

However, the question is whether to show decreasing after e, or show by induction? Induction, technically will only be valid for integers, rather than for all numbers. You can wave your arms about continuity, but indeuction won't guarantee that it doesn't do something funny between the integers.

So, I guess the question is whether you want if decreasing on integers or also between them.

Btw, the general domain for this function IS all positive reals, unless otherwise restrcited.

 

[Maxter]

I got to read more carefully, the "x" threw me, rather than an "n" (creature of habit), guess the problem is for integers. You can still prove for all numbers greater than e, which includes the integers. Then you can use the derivative that will work if you decide to take a different path.

 

[Dschumanji]

I got to read more carefully, the "x" threw me, rather than an "n" (creature of habit), guess the problem is for integers. You can still prove for all numbers greater than e, which includes the integers. Then you can use the derivative that will work if you decide to take a different path.

Would I just have to show that the derivative is negative for x > e?

 

[Nebuchadnezza]

Yes.

 

[Dschumanji]

Yes.

Well that is definitely good to know. However, it still doesn't help with the original question of using induction. Is it perhaps possible that this cannot be proved with induction?

 

[Dschumanji]

It looks like no one else has any input on how to solve this problem. I would like to thank everyone who helped with the alternative proof!

It is so interesting that this problem is hard to solve when restricted to the simple structure of the integers and the methods applicable to it, but easy to solve when the more complex structure of the reals and the methods applicable to it are used.

 

[Petr Mugver]

I have recently been studying the following function for fun (math nerd):

f(x) = x^(1/x) where x is an integer greater than 0

I want to prove that the function is decreasing for x > 2. It seems that the straight forward approach to this proof is to use induction. It can easily be shown that 3^(1/3) > 4^(1/4), which becomes my base case. I then assume that the following inequality holds:

(k-1)^(1/(k-1)) > k^(1/k)

Using the above inequality I must show that:

k^(1/k) > (k+1)^(1/(1+k))

I have tried everything in my mathematical toolkit to rewrite the first inequality into something that looks like the second but to no avail. My last attempt involved taking the logarithm to the base k-1 on the left side and the logarithm to the base k on the right side. It seems to work but I run into some technicalities that I can't seem to get around when I try to raise each side to a different base. Someone please help!

Ok I got it.

elevate the equation k^(1/k) > (k+1)^(1/(1+k)) to k(k+1), and then divide by k to obtain

(1 + 1/k)^k < k (*)

we want to show this is true for k > 2 by induction. It is true for k = 3. Then write

(1 + 1/(k+1))^(k+1) < (1 + 1/k)^(k+1) = (1 + 1/k)(1 + 1/k)^k

which is obviously true. Then, uning the inductive hypothesis (*),

(1 + 1/k)(1 + 1/k)^k < (1 + 1/k)k = k + 1

So

(1 + 1/(k+1))^(k+1) < k + 1

which is what we wanted to proove.

 

[Dschumanji]

Ok I got it.

elevate the equation k^(1/k) > (k+1)^(1/(1+k)) to k(k+1), and then divide by k to obtain

(1 + 1/k)^k < k (*)

we want to show this is true for k > 2 by induction. It is true for k = 3. Then write

(1 + 1/(k+1))^(k+1) < (1 + 1/k)^(k+1) = (1 + 1/k)(1 + 1/k)^k

which is obviously true. Then, uning the inductive hypothesis (*),

(1 + 1/k)(1 + 1/k)^k < (1 + 1/k)k = k + 1

So

(1 + 1/(k+1))^(k+1) < k + 1

which is what we wanted to proove.

That was brilliant, Petr Mugver! Thank you for the excellent solution (I hope you didn't spend to long on it). I feel so stupid considering how straight forward it is. I really need to practice proofs concerning inequalities and induction, they seem to always require applying strange operations and noticing certain subtleties.