Proving one function is greater than other?

I want to see where is this inequality true

where x in (e^e,∞).

[Torsten]

Prove that

f(x) = 53.989/21.233 * (log(log(x)))^(4/3)/(log(x))^(1/3) 

is monotonically increasing (i.e. by showing that its derivative is > 0) and f(e^e) > 1.

[Sam Chak]

You want to prove it rigorously by using the Symbolic Math Toolbox, or by graphical approach?

[Fatima Majeed]

@Torsten but this f(x) is not monotonically increasing.By the way I try this code but it is not true.

% Define the function

syms x

f = 53.989/21.233 * (log(log(x))).^(4/3)./(log(x)).^(1/3);

% Compute the first derivative

df = diff(f, x);

disp(df)

syms x

expr = (215956*log(log(x))^(1/3))/(63699*x*log(x)^(4/3)) - (53989*log(log(x))^(4/3))/(63699*x*log(x)^(4/3));

% Solve the inequality

solution = solve(expr > 0, x);

disp(solution);

[Fatima Majeed]

@Sam Chak any of them but I want to know exactly where is the above inequality true

[Matt J]

The notation is ambiguous. We must know whether is to be interpreted as or as .

[Fatima Majeed]

@Matt J It is as (log(x))^y

Answer 1

[Matt J]

Make the change of variables and rearrange the inequality as,

Since is a convex function on such that >0 and , it readily follows that for all . QED.

[Fatima Majeed]

Could you explain how to rearrange the inequality?

[Matt J]

Here are the steps,

[Fatima Majeed]

But here (log z)^4 not log(z^4)

[Sam Chak]

Hi @Fatima Majeed

Maybe something like this?

syms z real
assume(z > exp(1))
sol = solve((log(z))^4 - ((21.233/53.989)^3)*z > 0, z, 'ReturnConditions', true)

sol = struct with fields:
             z: x
    parameters: x
    conditions: x < (36893488147419103232*lambertw(-1, -(8766535218140255^(1/4)*144115188075855872^(3/4))/576460752303423488)^4)/8766535218140255 & 6121026514868073/22517998136...

vpa(sol.conditions)

ans = 

[Matt J]

But here (log z)^4 not log(z^4)

Correct, I never said anything about log(z^4). However, I've been using the notation (log z)^4= log (z)^4 interchangeably, because you have.

We can also get some visual evidence both for the convexity and positivity of f(z) from the plot below.

fplot(  @(z) log(z).^4-0.0608*z , [exp(1),10])

[Fatima Majeed]

When writing the function in MATLAB, please use `log(z)^4` instead of `(log(z))^4`. Additionally, your work should apply to all \( x \) in the range \((e^e, \infty)\), not just a specific region. I believe the end of your work should align with Sam's work. I want to thank both of you for your efforts.

[Sam Chak]

Hi @Fatima Majeed

If we assume , we get this results:

syms z real
assume(z > 0)

fun = (log(z))^4 - ((21.233/53.989)^3)*z;
sol = solve(fun > 0, z, 'ReturnConditions', true)

sol = struct with fields:
             z: [2x1 sym]
    parameters: x
    conditions: [2x1 sym]

vpa(sol.conditions)

ans = 

figure(1)
fplot(fun, [0 3]), ylim([-0.1 1]), yline(0, '--'), grid on

figure(2)
fplot(fun, [0 5e5]), yline(0, '--'), grid on, xlabel('z'), ylabel('f(z)'), xlabel('z'), ylabel('f(z)')

[Torsten]

... and in order to return to x, you will have to substitute x = exp(z).

[Fatima Majeed]

@Sam Chak that is stunning

[Matt J]

I want to thank both of you for your efforts.

You're welcome, but please Accept-click the answer if it has resolved your question.

Answer 2

[Torsten]

syms x y
f = 53.989/21.233 * (log(log(x))).^(4/3)./(log(x)).^(1/3);
%x is solution where f starts getting greater than 1
xstart = vpasolve(f==1,x,5)

xstart = 

5.8933180867157239497727768797558

log(log(xstart))/(21.233*log(xstart)) 

ans = 

0.01521725041175722892398806828762

1/(53.989*log(xstart)^(2/3)*log(log(xstart))^(1/3))

ans = 

0.01521725041175722892398806828762

ftrans = subs(f,x,exp(exp(y)));
%exp(exp(y)) is solution where f ends being greater than 1
yend = vpasolve(ftrans==1,y,13)

yend = 

13.085773977624348668463856655909

log(log(exp(exp(yend))))/(21.233*log(exp(exp(yend))))

ans = 

0.0000012785232351771168083804508742678

1/(53.989*log(exp(exp(yend)))^(2/3)*log(log(exp(exp(yend))))^(1/3))

ans = 

0.0000012785232351771168083804508742678

[Fatima Majeed]

While your analysis is thorough, it appears that you've mainly focused on situations where the inequality can be replaced by an equation, rather than identifying where the inequality holds true. If we could also address those instances, it would provide a more comprehensive understanding of the scenario.Also ,The paper discussing this problem states that the inequality does not hold for \( x \) in the interval \((e^{482036}, \infty)\).

[Sam Chak]

Hi Fatima, can you provide the paper or link, or a cropped section for study purposes? Sounds like an interesting problem.

While I know what a log function is, I never use log(log(x)) or exp(exp(x)) in this approach.

[Fatima Majeed]

https://doi.org/10.48550/arXiv.2402.04272

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