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7:36 AM
@MrPie Hi
 
 
1 hour later…
8:38 AM
hello
@TheSimpliFire that's too hard for me lmfao
 
8:50 AM
@MrPie It looks way harder than it actually is
For part a) all you need is elementary differentiation
 
i've never seen two mins together before
so i don't even understand the question :P
 
Consider the function $f(\alpha,x)=x^\alpha-\log x$
Minimise this function with respect to $x$ first, then with respect to $\alpha$
 
ok
 
I.e. treat $\alpha$ as a constant first, then as a variable once you've found $x$
 
hang on im a bit busy
besides, i think this is a bit of a jump, no disrespect. i don't know how to minimise
 
8:54 AM
@MrPie It's ok :)
Remember that minimum/max occur when f'(x)=0
 
i mean, im familiar with it, but application-wise? meh
 
Essentially, it's just $\frac d{dx}(x^\alpha-\log x)=0$
 
oh
 
(or more correctly $\partial/\partial x$ as there's more than one variable but it doesn't affect the result)
 
well, why didn't you say that!
 
8:56 AM
You have to remember that critical points are when the gradient is zero @MrPie
 
yes
 
So once you differentiate and set to zero, you will have an expression for $x$ in terms of $\alpha$
 
$\alpha x^{\alpha } = 1$
 
i don't really know how to solve for this
but
 
9:00 AM
Solve for $x$, not $\alpha$ :P
 
oh well that's just $x=(1/a)^{1/a}$
 
That's correct
Now plug that into $f(x,\alpha)=x^\alpha-\log x$
 
but i don't know how to find two independent non-trivial values where this is true
 
@MrPie You don't need to yet
I'll explain what you have done once you've substituted into f(x,a)
 
$f(x,\alpha)=\frac 1{\alpha}(1+\log \alpha)$
 
9:03 AM
Yes
So this is what you've done
 
so now it's just $f(\alpha)$
 
We have $\min\limits_\alpha\min\limits_x\{x^\alpha-\log x\}=\min\limits_\alpha\left(\min\limits_x\{x^\alpha-\log x\}\right)$. In $\min\limits_x\{x^\alpha-\log x\}$, we treat $\alpha>0$ as a constant, like $2.5$ or $\pi$ and we set the derivative wrt $x$ to zero. This minimises $x^\alpha-\log x$ for any $\alpha$.
 
i see now
so it's just a double derivative?
 
Therefore, the value of $x$ that minimises $x^\alpha-\log x$ will be dependent on $\alpha$. Consequently $\min\limits_x\{x^\alpha-\log x\}$ will be a function of $\alpha$, which we call $f^\ast(\alpha)$. You have evaluated $f^\ast(\alpha)=(1+\log\alpha)/\alpha$.
Hence $\min\limits_\alpha\min\limits_x\{x^\alpha-\log x\}=\min\limits_\alpha(1+\log\alpha)/\alpha$.
@MrPie In some sense, you are differentiating wrt each variable once so a total of two derivatives
Now you just need to minimise again using the same technique but for $\alpha$.
 
well, if i have learnt correctly, $\min\limits_\alpha f^\ast (\alpha) = -\cfrac{\log \alpha}{\alpha^2}$ and the point where this is $0$ is obviously $\alpha=1$. That is the unique value.
 
9:09 AM
Not quite
Do you know the quotient rule for differentiation?
 
$dy/dx = dy/du \cdot du/dx$?
 
Ah that's the chain rule
The quotient rule is for $\frac d{dx}\frac{f(x)}{g(x)}$ (a quotient of functions)
 
uhh...
maybe i don't
 
or the product rule
They are equivalent
 
i know the product rule
 
9:11 AM
Yay
 
that's how integration by parts is formed
 
Excellent, you can use the product rule by noting that $(1+\log\alpha)/\alpha=(1/\alpha)+(1/\alpha)\cdot\log\alpha$
 
$(f\cdot g)' = f'g + fg'$
 
Correct
 
$d/d\alpha (1+\log\alpha)/\alpha = -\frac 1{\alpha^2} -\frac{\log \alpha}{\alpha^2} + \frac{1}{\alpha^2} = -\frac{\log \alpha}{\alpha^2}$
it's the same :P
 
9:16 AM
Shoot I wrote the problem wrong xd
Oh well, let's continue from $\alpha=1$
 
so we plug that into the integral?
 
Sorry about that, I wrote $-\frac1{\alpha^2}-\frac{\log\alpha}{\alpha^2}+1$ for some reason
@MrPie Yes, though the inequality should now reverse direction
 
I really need to work on my maths...
I just went "ah, we're adding $1$ to $\log \alpha$ when we derive values that add a constant, they disappear. Also the derivative of $e^x$ is $e^x$ so that log won't change. It's just a matter of deriving $1/a$ and multiplying it by $\log$"
i have to go though, im a bit busy
 
@MrPie No, I do. I thought $(\log\alpha)'=\alpha$ :P
@MrPie Cya
 
 
2 hours later…
11:19 AM
@Haran Hi, Do you know a databse of the Mordell-equation, allowing to look for large solutions explicitely ?
 
12:16 PM
@DanielFischer , Hi, I search a database for the Mordell-equation where I can look up large solutions. Do you know such a database ?
 
12:34 PM
@Peter No, sorry. Do the search engines only turn up databases with only small solutions?
 
12:54 PM
I did not find interactive databases, only a raw text.
 
1:10 PM
@Peter Nope
 
No further solution upto $\ x=3\cdot 10^7\ $. — Peter 4 hours ago
Any ideas for this question ?
According to my calculations, there is no further solution upto $n=10^8$
 
1:47 PM
I don't know if it achieves anything but note $\sigma(\sigma(x^2))>\dfrac{216x^3}{\pi^4\phi(x)\phi(\sigma(x^2))}>\dfrac{216x^3}{\pi^4(x-\sqrt x)(\sigma(x^2)-\sqrt{\sigma(x^2)})}$ and $2x\sigma(x)<2e^\gamma x^2\log\log x$
Perhaps certain $x$ contradict the two inequalities
Further note that it's generally very hard to get tighter lower bounds so the expression is more complicated due to nesting
 
2:01 PM
OK, I did a quick browse on Wikipedia and I don't see how Lagarias' bound improves on Robin's as $H_x+e^{H_x}\log H_x>e^\gamma x\log\log x$ for all positive $x$.
 

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