Proving a=b=2: Simple yet Annoying Homework Equation Solution

[Hurkyl]

Anything difficult here so as to go for the second derivative of f, instead?

I don't know -- the argument is imprecise enough that I can't tell if you've said anything correct!

e.g. I'm not sure what you might mean by "the rate of change of (x-1) w.r.t. the change of x" such that it is both something different than the corresponding thing for (x-2), and the difference could be said to be a constant angle.


The advantage to the derivative is that it turns everything into simple arithmetic -- one can eliminate (or drastically reduce) the need for imprecision, and it's usually much easier to spot and check all of the edge cases.


e.g. I could have just asserted f'(x) = 2/x - log(1 - 1/x) is positive on (2,+infty). After all, log(1-1/x) looks like -1/x, which is half the size of 2/x! But there is the edge case to consider -- is 2 really close enough to +infty for us to rely on that approximation?

I could try to puzzle it out, but it's so much easier and clearer (both to myself and for the reader) to simply observe that f''(x) = -(x-2)/(x² (x-1)) is somewhat more obviously negative on (2, +infty).

 

[Altabeh]

I don't know -- the argument is imprecise enough that I can't tell if you've said anything correct!

e.g. I'm not sure what you might mean by "the rate of change of (x-1) w.r.t. the change of x" such that it is both something different than the corresponding thing for (x-2), and the difference could be said to be a constant angle.


The advantage to the derivative is that it turns everything into simple arithmetic -- one can eliminate (or drastically reduce) the need for imprecision, and it's usually much easier to spot and check all of the edge cases.


e.g. I could have just asserted f'(x) = 2/x - log(1 - 1/x) is positive on (2,+infty). After all, log(1-1/x) looks like -1/x, which is half the size of 2/x! But there is the edge case to consider -- is 2 really close enough to +infty for us to rely on that approximation?

I could try to puzzle it out, but it's so much easier and clearer (both to myself and for the reader) to simply observe that f''(x) = -(x-2)/(x² (x-1)) is somewhat more obviously negative on (2, +infty).

I think you got me wrong. If something sounds meaningless to you, I assume, it is not false or "imprecise" in any sense. What I meant is that, simply, while the slopes of x-1 and x-2 are the same, but the first one lies above the latter one, so x-1 always is greater than x-2 by a constant difference. (Maybe I was blurry or evern wrong on this one in my early post.) And about log(x-1) and log(x) which everything about the growth of f(x) depends only on 'em... these two for x>=2 are again in a similar situation as log(x) lies right above log(x-1) but not by a varying difference (as x-->oo log(x)=log(x-1)) and since they are "strictly increasing" as are x-1 and x-2, so by analogy, you can see that linear functions are faster than logarithms in their growth, so (x-1)*log(x-1) is ALWAYS larger than (x-2)*log(x) for x>=2, so strictly increasing. This, I think, would shed more light on the whole problem besides your mathematical reasoning.

Anyways, I must say that your idea of proof is brilliant...
AB

 

[Hurkyl]

Maybe that will click if I spend some time thinking it through.



I can't claim credit for discovering my method of proof -- I actually think it's a standard trick that, for some reason, isn't really made to sink in after your first calculus class.

It's useful enough that it's usually one of the first few things I try on a problem like this. Though, as it turns out, in this particular case, I computed everything because I was trying to "graph" the function. (I don't have a calculator handy)

I use scare quotes because I didn't actually plot it -- I just wanted broad strokes like "f''(x) is increasing from -infty to 0 over (2, +infty)".