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

[Hurkyl]

Now, correct me if I'm wrong, but I think this implies that there are only two entries in the (b-1)th row in pascals triangle, so b-1=1 which implies b = 2.

What is your rationale? I'm guessing you're implicit argument is that the rest of terms have to sum to zero, therefore they must individually be zero. But that argument clearly doesn't work -- it's easy to construct a counterexample. The argument would work if we had a sum of nonnegative terms, but that isn't the case.

Also, you've assumed b is an integer.

 

[Hurkyl]

Oh, bah, my calculus proof can be salvaged.

f''(x) < 0 on (2, +infty)

Therefore, f'(x) >0 on (2, +infty) (it decreases to zero)

Therefore, f(x) is increasing on (2, +infty) -- in particular it has no zeroes in that interval.

Therefore, b <= 2. Furthermore, this implies a >= 2.

By symmetry^*, we also deduce a <= 2 and b >= 2.

Therefore a=b=2.

*: Recall we've already shown a+b=ab=a^b=b^a

 

[Altabeh]

Oh, bah, my calculus proof can be salvaged.

f''(x) < 0 on (2, +infty)

Therefore, f'(x) >0 on (2, +infty) (it decreases to zero)

Therefore, f(x) is increasing on (2, +infty) -- in particular it has no zeroes in that interval.

Therefore, b <= 2. Furthermore, this implies a >= 2.

By symmetry^*, we also deduce a <= 2 and b >= 2.

Therefore a=b=2.

*: Recall we've already shown a+b=ab=a^b=b^a

You don't need to go further from (x-1)*log(x-1)-(x-2)*log(x)=f(x): This function is strictly increasing over [2,+oo) and the only solution to f(x)=0 is x=2=b (remember for x between 1 and 2 the function does not hit the x-axis). Thus plugging this into a+b=ab gives a=2. Period

AB

 

[Hurkyl]

You don't need to go further from (x-1)*log(x-1)-(x-2)*log(x)=f(x): This function is strictly increasing over [2,+oo) and the only solution to f(x)=0 is x=2=b (remember for x between 1 and 2 the function does not hit the x-axis).

Both of those facts must be argued, not just assumed.

I and found the symmetry argument simpler than arguing f(x) has no zeroes on (1,2).

 

[Altabeh]

Both of those facts must be argued, not just assumed.

I and found the symmetry argument simpler than arguing f(x) has no zeroes on (1,2).

You can check the first claim by a simple analytical comparision between terms:

1-(x-1)*log(x-1)-(x-2)*log(x)=f(x); Here we have two factors (x-1) and (x-2) s.t. for x>=2, the rate of change of (x-1) w.r.t. the change of x is clearly faster than that of (x-2) by a constant angle. Also the rate of growth of log(x-1) is so much slower than log(x) but one knows that all logarithm functions grow slower than linear functions, leading to a faster-growing term (x-1)*log(x-1) compared to (x-2)*log(x). So the function is strictly increasing on [2,+oo).

2- From 1, it is too easy to show that a strictly increasing function hits the x-axis once and only once. (Imagine how a line can only have ONE x-intercept.)

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

AB

 

[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)".