Proof

LeixlipRed

Prove that there do NOT exist polynomials f(x) and g(x) st:

e^x=f(x)/g(x) for all x

Was just looking over old intervarsity papers and came across this. I have some ideas but I'm trying to adapt a slightly different proof to fit this one. Interested to see if any heads on here have ideas. If it's easy use spoiler tags. Want to think about it more when I have energy.

LeixlipRed

I think I figured it out. Simple I believe. So simple I feel silly for posting it. Just use the fact that

the deriv of both sides must be equal to what you start with

. Then you can derive a contradiction.

Fremen

Yeah, you end up with fg = f'g - gf'
which can't be true since if the LHS is of order N, the RHS must be at most of order N-1

LeixlipRed

Yep. I was thinking more loosely at the end there but that's more concise.

Michael Collins

Nice proof there.

Here's another one involving the exponential function:

Prove that e^x != 0, for any x ε R

LeixlipRed

Is x! not defined only for when x is an element of N? Or am I reading that wrong? (e^x)! or e^(x!)?

Michael Collins

Sorry should have explained that I'm using computer science notation. != means "not equal to".

(Although x! is defined for most real or even complex values of x by the Gamma Function)

Fremen

Michael Collins wrote: »

Prove that e^x != 0, for any x ε R

Weird question, that's something I never even thought about. It's just something I "know" without needing to prove. A proof wasn't immediately obvious though.

Suppose there exists "a" such that e^a = 0.
Clearly a is nonzero.
Then (e^a)^(1/a) = 0^(1/a) = 0

but (e^a)^(1/a) = e, giving a contradiction.

LeixlipRed

Blonde moment there Fremen beat me to it as well.

LeixlipRed

Ok, here's an easy one. Use spoiler tags from now on as some readers might like to try prove themselves.

Suppose f is a differentiable function such that f'=f, for all x in R. Prove that there is a constant c such that

f(x)=ce^x

Michael Collins

Fremen wrote: »

Weird question, that's something I never even thought about. It's just something I "know" without needing to prove. A proof wasn't immediately obvious though.

Suppose there exists "a" such that e^a = 0.
Clearly a is nonzero.
Then (e^a)^(1/a) = 0^(1/a) = 0

but (e^a)^(1/a) = e, giving a contradiction.

Excellent! It's different to what I had in mind, but equally as valid. You could also notice:

e^x . e^(-x) = e^(0) = 1, hence since two numbers multiplied together equals one, for all x => neither of them can be zero, for any x

LeixlipRed wrote: »

Ok, here's an easy one. Use spoiler tags from now on as some readers might like to try prove themselves.

Suppose f is a differentiable function such that f'=f, for all x in R. Prove that there is a constant c such that

f(x)=ce^x

Well can you not just say:

f'(x) = d/dx (ce^x) = c d/dx (e^x) = ce^x

?

LeixlipRed

Hmmm, can you? You're assuming c exists and proceeding from there.

Michael Collins

LeixlipRed wrote: »

Hmmm, can you? You're assuming c exists and proceeding from there.

Well I think if you need to prove something is true for a specific c, and can come up with one that works, surely that proves it? I'm not saying I'm sure I'm right like, but I can't see what's wrong with it!

You obviously have a different proof in mind, how about this (somewhat brute force) method:

df/dx = f(x)
df/f(x) = dx
ln(f(x)) = x + c' ...integrating both sides (c is some constant)
f(x) = ce^x ...taking natural antilog of both sides, note c != 0 by the proof of post 5 above!

Although I bet there's a better, sexier proof!

LeixlipRed

Yeh, I had something else in mind. That second proof works anyway. The one I had in mind was

Let g(x)=f(x)/e^x, so g'=[e^x(f'-f)]/e^2x. g'=0 (f'-f=0 and e^2x != 0), hence g is some constant c.

. It's possible I've remembered the question wrong and hence the confusion over your first effort. I'll think about it anyway *strokes chin*

Fremen

I think there's a problem with your proof MC.

"e^x . e^(-x) = e^(0) = 1, hence since two numbers multiplied together equals one, for all x => neither of them can be zero, for any x"

But if e^x *does* equal zero, then multiplication by e^(-a) is undefined. By the same logic, let f(x) be an arbitrary function.
Then f(x)*(1/f(x)) = 1 for all x, so f is always nonzero, which is clearly false since f is arbitrary.

Michael Collins

Fremen wrote: »

I think there's a problem with your proof MC.

"e^x . e^(-x) = e^(0) = 1, hence since two numbers multiplied together equals one, for all x => neither of them can be zero, for any x"

But if e^x *does* equal zero, then multiplication by e^(-a) is undefined. By the same logic, let f(x) be an arbitrary function.
Then f(x)*(1/f(x)) = 1 for all x, so f is always nonzero, which is clearly false since f is arbitrary.

:eek: Interesting...I didn't consider that before, but I think you can protect the proof from such problems by noticing that

The exponential function exp(x) is defined for every x, as its power series converges for every x on the real line

MathsManiac

Well, whether or not any of these proofs is valid surely depends on how you've chosen to define the exponential function in the first place, and which other results you have proved first.

I would certainly have thought that exp(x)>0 for all x would come before exp(x)exp(y) = exp(x+y).

The most common development would suggest that it's an immediate consequence of the definition of exp:
exp is defined as the inverse of ln.
Since ln is defined only for x>0, it follows that exp(x)>0 for all x.

LeixlipRed

Michael, I'm fairly sure I was right on that proof. You're trying to prove c exists. So you can't assume it does and base your conclusion on that. Uniqueness is what I'm talking about here. Obviously f'=f for f=ce^x but is the only case? And you cant say it is if you use your proof.

Michael Collins

MathsManiac wrote: »

Well, whether or not any of these proofs is valid surely depends on how you've chosen to define the exponential function in the first place, and which other results you have proved first.

I would certainly have thought that exp(x)>0 for all x would come before exp(x)exp(y) = exp(x+y).

The most common development would suggest that it's an immediate consequence of the definition of exp:
exp is defined as the inverse of ln.
Since ln is defined only for x>0, it follows that exp(x)>0 for all x.

Yeh, I guess it does depend completely on the definition. I saw the proof of it in a complex analysis text (Rudin actually), which defines the exponential in the usual way - an absolutely convergent power series on the whole complex plane. So then you can tell that e^z is defined for every z ε C, which is why Leixlip's problem with the proof can be overcome, and you also know that e^(z1) . e^(z2) = e^(z1 + z2) from the series itself.

LeixlipRed wrote: »

Michael, I'm fairly sure I was right on that proof. You're trying to prove c exists. So you can't assume it does and base your conclusion on that. Uniqueness is what I'm talking about here. Obviously f'=f for f=ce^x but is the only case? And you cant say it is if you use your proof.

Are we actually assuming anything tho, are we not just trying possibilities? Like what if I say, d/dx (1.e^x) = 1.e^x, so it works. So now have we not found a constant value (=1) for which it works? Hence there exists at least one c with that property...

We haven't proven anything related to uniqueness alright, there could be many functions that satisfy f'=f, but we don't care. We've found one example that does work.

LeixlipRed

What the proof is asking is given that f'=f show that there exists a constant c st f(x)=ce^x. So you can only use the assumed fact. I really think there's a hole in your deductive reasoning there. You can only assume that f'=f. Nothing more. I'm asking you to assume statement A implies B and you're showing B implies A. Which is true. But not what the proof asks.