Taylor's theorem and Constancy theorem

alexmufc1995

Corollary 12.8 is this:
If f'(t)=0 for all t in (a,b) then f is constant on (a,b).

In essence I think I have to do something like this:

Let f(x) = LHS and g(x) = RHS. f is differentiable on (-1,1) by properties of the root function, whilst g is differentiable on (-1,1) by the ratio test. I'm okay up to here.

Show that f'(x) - g'(x) = 0. f(0)-g(0) = 0 so conclude that f(x)=g(x) on (-1,1). Then show f(1) = g(1) separately. I'm having problems with these parts, particularly as I can't use Binomial theorem.

Any advice with the manipulation? Thanks.

(edited 9 years ago)

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OP

Bump?

Original post by alexmufc1995

Corollary 12.8 is this:
If f'(t)=0 for all t in (a,b) then f is constant on (a,b).

In essence I think I have to do something like this:

Let f(x) = LHS and g(x) = RHS. f is differentiable on (-1,1) by properties of the root function, whilst g is differentiable on (-1,1) by the ratio test. I'm okay up to here.

Show that f'(x) - g'(x) = 0. f(0)-g(0) = 0 so conclude that f(x)=g(x) on (-1,1). Then show f(1) = g(1) separately. I'm having problems with these parts, particularly as I can't use Binomial theorem.

Any advice with the manipulation? Thanks.

You're allowed to use the binomial theorem as long as you prove it. Can you prove it?
You can differentiate a power series by term-by-term differentiation (that's a theorem).

Reply 3

Smaug123

15

Original post by alexmufc1995

Bump?

It's been pointed out to me that if you differentiate both sides, you can form a differential equation kind of thing by multiplying through by a certain expression. You're probably working off Proposition 15.8 from http://www0.maths.ox.ac.uk/system/files/coursematerial/2014/2645/31/anal2notes.pdf already, but the method is similar.

EDIT: in fact, that method is pretty much just the proof of the Binomial Theorem from Chapter 16, in the particular case p=1/2.

(edited 9 years ago)

Reply 4

alexmufc1995

OP

8

Original post by Smaug123

It's been pointed out to me that if you differentiate both sides, you can form a differential equation kind of thing by multiplying through by a certain expression. You're probably working off Proposition 15.8 from http://www0.maths.ox.ac.uk/system/files/coursematerial/2014/2645/31/anal2notes.pdf already, but the method is similar.

EDIT: in fact, that method is pretty much just the proof of the Binomial Theorem from Chapter 16, in the particular case p=1/2.

I am indeed working off 15.8. I can't see where Taylor's theorem is coming in, and I'm reluctant to use the Binomial Theorem at all, since it states not to quote it (I'm unsure over the distinction between using and quoting - seems subjective!)

Reply 5

Smaug123

15

Original post by alexmufc1995

I am indeed working off 15.8. I can't see where Taylor's theorem is coming in, and I'm reluctant to use the Binomial Theorem at all, since it states not to quote it (I'm unsure over the distinction between using and quoting - seems subjective!)

Fair enough. To be honest, I'm not sure where Taylor comes in either.

Reply 6

davros

16

Original post by Smaug123

Fair enough. To be honest, I'm not sure where Taylor comes in either.

I haven't done this stuff formally for many years, so feel free to jump on me, but won't Taylor's theorem give you precisely the answer you'd like to get via the Binomial Theorem, purely by the uniqueness of power series (within the appropriate region of convergence)? Or is that there are various versions of Taylor's theorem with remainder, and you still have some work to do to show that the series converges as required?