I'd like to prove e^x is differentiable everywhere.
Obviously, a start would be defining e^x which I can do as a series. starting with (1 + 1/n)^nx and expanding using the binomial theorem. Then, I would want to show that for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h exists.
The last part I'm unsure how to approach. In particular, how to capture that the difference quotient exists for all x.
On Nov 9, 2:45 pm, aegis <ae...@mad.scientist.com> wrote:
> I'd like to prove e^x is differentiable everywhere.
> Obviously, a start would be defining e^x > which I can do as a series. starting with > (1 + 1/n)^nx and expanding using the binomial > theorem. Then, I would want to show that > for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h > exists.
> The last part I'm unsure how to approach. > In particular, how to capture that the difference > quotient exists for all x.
> Thoughts?
e^x is the inverse of ln(x); you can define ln(x) as the integral from 1 to x of (1/t) with respect to t. Show that ln(x) is differentiable at every x>0, and the derivative is never 0. Then use the Inverse Function Theorem.
Alternatively, define e^x as a power series,
e^x = 1 + x + x^2/2 + x^3/3! + ... + x^n/n! + ...
and show that it coverges everywhere, and then apply theorems about differentiating power series term-by-term.
On Mon, 09 Nov 2009 12:45:25 -0800, aegis wrote: > I'd like to prove e^x is differentiable everywhere.
> Obviously, a start would be defining e^x > which I can do as a series. starting with > (1 + 1/n)^nx and expanding using the binomial > theorem. Then, I would want to show that > for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h > exists.
> The last part I'm unsure how to approach. > In particular, how to capture that the difference > quotient exists for all x.
> Thoughts?
you could show that the series representation is uniformly continuous for all z (via say Wierstrass well known test), and so the function is continuous for all z.
you could also derive the exp(z+h)=exp(z)exp(h) property directly from the series using the fact that the series for exp is absolutely convergent.
you can show (exp(h)-1)/h = 1+ h/2! +h^2/3!+...
and show that this series is also continuous for all values of h via the same arguments as for exp(z).
So lim(h->0) (exp(h)-1)/h =1 from which dexp(z)/dz= lim (exp(z+h)-exp(z))/h = exp(z) follows for all values of z.
In article <d0e358b3-4186-4df8-b82a-0d1620875...@v30g2000yqm.googlegroups.com>,
aegis <ae...@mad.scientist.com> wrote: > I'd like to prove e^x is differentiable everywhere.
> Obviously, a start would be defining e^x > which I can do as a series. starting with > (1 + 1/n)^nx and expanding using the binomial > theorem. Then, I would want to show that > for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h > exists.
> The last part I'm unsure how to approach. > In particular, how to capture that the difference > quotient exists for all x.
> Thoughts?
If you prove the functional equation, then differentiability at 0 implies differentiability everywhere.
<ae...@mad.scientist.com> wrote: >I'd like to prove e^x is differentiable everywhere.
>Obviously, a start would be defining e^x >which I can do as a series. starting with >(1 + 1/n)^nx and expanding using the binomial >theorem.
You _need_ to start with the definition. It's not clear to me what definition you have in mind: are you defining it by the power series or as the limit of (1+1/n)^(nx)?
If you define it by the power series you can use this fact from calculus: If f_n -> f on (a,b), f_n is differentiable on (a,b) and f_n' -> g uniformly on (a,b) then f is differentiable and f' = g.
>Then, I would want to show that >for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h >exists.
>The last part I'm unsure how to approach. >In particular, how to capture that the difference >quotient exists for all x.
>Thoughts?
David C. Ullrich
"Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to." (John Jones, "My talk about Godel to the post-grads." in sci.logic.)
On Mon, 9 Nov 2009, aegis wrote: > I'd like to prove e^x is differentiable everywhere.
> Obviously, a start would be defining e^x > which I can do as a series. starting with > (1 + 1/n)^nx and expanding using the binomial > theorem. Then, I would want to show that > for all x, d(e^x)/dx = lim(h->0) (e^(x + h) - e^x)/h > exists.
> The last part I'm unsure how to approach. > In particular, how to capture that the difference > quotient exists for all x.
> Thoughts?
With a simple twist of definition, you can also stick to polynomials: e^x is the limit (locally uniformly convergent) of
p_n(x) = (1+x/n)^n
and the sequence of derivatives
p_n'(x) = (1+x/n)^(n-1)
converges locally uniformly to the same limit.
The theorem needed here is (and can be considered a form of closed-graph statement for the operator of the derivative):
If f_n(x) converge uniformly to f(x) for -K<=x<=K and f_n'(x) converge uniformly to g(x) for -K<x<K then f'(x) = g(x) for -K<x<K. (Proof uses Mean Value Theorem).
