proving e^x is differentiable everywhere
aegis
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?
Arturo Magidin
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.
--
Arturo Magidin
rancid moth
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.
A N Niel
<d0e358b3-4186-4df8...@v30g2000yqm.googlegroups.com>,
aegis <ae...@mad.scientist.com> wrote:
If you prove the functional equation, then differentiability at 0
implies differentiability everywhere.
David C. Ullrich
<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.)
Zdislav V. Kovarik
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).
Cheers, ZVK(Slavek).