(29.4) Let and be differentiable functions on an open interval . Suppose there are with and . Show that there exists so that
Hint. Consider .
Proof. Let , so that . By Rolle’s theorem, as , there exists between so that . But as always, this implies that .
(31.2) Consider the hyperbolic sine and cosine functions, and .
Find the Taylor series for and show that it converges to .
Find the Taylor series for and show that it converges to .
We have that and that by basic calculation. Let be where we center our Taylor series, and notice that and . So using the formula for Taylor series, we get that the Taylor series for is,
Let and consider the interval . Notice that on the interval . By a similar argument, on . By the Corollary to Taylor’s theorem, we deduce that converges to its Taylor series on . is arbitrary, so we may in fact deduce that converges to its Taylor series on all of .
The Taylor series for is,
and the argument for showing that it converges to on is identical to part (a).
(31.6) We’re going to work through an alternative proof of Taylor’s theorem, for Taylor series centered at .
Let be defined on with and assume that, for given , exists on . Let , and take to be the unique number so that
is valid. Define
for . An earlier version had a typo, saying that was a function of , instead of . The version now is correct.
Show that is differentiable on and that furthermore
Show .
Apply Rolle’s Theorem to obtain so that .
This is a simple calculation. With care, you will notice that all interior terms in the sum cancel to yield the desired expression.
A calculation shows that .
As there exists for which by Rolle’s theorem. implies that , from which we conclude that .
Let be differentiable on with .
Select and define for .
Prove that is a convergent sequence. Hint. To show that is Cauchy, first show that for .
Show that has a fixed point. In other words, there is a number so that .
Proof. Consider the interval from to . The Mean Value Theorem says that there exists a number between and so that
In particular, we have that . From this, iterated, we see that for all , . The series is convergent as , and say its value is .
Let and consider a natural number . Then for we have that
So converges.
Let . Then notice that as is differentiable, hence continuous, but also that . So .