Limit of the remainder of Taylor polynomial of composite functions

  • #1
123
10

Homework Statement:

(Spivak's Calculus, 20-9) This is a part of a problem to express the Taylor polynomial of a composite function. Let ##f(x)=P_{n,0,f}(x)+R_{n,0,f}(x)## and ##g(x)=P_{n,0,g}(x)+R_{n,0,g}(x)## where ##P_{n,0,f},P_{n,0,g}## are the Taylor polynomials of degree ##n## at ##0## for ##f## and ##g##, ##R_{n,0,f},R_{n,0,g}## are the corresponding remainders, and ##g(0)=0##. In part of the problem I need to show that $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(g(x))} {x^n}=0$$

Relevant Equations:

If ##p## and ##q## are polynomials in ##x-a## and ##\lim_{x \rightarrow 0} \frac {R(x)} {(x-a)^n}=0## then ##p(q(x)+R(x))=p(q(x))+\bar R(x)## where ##\lim_{x \rightarrow 0} \frac {\bar R(x)} {(x-a)^n}=0##
Since $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(x)} {x^n}=0,$$ ##P_{n,0,g}(x)## contains only terms of degree ##\geq 1## and ##R_{n,0,g}## approaches ##0## as quickly as ##x^n##, I can most likely prove this using ##\epsilon - \delta## arguments, but that seems overly complicated. I also can't use Taylor's Theorem since I don't know if ##f^{(n+1)}## exists. L'hopital's rule also didn't seem to do much since I know pretty much nothing about the derivatives of ##g## or of ##R_{n,0,f}##, so I don't know if ##R_{n,0,f}(g(x))^{(n)}## approaches #0# at #0#, as a result I'm not really sure how I'm supposed to do this. Help would be appreciated.

Also this is in the chapter that introduces Taylor Polynomials, I've not reached anything about infinite series yet so please no solutions which involve infinite sums or expansions. Thanks in advance to the helpers :)
 

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336
Homework Statement: (Spivak's Calculus, 20-9) This is a part of a problem to express the Taylor polynomial of a composite function. Let ##f(x)=P_{n,0,f}(x)+R_{n,0,f}(x)## and ##g(x)=P_{n,0,g}(x)+R_{n,0,g}(x)## where ##P_{n,0,f},P_{n,0,g}## are the Taylor polynomials of degree ##n## at ##0## for ##f## and ##g##, ##R_{n,0,f},R_{n,0,g}## are the corresponding remainders, and ##g(0)=0##. In part of the problem I need to show that $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(g(x))} {x^n}=0$$
Homework Equations: If ##p## and ##q## are polynomials in ##x-a## and ##\lim_{x \rightarrow 0} \frac {R(x)} {(x-a)^n}=0## then ##p(q(x)+R(x))=p(q(x))+\bar R(x)## where ##\lim_{x \rightarrow 0} \frac {\bar R(x)} {(x-a)^n}=0##

Since $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(x)} {x^n}=0,$$ ##P_{n,0,g}(x)## contains only terms of degree ##\geq 1## and ##R_{n,0,g}## approaches ##0## as quickly as ##x^n##, I can most likely prove this using ##\epsilon - \delta## arguments, but that seems overly complicated. I also can't use Taylor's Theorem since I don't know if ##f^{(n+1)}## exists. L'hopital's rule also didn't seem to do much since I know pretty much nothing about the derivatives of ##g## or of ##R_{n,0,f}##, so I don't know if ##R_{n,0,f}(g(x))^{(n)}## approaches #0# at #0#, as a result I'm not really sure how I'm supposed to do this. Help would be appreciated.

Also this is in the chapter that introduces Taylor Polynomials, I've not reached anything about infinite series yet so please no solutions which involve infinite sums or expansions. Thanks in advance to the helpers :)
I think you could write $$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(g(x))} {x^n}$$ as
$$\lim_{x \rightarrow 0} \frac {R_{n,0,f}(g(x))} {g^n(x)}\frac {g^n(x))} {x^n}$$
and then look at the limits of each term.
 
  • Like
Likes Delta2
  • #3
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
In addition to what's being said in #2, use L'hopital to find ##\lim_{x \rightarrow 0}\frac{g(x)}{x}## and then you can easily find ##\lim_{x\to 0}(\frac{g(x)}{x})^n##.

Also assuming the Peano form of the remainder I believe it is relatively easy to show that ##\lim_{x\to 0}\frac{R_{n,0,f}(g(x))}{g^n(x)}=0## using the fact that g is continuous at 0 and ##g(0)=0##.
 

Related Threads on Limit of the remainder of Taylor polynomial of composite functions

  • Last Post
Replies
2
Views
281
Replies
2
Views
469
Replies
8
Views
890
Replies
4
Views
5K
Replies
3
Views
420
Replies
3
Views
4K
  • Last Post
Replies
2
Views
960
  • Last Post
Replies
4
Views
1K
Replies
8
Views
7K
Replies
5
Views
3K
Top