Limit of the remainder of Taylor polynomial of composite functions

In summary, the problem involves finding the Taylor polynomial of a composite function and showing that its remainder approaches 0 as quickly as ##x^n##. This can be proven using ##\epsilon - \delta## arguments or L'hopital's rule, but the latter may be complicated due to lack of information about the derivatives of the functions involved. To solve this, one can express the limit as a product of two other limits and use L'hopital's rule to evaluate them. Additionally, it is possible to use the Peano form of the remainder to show that the limit is equal to 0.
  • #1
Adgorn
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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 :)
 
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  • #2
Adgorn said:
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.
 
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  • #3
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##.
 

FAQ: Limit of the remainder of Taylor polynomial of composite functions

What is the "limit of the remainder" in the context of Taylor polynomials of composite functions?

The limit of the remainder refers to the difference between the actual value of a function and the approximation given by a Taylor polynomial of a composite function. It represents the error or leftover terms in the approximation.

Why is the "limit of the remainder" important in mathematics?

The limit of the remainder allows us to determine the accuracy of a Taylor polynomial approximation. By understanding the behavior of the remainder as the degree of the polynomial increases, we can determine how many terms are needed to achieve a desired level of accuracy.

How is the "limit of the remainder" calculated?

The limit of the remainder is calculated using the Lagrange form of the remainder, which involves taking the derivative of the function and evaluating it at a point within the interval of the Taylor polynomial. This value is then divided by the factorial of the degree of the polynomial.

What is the relationship between the "limit of the remainder" and the degree of the Taylor polynomial?

As the degree of the Taylor polynomial increases, the limit of the remainder tends to decrease. This means that the accuracy of the approximation improves as we include more terms in the polynomial.

How can the "limit of the remainder" be used to improve the accuracy of an approximation?

By understanding the behavior of the remainder and its relationship to the degree of the polynomial, we can determine the minimum number of terms needed to achieve a desired level of accuracy. We can also use techniques such as alternating series and error bounds to further improve the accuracy of the approximation.

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