Taylor's inequality for the remainder of a series

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Taylor'southward inequality tells us the maximum rest of the series

Taylor'south inequality states that, for a office ???f(x)???,

taylor's inequality

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This theorem looks elaborate, merely it'southward nothing more than a tool to find the remainder of a serial.

For example, often we're asked to find the ???n???th-degree Taylor polynomial that represents a function ???f(x)???. The sum of the terms later on the ???n???th term that aren't included in the Taylor polynomial is the remainder. We can utilise Taylor's inequality to detect that remainder and say whether or non the ???due north???th-caste polynomial is a adept approximation of the function's actual value.

Sometimes nosotros can use Taylor's inequality to show that the balance of a power series is ???R_n(x)=0???. If the remainder is ???0???, and then we know that the series representation of the office is equal to the exact value of the original part.

If we desire to use the theorem to show that the power series representation of the role is equal to the the function itself, then we'll need to show that both parts of Taylor's inequality are true and that the remainder is ???0???.

How to utilise Taylor'southward inequality to find the balance

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Showing that the function's representation is equal to the original function

Instance

Use Taylor's inequality to show that the the Maclaurin series representation of the function is equal to the original function.

???f(ten)=\sin{x}???

Using a table of common Maclaurin serial, we know that the power serial representation of the Maclaurin serial for ???f(ten)=\sin{10}??? is

???\sin{x}=\sum^{\infty}_{north=0}\frac{(-1)^nx^{2n+1}}{(2n+ane)!}???

In club to bear witness that this equation is true, that the sum of the Maclaurin series is in fact equal to the original office, we'll need to use Taylor's inequality to show that the balance of the power series is ???0???.

Since we're dealing with a Maclaurin series, ???a=0???, and we can adjust the inequalities from the theorem from

inequalities in taylor's theorem

changing the inequalities

rewriting taylor's inequality

Adjacent, we'll need to accept the first few derivatives of ???f(x)=\sin{x}??? in guild to plug in a value for ???f^{due north+1}(x)???.

table of derivatives for the maclaurin series

Since we'll be taking the absolute value of ???f^{n+1}(x)???, we tin can say that

???\left|f^{n+i}(x)\right|=\cos{10}???

or

???\left|f^{n+1}(ten)\right|=\sin{10}???

We know that ???\cos{x}??? and ???\sin{x}??? only exist between ???-1??? and ???one???, and so we could say

???-i\leq\left|f^{n+1}(10)\right|\leq 1???

However, since nosotros're dealing with absolute value, ???\left|f^{n+i}(ten)\right|??? can't exist negative, and then

???0\leq\left|f^{northward+1}(10)\correct|\leq 1???

This inequality tells united states of america that the value of ???|f^{n+1}(10)|??? is somewhere on the interval ???[0,1]???. Let's pick a few values in the interval and plug them into the first inequality from Taylor'southward inequality.

plugging values into the taylor's inequality

Taylor's inequality for Sequences and Series

In social club to evidence that this equation is true, that the sum of the Maclaurin series is in fact equal to the original office, we'll demand to use Taylor'due south inequality to show that the remainder of the power series is 0.

What we can meet is that, if we pick any value ???\left|f^{n+ane}(x)\right|<1???, then we won't be including the whole interval ???[0,1]???. But if we pick ???\left|f^{n+ane}(10)\right|=1???, then we know that ???M??? volition ever be greater than or equal to whatever value in the interval. Therefore, equally a rule, nosotros'll always choice the right-hand side of the interval. In this case, that's ???M=i???, which we'll plug into our already simplified version of Taylor's inequality.

simplified version of taylor's inequality

With Taylor's inequality simplified after plugging in ???a=0??? and ???M=1???, we'll use squeeze theorem to evaluate the remainder inequality and try to show that the remainder is ???0???.

???\left|R_n(x)\right|\le{\frac{1}{(n+i)!}}\left|ten\right|^{n+one}???

???\lim_{n\to\infty}\left|R_n(10)\right|\le\lim_{due north\to\infty}{\frac{1}{(due north+i)!}}\left|x\right|^{n+1}???

???\lim_{n\to\infty}\left|R_n(x)\right|\le{\frac{ane}{(\infty+1)!}}\left|10\right|^{\infty+1}???

???\lim_{n\to\infty}\left|R_n(ten)\correct|\le0\cdot\left|x\correct|^{\infty}???

???\lim_{n\to\infty}\left|R_n(x)\right|\le0???

Past definition, it'southward impossible for a residual to e'er be negative, so it must be true that

???\lim_{northward\to\infty}\left|R_n(x)\correct|=0???

Since the remainder is ???0???, we know that the power series representation of the Maclaurin series of the part is exactly equal to the original function ???f(x)=\sin{x}???.

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