Infinite Sequences and Series: Formulas for the Remainder Term in Taylor Series

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(Edited)

In this video I expand upon my previous video on Taylor and Maclaurin series by going over a few alternative forms of the Remainder Term. Recall that the Remainder Term is just that, a remainder term, and is what is left over when approximating a function with its Taylor series. In the previous video I showed that we can use Taylor’s Inequality to determine the remainder term, but in this video, I show that we can also use an integral form as well as a form called Lagrange’s form of the remainder term. I prove both of these forms of the remainder as well as going over some examples to compare with the previous video’s use of Taylor’s Inequality.

The topics covered in this video are listed below with their time stamps.

  1. @ 2:36 - Recap on Taylor and Maclaurin Series
  2. @ 8:26 - Formulas for the Remainder Term in Taylor Series
  3. @ 11:32 - Theorem 1: Integral Form of the Remainder Term
    • @ 40:48 - Recap of Example 4 from Previous Video
    • @ 43:44 - Example 1: sin x
      • @ 52:24 - Absolute Value Integral Property
  4. @ 1:08:42 - Lagrange's Form of the Remainder Term
    • @ 1:13:38 - Theorem 2: Weighted Mean Value Theorem for Integrals
    • @ 1:28:20 - Theorem 3: Lagrange's Form of the Remainder Term
    • @ 1:40:36 - Example 2: sin x
    • @ 1:44:35 - Example 3: x1/3

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Infinite Sequences and Series: Formulas for the Remainder Term in Taylor Series

Taylor Series Remainder Term.jpeg

Calculus Book Reference

Note that I mainly follow along the following calculus book:

Calculus: Early Transcendentals Sixth Edition by James Stewart
Additional Topics: Formulas for the Remainder Term in Taylor series: https://www.stewartcalculus.com/data/CALCULUS%206E%20Early%20Transcendentals/upfiles/topics/6et_at_02_frtts.pdf

Archive: https://web.archive.org/web/20200225055525/https://www.stewartcalculus.com/data/CALCULUS%206E%20Early%20Transcendentals/upfiles/topics/6et_at_02_frtts.pdf
Local PDF Download: https://1drv.ms/b/s!As32ynv0LoaIiIA4X5OFzeEgWP4q6g?e=Bh4RLu


Topics to Cover

  1. Recap on Taylor and Maclaurin Series
  2. Formulas for the Remainder Term in Taylor Series
  3. Theorem 1: Integral Form of the Remainder Term
    • Recap of Example 4 from Previous Video
    • Example 1: sin x
      • Absolute Value Integral Property
  4. Lagrange's Form of the Remainder Term
    • Theorem 2: Weighted Mean Value Theorem for Integrals
    • Theorem 3: Lagrange's Form of the Remainder Term
    • Example 2: sin x
    • Example 3: x1/3

Recap on Taylor and Maclaurin Series

First a quick recap from my earlier video on Taylor series.

https://peakd.com/mathematics/@mes/infinite-sequences-and-series-taylor-and-maclaurin-series

Retrieved: 20 February 2020
Archive: https://archive.ph/wip/8CV5z

image.png

Taylor series of the function f at a (or about a or centered at a):

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Maclaurin series is a Taylor series with a = 0:

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A function is equal to the sum of its Taylor series if the limit of the Remainder approaches zero:

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Taylor's Inequality is useful in finding the limit of the remainder.

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I also alluded to alternatives to Taylor's Inequality:

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Formulas for the Remainder Term in Taylor Series

In my earlier video, recapped above, we considered functions f with derivatives of all orders and their Taylor series:

image.png

The n-th partial sum of this Taylor series is the n-th degree Taylor polynomial of f at a:

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We can write:

image.png

where Rn(x) is the remainder of the Taylor series.

We know that f is equal to the sum of its Taylor series on the interval |x - a| < R, where R is the radius of convergence, if we can show that limn→∞ Rn(x) = 0 for |x - a| < R.


Here we derive formulas for the remainder term Rn(x).

The first such formula involves an integral.

Theorem 1: Integral Form of the Remainder Term

If f(n+1) is continuous on an open interval I that contains a, and x is in I, then:

image.png


Proof:

We use mathematical induction, referenced below from my earlier video.

https://youtu.be/WdIr_onvUtE

Retrieved: 2 July 2019
Archive: http://archive.fo/4lslc

image.png

For n = 1:

image.png

and the integral in the theorem is:

image.png

To evaluate this integral we integrate by parts, referenced from my earlier video:

https://peakd.com/mathematics/@mes/integration-by-parts-proof

Retrieved: 21 February 2020
Archive: https://archive.ph/wip/3AkNY

image.png

https://peakd.com/mathematics/@mes/integration-by-parts-example-5-definite-integrals

Retrieved: 21 February 2020
Archive: https://archive.ph/wip/lYdwO

image.png

Note the Fundamental Theorem of Calculus for reference.

https://youtu.be/3o8Q6UJzJyk

Retrieved: 20 December 2019
Archive: https://archive.ph/wip/Bd7wE

image.png

https://youtu.be/yuIl-BPQHss

Retrieved: 20 December 2019
Archive: https://archive.ph/wip/AbsvX

image.png

Thus, we have:

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Thus:

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The theorem is therefore proved for n = 1.

Now we suppose that Theorem 1 is true for n = k, that is:

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We want to show that it's true for n = k + 1, that is:

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Again we use integration by parts, this time with:

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Thus:

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Therefore, Theorem 1 is true for n = k + 1 when it is true for n = k.

Thus, by mathematical induction, it is true for all n.


To illustrate Theorem 1 we use it to solve Example 4 from my earlier video and referenced below.

https://peakd.com/mathematics/@mes/infinite-sequences-and-series-taylor-and-maclaurin-series

image.png

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Also note Equation 1 which I will be referencing later in this video.

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Also note the Squeeze Theorem which was used in the above Example 4 from my earlier video, and will be used in Example 1 of this video.

https://youtu.be/vzJ7SEBLuWo

Retrieved: 21 January 2020
Archive: https://archive.ph/wip/IyP35

image.png

image.png

https://peakd.com/mathematics/@mes/infinite-sequences-limits-squeeze-theorem-fibonacci-sequence-and-golden-ratio-more

Retrieved: 21 January 2020
Archive: https://archive.ph/wip/kJR4R

image.png

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The sequence {bn} is squeezed between the sequences {an} and {cn}.


Example 1

Find the Maclaurin series for sin x and prove that it represents sin x for all x.

Solution:

We arrange our computation in two columns as follows:

image.png

Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows:

image.png

With a = 0 in Theorem 1, the remainder is:

image.png

Since f(n+1)(t) is ±sin t or ±cos t, we know that |f(n+1)(t)| ≤ 1 for all t.

image.png


Before continuing further, I will prove the following integral property which I should've proved in an earlier video years ago…

Absolute Value Integral Property

If f is continuous on [a, b], that is a ≤ b, then:

image.png


Proof:

We know that:

image.png


Thus, in our case and for x > 0, we have:

image.png

For x < 0 we can write:

image.png

So:

image.png

Thus, in any case, we have:

image.png

The right side of this inequality approaches 0 as n → ∞, as per Equation 1 from my earlier video referenced earlier above.

Thus |Rn(x)| → 0 as n → ∞ by the Squeeze Theorem.

image.png

It follows that Rn(x) → 0 as n → ∞, so sin x is equal to the sum of its Maclaurin series.


Lagrange's Form of the Remainder Term

For some purposes the integral formula in Theorem 1 is awkward to work with, so we are going to establish another formula for the remainder term.

To that end we need to prove the following generalization of the Mean Value Theorem for Integrals, but first a quick recap on the Mean Value Theorem for Derivatives and Integrals from my earlier videos.

https://youtu.be/x-2MiiG2E38

Retrieved: 24 February 2020
Archive: https://archive.ph/xRpCj

image.png

image.png

https://peakd.com/mathematics/@mes/mean-value-theorem-for-integrals-proof

Retrieved: 23 February 2020
Archive: https://archive.ph/wip/0pNjD

image.png

image.png

image.png


Theorem 2: Weighted Mean Value Theorem for Integrals

If f and g are continuous on [a, b] and g does not change sign in [a, b], then there exists a number c in [a, b] such that:

image.png


Proof:

Because g doesn't change sign, either:

image.png

For the sake of definiteness, let's assume that g(x) ≥ 0.

By the Extreme Value Theorem, referenced below from my earlier video, f has an absolute minimum value m and an absolute maximum value M.

https://youtu.be/UVVJPzce6M8

Retrieved: 23 February 2020
Archive: https://archive.ph/wip/f4MRq

image.png

image.png

Thus, by the Extreme Value Theorem we have:

image.png

Since g(x) ≥ 0, we have:

image.png

And so:

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If:

image.png

then these inequalities show that:

image.png

and so, for this case, Theorem 2 is true because both sides of the equation are 0, that is:

image.png

If:

image.png

then it must be positive and we can divide the earlier inequality by it:

image.png

First recall the Intermediate Value Theorem from my earlier video:

https://youtu.be/SYrfxAWtOOo

Retrieved: 23 February 2020
Archive: https://archive.ph/wip/VJImY

image.png

image.png

Then, by the Intermediate Value Theorem, there exists a number c in [a, b] such that:

image.png

And this proves the Weighted Mean Value Theorem.


Theorem 3: Lagrange's Form of the Remainder Term

If f(n+1) is continuous on an open interval I that contains a, and x is in I, then there exists a number c between a and x such that:

image.png


Note: This is an extension of the Mean Value Theorem (for Derivatives) which is the case n = 0:

image.png


Proof:

The function g(t) = (x - t)n doesn't change sign in the interval from a to x, so the Weighted Mean Value Theorem for Integrals gives a number c between a and x such that:

image.png

Then, by Theorem 1,

image.png

The formula for the remainder term in Theorem 3 is called Lagrange's form of the remainder term.


Notice that this expression:

image.png

is very similar to the terms in the Taylor series except that f(n+1) is evaluated at c instead of at a.

image.png

All we can say about the number c is that it lies somewhere between a and x.


In the following example we show how to use Lagrange's form of the remainder term as an alternative to the integral form in Example 1.

Example 2

Prove the Maclaurin series for sin x represents sin x for all x.

Solution:

Using the Lagrange form of the remainder term with a = 0, we have:

image.png

where f(x) = sin x and c lies between 0 and x.

But f(n+1)(c) is ±sin c or ±cos c.

In any case, |f(n+1)(c)| ≤ 1 and so:

image.png

By Equation 1 referenced earlier, the right side of this inequality approaches 0 as n → ∞, so |Rn(x)| → 0 by the Squeeze Theorem, as in Example 1.

It follows that the Rn(x) → 0 as n → ∞, so sin x is equal to the sum of its Maclaurin series.


Example 3

(a) Approximate the function f(x) = x1/3 by a Taylor polynomial of degree 2 at a = 8.

(b) How accurate is this approximation when 7 ≤ x ≤ 9?

Solution:

(a)

Differentiating and creating two columns we have:

image.png

Thus the second-degree Taylor polynomial is:

image.png

The desired approximation is:

image.png

(b) Using the Lagrange form of the remainder we can write:

image.png

where c lies between 8 and x.

In order to estimate the error we note that if 7 ≤ x ≤ 9, then:

image.png

Also, since c > 7, we have:

image.png

Calculation Check: 7^(8/3) = 179.306

And so:

image.png

Calculation Check: 5/(81·179)

Thus if 7 ≤ x ≤ 9, the approximation in part (a) is accurate to within 0.0004.



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4 comments
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Why do You use tag "steemiteducation" on hive :)?

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