Infinite Sequences and Series: Formulas for the Remainder Term in Taylor Series
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.
- @ 2:36 - Recap on Taylor and Maclaurin Series
- @ 8:26 - Formulas for the Remainder Term in Taylor Series
- @ 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
- @ 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
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.pdfArchive: 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
- Recap on Taylor and Maclaurin Series
- Formulas for the Remainder Term in Taylor Series
- Theorem 1: Integral Form of the Remainder Term
- Recap of Example 4 from Previous Video
- Example 1: sin x
- Absolute Value Integral Property
- 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
Taylor series of the function f at a (or about a or centered at a):
Maclaurin series is a Taylor series with a = 0:
A function is equal to the sum of its Taylor series if the limit of the Remainder approaches zero:
Taylor's Inequality is useful in finding the limit of the remainder.
I also alluded to alternatives to Taylor's Inequality:
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:
The n-th partial sum of this Taylor series is the n-th degree Taylor polynomial of f at a:
We can write:
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:
Proof:
We use mathematical induction, referenced below from my earlier video.
Retrieved: 2 July 2019
Archive: http://archive.fo/4lslc
For n = 1:
and the integral in the theorem is:
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
https://peakd.com/mathematics/@mes/integration-by-parts-example-5-definite-integrals
Retrieved: 21 February 2020
Archive: https://archive.ph/wip/lYdwO
Note the Fundamental Theorem of Calculus for reference.
Retrieved: 20 December 2019
Archive: https://archive.ph/wip/Bd7wE
Retrieved: 20 December 2019
Archive: https://archive.ph/wip/AbsvX
Thus, we have:
Thus:
The theorem is therefore proved for n = 1.
Now we suppose that Theorem 1 is true for n = k, that is:
We want to show that it's true for n = k + 1, that is:
Again we use integration by parts, this time with:
Thus:
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
Also note Equation 1 which I will be referencing later in this video.
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.
Retrieved: 21 January 2020
Archive: https://archive.ph/wip/IyP35
Retrieved: 21 January 2020
Archive: https://archive.ph/wip/kJR4R
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:
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows:
With a = 0 in Theorem 1, the remainder is:
Since f(n+1)(t) is ±sin t or ±cos t, we know that |f(n+1)(t)| ≤ 1 for all t.
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:
Proof:
We know that:
Thus, in our case and for x > 0, we have:
For x < 0 we can write:
So:
Thus, in any case, we have:
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.
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.
Retrieved: 24 February 2020
Archive: https://archive.ph/xRpCj
https://peakd.com/mathematics/@mes/mean-value-theorem-for-integrals-proof
Retrieved: 23 February 2020
Archive: https://archive.ph/wip/0pNjD
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:
Proof:
Because g doesn't change sign, either:
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.
Retrieved: 23 February 2020
Archive: https://archive.ph/wip/f4MRq
Thus, by the Extreme Value Theorem we have:
Since g(x) ≥ 0, we have:
And so:
If:
then these inequalities show that:
and so, for this case, Theorem 2 is true because both sides of the equation are 0, that is:
If:
then it must be positive and we can divide the earlier inequality by it:
First recall the Intermediate Value Theorem from my earlier video:
Retrieved: 23 February 2020
Archive: https://archive.ph/wip/VJImY
Then, by the Intermediate Value Theorem, there exists a number c in [a, b] such that:
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:
Note: This is an extension of the Mean Value Theorem (for Derivatives) which is the case n = 0:
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:
Then, by Theorem 1,
The formula for the remainder term in Theorem 3 is called Lagrange's form of the remainder term.
Notice that this expression:
is very similar to the terms in the Taylor series except that f(n+1) is evaluated at c instead of at a.
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:
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:
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:
Thus the second-degree Taylor polynomial is:
The desired approximation is:
(b) Using the Lagrange form of the remainder we can write:
where c lies between 8 and x.
In order to estimate the error we note that if 7 ≤ x ≤ 9, then:
Also, since c > 7, we have:
Calculation Check: 7^(8/3) = 179.306
And so:
Calculation Check: 5/(81·179)
Thus if 7 ≤ x ≤ 9, the approximation in part (a) is accurate to within 0.0004.
Hey @mes: time to update your witnesses votes?
https://steempeak.com/steem/@ura-soul/an-update-on-the-state-of-steem-from-my-pov
Done and done! Thanks for the headsup!
Well done, mate! ;)
Why do You use tag "steemiteducation" on hive :)?