# 5.6.E: Problems on Tayior's Theorem

Exercise (PageIndex{1})

Complete the proofs of Theorems (1,1^{prime},) and 2.

Exercise (PageIndex{2})

Verify Note 1 and Examples (b) and (left(mathrm{b}^{prime prime} ight)).

Exercise (PageIndex{3})

Taking (g(t)=(a-t)^{s}, s>0,) in ((6),) find
[
R_{n}=frac{f^{(n+1)}(q)}{n ! s}(x-p)^{s}(x-q)^{n+1-s} quad( ext { Schloemilch-Roche remainder }).
]
Obtain (left(5^{prime} ight)) and (left(5^{prime prime} ight)) from it.

Exercise (PageIndex{4})

Prove that (P_{n}) (as defined) is the only polynomial of degree (n) such that
[
f^{(k)}(p)=P_{n}^{(k)}(p), quad k=0,1, ldots, n .
]
[Hint: Differentiate (P_{n} n) times to verify that it satisfies this property.
For uniqueness, suppose this also holds for
[
P(x)=sum_{k=0}^{n} a_{k}(x-p)^{k} .
]
Differentiate (P n) times to show that
[
P^{(k)}(p)=f^{(k)}(p)=a_{k} k ! ,
]
(left. ext { so } P=P_{n} . ext { (Why?) } ight])

Exercise (PageIndex{5})

With (P_{n}) as defined, prove that if (f) is (n) times differentiable at (p,) then
[
f(x)-P_{n}(x)=oleft((x-p)^{n} ight) ext { as } x ightarrow p
]
(Taylor's theorem with Peano remainder term).
[Hint: Let (R(x)=f(x)-P_{n}(x)) and
[
delta(x)=frac{R(x)}{(x-p)^{n}} ext { with } delta(p)=0 .
]
Using the "simplified" L'Hôpital rule (Problem 3 in (§3) ) repeatedly (n) times, prove (left. ext { that } lim _{x ightarrow p} delta(x)=0 . ight])

Exercise (PageIndex{6})

Use Theorem (1^{prime}) with (p=0) to verify the following expansions, and prove that (lim _{n ightarrow infty} R_{n}=0).
(a) (sin x=x-frac{x^{3}}{3 !}+frac{x^{5}}{5 !}-cdots-frac{(-1)^{m} x^{2 m-1}}{(2 m-1) !}+frac{(-1)^{m} x^{2 m+1}}{(2 m+1) !} cos heta_{m} x)
for all (x in E^{1}) ;
(b) (cos x=1-frac{x^{2}}{2 !}+frac{x^{4}}{4 !}-cdots+frac{(-1)^{m} x^{2 m}}{(2 m) !}-frac{(-1)^{m} x^{2 m+2}}{(2 m+2) !} sin heta_{m} x) for
all (x in E^{1} .)
[Hints: Let (f(x)=sin x) and (g(x)=cos x .) Induction shows that
[
f^{(n)}(x)=sin left(x+frac{n pi}{2} ight) ext { and } g^{(n)}(x)=cos left(x+frac{n pi}{2} ight) .
]
Using formula (left(5^{prime} ight),) prove that
[
left|R_{n}(x) ight| leqleft|frac{x^{n+1}}{(n+1) !} ight| ightarrow 0 .
]
Indeed, (x^{n} / n !) is the general term of a convergent series
[
left.sum frac{x^{n}}{n !} quad ext { (see Chapter } 4, §13, ext { Example }(mathrm{d}) ight).
]
(left. ext { Thus } x^{n} / n ! ightarrow 0 ext { by Theorem } 4 ext { of the same section. } ight])

Exercise (PageIndex{7})

For any (s in E^{1}) and (n in N,) define
[
left(egin{array}{l}{s} {n}end{array} ight)=frac{s(s-1) cdots(s-n+1)}{n !} ext { with }left(egin{array}{l}{s} {0}end{array} ight)=1 .
]
Then prove the following.
(i) (lim _{n ightarrow infty} nleft(egin{array}{l}{s} {n}end{array} ight)=0) if (s>0),
(ii) (lim _{n ightarrow infty}left(egin{array}{l}{s} {n}end{array} ight)=0) if (s>-1),
(iii) For any fixed (s in E^{1}) and (x in(-1,1)).
[
lim _{n ightarrow infty}left(egin{array}{l}{s} {n}end{array} ight) n x^{n}=0 ;
]
hence
[
lim _{n ightarrow infty}left(egin{array}{c}{s} {n}end{array} ight) x^{n}=0 .
]
(left[ ext { Hints: }(mathrm{i}) ext { Let } a_{n}=left|nleft(egin{array}{l}{s} {n}end{array} ight) ight| . ext { Verify that } ight.)
[
a_{n}=|s|left|1-frac{s}{1} ight|left|1-frac{s}{2} ight| cdotsleft|1-frac{s}{n-1} ight| .
]
If (s>0,left{a_{n} ight} downarrow) for (n>s+1,) so we may put (L=lim a_{n}=lim a_{2 n} geq 0 .) (Explain!)
Prove that
[
frac{a_{2 n}}{a_{n}}]
so for large (n),
[
frac{a_{2 n}}{a_{n}}]
With (varepsilon) fixed, let (n ightarrow infty) to get (L leqleft(e^{-frac{1}{2} s}+varepsilon ight) L .) Then with (varepsilon ightarrow 0,) obtain (L e^{frac{1}{2} s} leq L).
(left. ext { As } e^{frac{1}{2} s}>1 ext { (for } s>0 ight),) this implies (L=0,) as claimed.
(ii) For (s>-1, s+1>0,) so by ((i)),
[
(n+1)left(egin{array}{c}{s+1} {n+1}end{array} ight) ightarrow 0 ; ext { i.e., }(s+1)left(egin{array}{c}{s} {n}end{array} ight) ightarrow 0 . quad( ext { Why? })
]
(iii) Use the ratio test to show that the series (sumleft(egin{array}{l}{s} {n}end{array} ight) n x^{n}) converges when (|x|<1).
( ext { Then apply Theorem } 4 ext { of Chapter } 4, §13 .])

Exercise (PageIndex{8})

Continuing Problems 6 and (7,) prove that
[
(1+x)^{s}=sum_{k=0}^{n}left(egin{array}{l}{s} {k}end{array} ight) x^{k}+R_{n}(x) ,
]
where (R_{n}(x) ightarrow 0) if either (|x|<1,) or (x=1) and (s>-1,) or (x=-1) and (s>0 .)
[Hints: (a) If (0 leq x leq 1,) use (left(5^{prime} ight)) for
[
R_{n-1}(x)=left(egin{array}{c}{s} {n}end{array} ight) x^{n}left(1+ heta_{n} x ight)^{s-n}, quad 0< heta_{n}<1 . ext { (Verify!) }
]
Deduce that (left|R_{n-1}(x) ight| leqleft|left(egin{array}{c}{s} {n}end{array} ight) x^{n} ight| ightarrow 0 .) Use Problem 7(( ext { iii) if }|x|<1 ext { or Problem } 7( ext { ii })) if (x=1).
(b) If (-1 leq x<0,) write (left(5^{prime prime} ight)) as
[
R_{n-1}(x)=left(egin{array}{c}{s} {n}end{array} ight) n x^{n}left(1+ heta_{n}^{prime} x ight) s^{-1}left(frac{1- heta_{n}^{prime}}{1+ heta_{n}^{prime} x} ight)^{n-1} . ext { (Check!) }
]
As (-1 leq x<0,) the last fraction is (leq 1 .) (Why?) Also,
[
left(1+ heta_{n}^{prime} x ight)^{s-1} leq 1 ext { if } s>1, ext { and } leq(1+x)^{s-1} ext { if } s leq 1 .
]
Thus, with (x) fixed, these expressions are bounded, while (left(egin{array}{c}{s} {n}end{array} ight) n x^{n} ightarrow 0) by Problem 7((mathrm{i})) (left. ext { or }( ext { iii }) . ext { Deduce that } R_{n-1} ightarrow 0, ext { hence } R_{n} ightarrow 0 . ight])

Exercise (PageIndex{9})

Prove that
[
ln (1+x)=sum_{k=1}^{n}(-1)^{k+1} frac{x^{k}}{k}+R_{n}(x) ,
]
where (lim _{n ightarrow infty} R_{n}(x)=0) if (-1[Hints: If (0 leq x leq 1,) use formula (left(5^{prime} ight)).
If (-1[
R_{n}(x)=frac{ln (1+x)}{(-1)^{n}}left(frac{1- heta_{n}}{1+ heta_{n} x} cdot x ight)^{n} .
]
( ext { Proceed as in Problem } 8 .])

Exercise (PageIndex{10})

Prove that if (f : E^{1} ightarrow E^{*}) is of class (mathrm{CD}^{1}) on ([a, b]) and if (-infty[
fleft(x_{0} ight)>frac{f(b)-f(a)}{b-a}left(x_{0}-a ight)+f(a) ;
]
i.e., the curve (y=f(x)) lies above the secant through ((a, f(a))) and ((b, f(b)) .)
[Hint: This formula is equivalent to
[
frac{fleft(x_{0} ight)-f(a)}{x_{0}-a}>frac{f(b)-f(a)}{b-a} ,
]
i.e., the average of (f^{prime}) on (left[a, x_{0} ight]) is strictly greater than the average of (f^{prime}) on ([a, b],) (left. ext { which follows because } f^{prime} ext { decreases on }(a, b) .( ext { Explain! }) ight])

Exercise (PageIndex{11})

Prove that if (a, b, r,) and (s) are positive reals and (r+s=1,) then
[
a^{r} b^{s} leq r a+s b .
]
(This inequality is important for the theory of so-called (L^{p}) -spaces.)
[Hints: If (a=b) , all is trivial.
Therefore, assume (a[
a=(r+s) a]
Use Problem 10 with (x_{0}=r a+s b in(a, b)) and
[
f(x)=ln x, f^{prime prime}(x)=-frac{1}{x^{2}}<0 .
]
Verify that
[
x_{0}-a=x_{0}-(r+s) a=s(b-a)
]
and (r cdot ln a=(1-s) ln a ;) hence deduce that
[
r cdot ln a+s cdot ln b-ln a=s(ln b-ln a)=s(f(b)-f(a)) .
]
After substitutions, obtain
[
left.fleft(x_{0} ight)=ln (r a+s b)>r cdot ln a+s cdot ln b=ln left(a^{r} b^{s} ight) . ight]
]

Exercise (PageIndex{12})

Use Taylor's theorem (Theorem 1') to prove the following inequalities:
(a) (sqrt[3]{1+x}<1+frac{x}{3}) if (x>-1, x eq 0).
(b) (cos x>1-frac{1}{2} x^{2}) if (x eq 0).
(c) (frac{x}{1+x^{2}}0).
(d) (x>sin x>x-frac{1}{6} x^{3}) if (x>0).

Apply Taylor’s Theorem to the function defined as to estimate the value of . Use . Estimate an upper bound for the error.

##### Solution

The Taylor approximation of the function around the point is given as follows:

If terms are used (including ), then the upper bound for the error is:

With , an upper bound for the error can be calculated by assuming . That’s because the absolute values of the derivatives of attain their maximum value in the interval at 4. The derivatives of have the following form:

The derivatives of the function evaluated at the point can be calculated as:

If two terms are used, then:

In this case, the upper bound for the error is:

Using Mathematica, the square root of approximated to 4 decimal places is . Therefore, the error when two terms are used is:

which satisfies that the actual error is less the upper bound:

If three terms are used, then:

In this case, the upper bound for the error is:

The actual error in this case is indeed less than the upper bound:

The following code provides a user-defined function for the Taylor series having the following inputs: a function , the value of , the value of , and the number of terms including the constant term.

You can download the MATLAB file below which provides the solution to this question.

#### Example 2

Apply Taylor’s Theorem to the function defined as to estimate the value of and . Use . Estimate an upper bound for the error.

##### Solution

First, we will calculate the numerical solution for the and :

The Taylor approximation around is given as:

If terms are used (including ) and if , then the upper bound of the error is:

If , then, the upper bound of the error is:

The derivatives of the function are given by:

when evaluated at these have the values:

For , and using two terms:

The terms are getting closer to each other and four terms provide a good approximation. The error term in the theorem gives an upper bound for as follows:

The maximum value would be obtained for . Therefore:

Indeed, the actual value of the error is less than the upper bound:

For , the Taylor series for this function around doesn’t give a very good approximation as will be shown here but rather keeps oscillating. First, using two terms:

In fact, the following table gives the values up to 21 terms. It is clear that the Taylor series is diverging.

The error term in the theorem gives an upper bound for when six terms are used as follows:

The maximum value will be obtained when :

This is a large upper bound and indicates that using six terms is not giving a good approximation. In general, the Taylor series works best if the distance between and is as small as possible. For some functions, like , , and , the Taylor series always converges. However, for functions with square roots, the Taylor series converges when is relatively close to . There are some analytical conditions that would indicate the radius of convergence of a Taylor series however, this is beyond the scope of this course!

The following code provides a user-defined function for the Taylor series having the following inputs: a function , the value of , the value of , and the number of terms including the constant term.

#### Example 3

Use zero through fourth order Taylor’s series expansion to approximate the value of the function defined as at . Use . Calculate the error associated with each expansion.

##### Solution

The true value of the function at is given by:

The zero order Taylor series expansion around has the following form:

The error in this case is given by:

The first order Taylor series expansion around has the following form:

The error in this case is given by:

The second order Taylor series expansion around has the following form:

The error in this case is given by:

The third order Taylor series expansion around has the following form:

The error in this case is given by:

The fourth order Taylor series expansion around has the following form:

The error in this case is given by:

It is important to note that the error reduces to zero when using the fourth order Taylor series approximation. That is because the fourth order Taylor series approximation of a fourth order polynomial function is identical to the function itself. You can think of this as follows, the zero order Taylor approximation provides a “constant” function approximation. The second order Taylor approximation provides a parabolic function approximation while the third order provides a cubic function approximation. The nth Taylor series approximation of a polynomial of degree “n” is identical to the function being approximated!

You can download the MATLAB file below which provides the solution to this question.

### Problems

1. Use the Taylor series for the function defined as to estimate the value of . Use once and for another time. Estimate an upper bound for the error for each approximation. Comment on the behaviour of the Taylor series of this function. (hint: For this particular function using a Taylor expansion around should not give a proper approximation for because 10 and 4 are far from each other)
2. Using the Taylor series and setting , derive the polynomial forms of the functions listed in the MacLaurin series section.
3. Use Taylor’s Theorem to find an estimate for at with . Employ the zero-, first-,
second-, and third-order versions and compute the truncation error for each case.
4. Using the MacLaurin series expansion for , find an approximation for as a function of the number of terms up to 5 terms. For each case, find the relative error and the relative approximate error .
5. Use zero- through third-order Taylor series expansions to predict for

## 5.6.E: Problems on Tayior's Theorem

p11 is not actually a polynomial because of the term at the end which is used to signal which polynomial is represented (in case some of the coefficients are 0). We can convert to a polynomial.

 > p11 := convert(p11,polynom)

The Taylor polynomials are usually good approximations to the function near a. Let's plot the first few polynomials for the sin function at x =0.

 > sinplot := plot(sin,-Pi..2*Pi,thickness=2):

 > tays:= plots[display](sinplot): for i from 1 by 2 to 11 do tpl := convert(taylor(sin(x), x=0,i),polynom): tays := tays,plots[display]([sinplot,plot(tpl,x=-Pi..2*Pi,y=-2..2, color=black,title=convert(tpl,string))]) od:

 > plots[display]([tays],view=[-Pi..2*Pi,-2..2])

Just how close the polynomials are to the function is determined using Taylor's theorem below.

Theorem: (Taylor's remainder theorem) If the (n+1)st derivative of f is defined and bounded in absolute value by a number M in the interval from a to x, then

This theorem is essential when you are using Taylor polynomials to approximate functions, because it gives a way of deciding which polynomial to use. Here's an example.

Problem Find the 2nd Taylor polynomial p[2] of at . Plot both the polynomial and f on the interval [.5,1.5]. Determine the maximum error in using p[2] to approximate ln(x) in this interval.

 > fplot := plot(f,.5..1.5,thickness = 2):

 > p[2] := x -> sum(([email protected]@i)(f)(1.)/i!*(x-1.)^i,i=0..2)

 > t2 := unapply( convert(taylor(f(x),x=1,3),polynom),x)

 > tplot := plot(t2,1..1.5,color=black):

 > plots[display]([fplot,tplot])

So the remainder is bounded by

We can see from the plot of f and the polynomial that the actual error is never more than about .1 on the interval [.5,1.5].

Which Taylor polynomial would you use to approximate the sin function on the interval from -Pi to Pi to within 1/10^6?

Well, 1 is a bound on any derivative of the sin on any interval. So we need to solve the inequality

for n. Solve will not be much help here because of the factorial, but we can find the smallest n by running through a loop.

 > n := 1: while evalf(1/n!*Pi^n) > 1/10^6 do n := n+1 od: print (take n to be ,n)

 > t17 := convert(taylor(sin(x),x=0,18),polynom)

Looks pretty much like the sin function.

Exercise: Show that is approximated to within 7 decimals by for all x in .

at x =0 on the interval [-2,2]

at x = 0 on the interval [-2..2]

Exercise: Write a procedure to compute sin(x) for any x by using p[5]. restricted to the interval [0,Pi/4].

Outline of solution : If x is negative, replace x with and use the oddness property. If x is greater than or equal to 2*Pi, then replace x with x-2*Pi and use the periodicity . Repeat this step until [0, 2*Pi ). If Pi/4 < x < Pi/2 , then use the trig indentity

We know that the higher the degree of an equation, the more "turning points" it may have. For example, a parabola has one "turning point."

(A parabola has an equation of the form $y=ax^2 + bx +c$.)

A cubic of the form $y=ax^3 + bx^2 +cx +d$ can have up to two "turning points," though it may have fewer. In general, an equation of degree $n$ may have up to $n-1$ turning points.

(Here is the polynomial $f(x) = 2x^4 - x^3 -3x^2 + 7x - 13$. It is degree 4 and it has the maximum number of turning points, 4-1=3. But, keep in mind, some degree 4 polynomials have only one or two turning points. The degree gives us the MAXIMUM number: $n-1$.)

This is important because, if you want to use a polynomial to approximate a function, you will want to use a polynomial of high enough degree to match the "features" of the function. The Taylor series will let you do this with functions that are "infinitely differentiable" since it uses the derivatives of the function to approximate the functions behavior.

Here are Taylor polynomials of increasing degree and the sine curve. Notice how they are "wrapping around" the sine curve, giving an approximation that fits better and better over more of the curve as the degree of the Taylor polynomial increases.

Since the sine curve has so many turning points it is easy to see that to match all of the features of the sine curve we will need to take the limit of the $n^$ degree Taylor polynomial as $n ightarrow infty$.*

That's the intuition behind the Taylor series. The higher the degree, the better the "fit." Why? Because higher degree curve have more "turning points" so they can better match the shape of things like the sine function. (As long as the function we are approximating is differentiable.)

*Side note: A function may have only a few turning points and still need infinitely many terms of the Taylor polynomial. Take the catenary, for example, which only has one turning point since it looks like a parabola. The Taylor series for the catenary will not have any terms where the coefficients are zero, since the derivatives of the catenary are hyperbolic sinusoidal functions.

But, even with the catenary, higher degree polynomials give a better approximation.

## 5.6.E: Problems on Tayior's Theorem

### Lecture Description

This video lecture, part of the series Calculus Videos: Series and Sequences by Prof. , does not currently have a detailed description and video lecture title. If you have watched this lecture and know what it is about, particularly what Mathematics topics are discussed, please help us by commenting on this video with your suggested description and title. Many thanks from,

- The CosmoLearning Team

### Course Index

1. Sequence: Convergence and Divergence (Part 1)
2. Sequences: Converging or Diverging (Part 2)
3. Series: Geometric Series and the Test for Divergence
4. Geometric Series: Fraction Representation
5. Examples of Geometric Series and the Test for Divergence (Part 1)
6. Examples of Geometric Series and the Test for Divergence (Part 2)
7. Telescoping Series
8. Series Diverges: Using Partial Sums
9. Integral Test for Series
10. Using the Integral Test for Series: Example 1
11. Using the Integral Test for Series: Example 2
12. Using the Integral Test for Series: Example 3
13. Series: Limit and Direct Comparison Test
14. Series: Limit and Direct Comparison Test Examples
15. Series: Limit and Direct Comparison Test Examples (cont.)
16. Alternating Series (Part 1)
17. Alternating Series (Part 2)
18. Alternating Series Estimation Theorem
19. Ratio Test to Determine if a Series Converges (Ex. 1)
20. Ratio Test to Determine if a Series Converges (Ex. 2)
21. Root Test for Series
22. Power Series Representation of Functions
23. Power Series: Interval of Convergence
24. Differentiating and Integrating Power Series
25. Power Series: Multiplying and Dividing
26. Taylor and MacLaurin Series (Ex. 1)
27. Taylor and MacLaurin Series (Ex. 2)
28. Taylor's Remainder Theorem or Taylor's Inequality
29. The Binomial Series (Ex. 1)
30. The Binomial Series (Ex. 2)

### Course Description

In this course, Calculus Instructor Patrick gives 30 video lessons on Series and Sequences. Some of the Topics covered are: Convergence and Divergence, Geometric Series, Test for Divergence, Telescoping Series, Integral Test, Limit and Direct Comparison Test, Alternating Series, Alternating Series Estimation Theorem, Ratio Test, Power Series, Taylor and MacLaurin Series, Taylor's Remainder Theorem (Taylor's Inequality), Binomial Series, and many more.

## 5.2 Coded demonstration

This theorem is reasonably intuitive. Suppose that the random variable (X_n) converges in distribution to a standard normal distribution (N(0,1)) . For part 1) of the Theorem, note that when we multiply a standard normal by a constant we “stretch” the distribution (assuming (|a|>1) , else we “compress” it). Recall from the discussion of the standard normal in Chapter 5 that (aN(0,1) = N(0,a^2)) . As (n) approaches infinity, therefore, by definition (A_n xrightarrow

a) , and so the degree to which the standard normal is stretched will converge to that constant too. To demonstrate this feature visually, consider the following simulation:

Here we have defined two random variables: X_n is a standard normal, and A_n converges in value to 2. Varying the value of n , I take (n) draws from a standard normal distribution and calculate the value the converging constant (A_n) . I then generate the product of these two variables. The figure plots the resulting distribution aX . We can see that as n increases, the distribution becomes increasingly normal, remains centred around 0 and the variance approaches 4 (since 95% of the curve is approximately bounded between (0 pm 2 imessqrt = 0 pm 2 imes2 = 0 pm 4) ).

Similarly, if we add the constant (a) to a standard distribution, the effect is to shift the distribution in its entirety (since a constant has no variance, it does not ‘’stretch’’ the distribution). As (A_n) converges in probability, therefore, the shift converges on the constant (a) . Again, we can demonstrate this result in R:

As n becomes larger, the resulting distribution becomes approximately normal, with variance of 1 and a mean value centred around (0 + a = 2) .

Slutsky’s Theorem is so useful precisely because it allows us to combine multiple random variables with known asymptotics, and retain this knowledge i.e. we know what the resultant distribution will converge to assuming (n o infty) .

## Main results

### The mean value theorem

The following result has been proved for example in [34], using Laplace transforms, and also in [26] using only the definition of AB derivatives and integrals.

### Theorem 2.1

AB integrals and derivatives of Caputo type satisfy the following inversion relation:

for (0<alpha<1) , (a< t< b) in (mathbb) , and (f:[a,b] ightarrow mathbb) differentiable such that (f') and (<>^>_<>D_^, f) are both in (L^<1>[a,b]) .

We can use this fact to prove the following analogue of the mean value theorem for fractional derivatives in the AB model.

### Theorem 2.2

Let (0<alpha<1) , (a< b) in (mathbb) , and (f:[a,b] ightarrowmathbb ) differentiable such that (f'in L^<1>[a,b]) and (<>^>_<>D_^, fin C[a,b]) . Then, for any (tin[a,b]) , there exists (xiin[a,t]) such that

### Proof

Now, by the integral mean value theorem, since (<>^>_<>D_^,f(x)) is continuous and ((t-x)^) is integrable and positive, there exists (xiin(a,t)) such that

For interest’s sake we also include the following corollary, another form of the ABC fractional mean value theorem in terms of an inequality.

### Corollary 2.1

With all notations and assumptions as in Theorem 2.2, if f is monotonic (increasing or decreasing), then

### Proof

We shall start from equation (2) to derive this inequality. Firstly, using the integral mean value theorem again, we can write the ABC derivative as

for some (cin(a,t)) , since (E_) is continuous and (f') is integrable and has constant sign. We substitute this into (2) to find

Since the Mittag–Leffler function on a negative argument is completely monotone [35], the result follows. □

### Taylor’s theorem

Before starting to prove analogues of Taylor’s theorem for fractional AB derivatives, we first establish the following lemma.

### Lemma 2.1

If (alphain(0,1)) and (a< b) in (mathbb) and (f:[a,b] ightarrow mathbb) is a differentiable function such that (f') and all functions of the form ((<>^>_<>D_^ )^f(t)) , (minmathbb) , are (L^<1>) functions, then

### Proof

So the left-hand side of equation (4) can be written as follows, where we denote (<>^>_<>I_^) and (<>^>_<>D_^) by simply (I^) and (D^) , respectively, for ease of notation:

where for the last step we used identity (5). Denoting the constant ((D^ )^f(a)) by A, we have

Now we are finally in a position to prove the following main result, our first analogue of Taylor’s theorem for fractional derivatives in the ABC model.

### Theorem 2.3

(AB Taylor series about (t=a) )

If (alphain(0,1)) and (ninmathbb) and (a< b) in (mathbb) and (f:[a,b] ightarrowmathbb) is a differentiable function such that (f') and all functions of the form ((<>^>_<>D_^ )^f(t)) , (minmathbb) , are (L^<1>) functions, then for all (tin[a,b]) ,

for some (xiin(a,t)) , where the function S is defined by

### Proof

The result of Lemma 2.1 can be rewritten as

valid for any (minmathbb) . Summing this identity over m to form a telescoping series, we get

Thus it will suffice to prove that

To establish (8), we use the mean value theorem for integrals once again, this time with one of the ‘functions’ involved being actually a distribution written in terms of the Dirac delta.

In order to get an infinite Taylor series expansion for a given function (f(t)) , it suffices to impose the following convergence condition on the remainder term:

where the norm used is the uniform norm on ([a,t]) .

One disadvantage of Theorem 2.3 is that for many functions f, the ABC fractional derivative (<>^>_<>D^_,f(t)) evaluated at the starting point (t=a) is zero. We can see this by considering the definition: since the ABC derivative is given by an integral from a to t, it will evaluate to zero given certain conditions on the behaviour of (f(t)) near (t=a) . Thus, we present the following generalisation of Theorem 2.3, inspired by the work of [36].

### Theorem 2.4

(AB Taylor series—general case)

If (alphain(0,1)) and (ninmathbb) and (a< b) in (mathbb) and (f:[a,b] ightarrowmathbb) is a differentiable function such that (f') and all functions of the form ((<>^>_<>D_^ )^f(t)) , (minmathbb) , are (L^<1>) functions, then for all (c,tin[a,b]) ,

where the sequence of functions (Delta_) is defined recursively by

and (Delta_=Delta_) , the functions (S_) being defined by (7), and the remainder term (R_) is a linear combination of terms of the form ((<>^>_<>D_^ )^f(xi)) for (xiin(a,b)) .

### Proof

We use formula (6) from Theorem 2.3 as our starting point, and apply it multiple times in different ways to derive (10).

Replacing t by c in equation (6), and replacing f by its ABC derivatives as appropriate, yields the following formulae for any fixed n (where we use the fact that (S_=1) ):

Substituting each of these equations in turn into (6) yields the following sequence of identities:

where the (Delta_) are defined by (11) and the successive remainders are given by

After n iterations of this process, we arrive at the final result:

Since (Delta_=Delta_) by definition, and letting (R_=R_) , we discover equation (10) as required. Note that (xiin(a,t)) and (xi_in(a,c)) for all m. □

Iterated ABC differintegrals to arbitrary order would be very difficult to compute directly. Fortunately, we can use the series formula from [26] to derive a significantly simpler expression for ((<>^>_<>D_^ )^f) as follows:

where this series is locally uniformly convergent in t. Using formula (12) for the iterated ABC derivative makes the Taylor series (6) and (10) easier to compute for specific individual functions f. See the next section for an example.

Unfortunately, given the complexity of the formula for the remainder term (R_) , it will be difficult to tell whether and when series (10) converges as n goes to infinity. But we certainly have a valid finite series result, which can be verified computationally even for large values of n.

## Increasingly Difficult Questions - Pythagoras/Pythagorean Theorem

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## 9.9 Taylor Polynomials

Consider a function y = f ⁢ ( x ) and a point ( c , f ⁢ ( c ) ) . The derivative, f ′ ⁢ ( c ) , gives the instantaneous rate of change of f at x = c . Of all lines that pass through the point ( c , f ⁢ ( c ) ) , the line that best approximates f at this point is the tangent line that is, the line whose slope (rate of change) is f ′ ⁢ ( c ) .

y f ⁢ ( 0 ) = 2 f ′′′ ⁢ ( 0 ) = - 1 f ′ ⁢ ( 0 ) = 1 f ( 4 ) ⁢ ( 0 ) = - 12 f ′′ ⁢ ( 0 ) = 2 f ( 5 ) ⁢ ( 0 ) = - 19 Figure 9.9.1: Plotting y = f ⁢ ( x ) and a table of derivatives of f evaluated at 0.

In Figure 9.9.1 , we see a function y = f ⁢ ( x ) graphed. The table below the graph shows that f ⁢ ( 0 ) = 2 and f ′ ⁢ ( 0 ) = 1 therefore, the tangent line to f at x = 0 is p 1 ⁢ ( x ) = 1 ⁢ ( x - 0 ) + 2 = x + 2 . The tangent line is also given in the figure. Note that “near” x = 0 , p 1 ⁢ ( x ) ≈ f ⁢ ( x ) that is, the tangent line approximates f well.

One shortcoming of this approximation is that the tangent line only matches the slope of f it does not, for instance, match the concavity of f . We can find a polynomial, p 2 ⁢ ( x ) , that does match the concavity without much difficulty, though. The table in Figure 9.9.1 gives the following information:

 f ⁢ ( 0 ) = 2 f ′ ⁢ ( 0 ) = 1 f ′′ ⁢ ( 0 ) = 2 .

Therefore, we want our polynomial p 2 ⁢ ( x ) to have these same properties. That is, we need

 p 2 ⁢ ( 0 ) = 2 p 2 ′ ⁢ ( 0 ) = 1 p 2 ′′ ⁢ ( 0 ) = 2 .

This is simply an initial-value problem. We can solve this using the techniques first described in Section 5.1 . To keep p 2 ⁢ ( x ) as simple as possible, we’ll assume that not only p 2 ′′ ⁢ ( 0 ) = 2 , but that p 2 ′′ ⁢ ( x ) = 2 . That is, the second derivative of p 2 is constant.

y Figure 9.9.2: Plotting f , p 2 , and p 4 . † † margin:

y Figure 9.9.3: Plotting f and p 13 .

If p 2 ′′ ⁢ ( x ) = 2 , then p 2 ′ ⁢ ( x ) = 2 ⁢ x + C for some constant C . Since we have determined that p 2 ′ ⁢ ( 0 ) = 1 , we find that C = 1 and so p 2 ′ ⁢ ( x ) = 2 ⁢ x + 1 . Finally, we can compute p 2 ⁢ ( x ) = x 2 + x + C . Using our initial values, we know p 2 ⁢ ( 0 ) = 2 so C = 2 . We conclude that p 2 ⁢ ( x ) = x 2 + x + 2 . This function is plotted with f in Figure 9.9.2 .

We can repeat this approximation process by creating polynomials of higher degree that match more of the derivatives of f at x = 0 . In general, a polynomial of degree n can be created to match the first n derivatives of f . Figure 9.9.2 also shows p 4 ⁢ ( x ) = - x 4 / 2 - x 3 / 6 + x 2 + x + 2 , whose first four derivatives at 0 match those of f . (Using the table in Figure 9.9.1 , start with p 4 ( 4 ) ⁢ ( x ) = - 12 and solve the related initial-value problem.)

As we use more and more derivatives, our polynomial approximation to f gets better and better. In this example, the interval on which the approximation is “good” gets bigger and bigger. Figure 9.9.3 shows p 13 ⁢ ( x ) we can visually affirm that this polynomial approximates f very well on [ - 2 , 3 ] . The polynomial p 13 ⁢ ( x ) is fairly complicated:

 16901 ⁢ x 13 6227020800 + 13 ⁢ x 12 1209600 - 1321 ⁢ x 11 39916800 - 779 ⁢ x 10 1814400 - 359 ⁢ x 9 362880 + x 8 240 + 139 ⁢ x 7 5040 + 11 ⁢ x 6 360 - 19 ⁢ x 5 120 - x 4 2 - x 3 6 + x 2 + x + 2 .

The polynomials we have created are examples of Taylor polynomials, named after the British mathematician Brook Taylor who made important discoveries about such functions. While we created the above Taylor polynomials by solving initial-value problems, it can be shown that Taylor polynomials follow a general pattern that make their formation much more direct. This is described in the following definition.

###### Definition 9.9.1 Taylor Polynomial, Maclaurin Polynomial

Let f be a function whose first n derivatives exist at x = c .

The Taylor polynomial of degree n of f at x = c is

 p n ⁢ ( x ) = f ⁢ ( c ) + f ′ ⁢ ( c ) ⁢ ( x - c ) + f ′′ ⁢ ( c ) 2 ! ⁢ ( x - c ) 2 + f ′′′ ⁢ ( c ) 3 ! ⁢ ( x - c ) 3 + ⋯ + f ( n ) ⁢ ( c ) n ! ⁢ ( x - c ) n = ∑ k = 0 n f ( k ) ⁢ ( c ) k ! ⁢ ( x - c ) k .

A special case of the Taylor polynomial is the Maclaurin polynomial, where c = 0 . That is, the Maclaurin polynomial of degree n of f is

 p n ⁢ ( x ) = f ⁢ ( 0 ) + f ′ ⁢ ( 0 ) ⁢ x + f ′′ ⁢ ( 0 ) 2 ! ⁢ x 2 + f ′′′ ⁢ ( 0 ) 3 ! ⁢ x 3 + ⋯ + f ( n ) ⁢ ( 0 ) n ! ⁢ x n = ∑ k = 0 n f ( k ) ⁢ ( 0 ) k ! ⁢ x k .

Generally, we order the terms of a polynomial to have decreasing degrees, and that is how we began this section. This definition, and the rest of this chapter, reverses this order to reflect the greater importance of the lower degree terms in the polynomials that we will be finding.

Watch the video:
Taylor Polynomial to Approximate a Function, Ex 3 from https://youtu.be/UINFWG0ErSA

We will practice creating Taylor and Maclaurin polynomials in the following examples.

###### Example 9.9.1 Finding and using Maclaurin polynomials

Find the n th Maclaurin polynomial for f ⁢ ( x ) = e x .

Use p 5 ⁢ ( x ) to approximate the value of e .

We start with creating a table of the derivatives of e x evaluated at x = 0 . In this particular case, this is relatively simple, as shown in Figure 9.9.4 . † † margin: f ⁢ ( x ) = e x ⇒ f ⁢ ( 0 ) = 1 f ′ ⁢ ( x ) = e x ⇒ f ′ ⁢ ( 0 ) = 1 f ′′ ⁢ ( x ) = e x ⇒ f ′′ ⁢ ( 0 ) = 1 ⋮ ⋮ f ( n ) ⁢ ( x ) = e x ⇒ f ( n ) ⁢ ( 0 ) = 1 Figure 9.9.4: The derivatives of f ⁢ ( x ) = e x evaluated at x = 0 . By the definition of the Maclaurin polynomial, we have

 p n ⁢ ( x ) = ∑ k = 0 n f ( k ) ⁢ ( 0 ) k ! ⁢ x k = ∑ k = 0 n 1 k ! ⁢ x k .

Using our answer from part 1, we have

 p 5 ⁢ ( x ) = 1 + x + 1 2 ⁢ x 2 + 1 6 ⁢ x 3 + 1 24 ⁢ x 4 + 1 120 ⁢ x 5 .

To approximate the value of e , note that e = e 1 = f ⁢ ( 1 ) ≈ p 5 ⁢ ( 1 ) . It is very straightforward to evaluate p 5 ⁢ ( 1 ) : † † margin:

y Figure 9.9.5: A plot of f ⁢ ( x ) = e x and its 5 th degree Maclaurin polynomial p 5 ⁢ ( x ) .

 p 5 ⁢ ( 1 ) = 1 + 1 + 1 2 + 1 6 + 1 24 + 1 120 = 163 60 ≈ 2.71667 .

This is an error of about 0.0016 , or 0.06 % of the true value.

A plot of f ⁢ ( x ) = e x and p 5 ⁢ ( x ) is given in Figure 9.9.5 .

###### Example 9.9.2 Finding and using Taylor polynomials

Find the n th Taylor polynomial of y = ln ⁡ x at x = 1 .

Use p 6 ⁢ ( x ) to approximate the value of ln ⁡ 1.5 .

Use p 6 ⁢ ( x ) to approximate the value of ln ⁡ 2 .

We begin by creating a table of derivatives of ln ⁡ x evaluated at x = 1 . While this is not as straightforward as it was in the previous example, a pattern does emerge, as shown in Figure 9.9.6 .

 p n ⁢ ( x ) = ∑ k = 0 n f ( k ) ⁢ ( c ) k ! ⁢ ( x - c ) k = ∑ k = 1 n ( - 1 ) k + 1 k ⁢ ( x - 1 ) k .

We can compute p 6 ⁢ ( x ) using our work above: † † margin: f ⁢ ( x ) = ln ⁡ x ⇒ f ⁢ ( 1 ) = 0 f ′ ⁢ ( x ) = 1 / x ⇒ f ′ ⁢ ( 1 ) = 1 f ′′ ⁢ ( x ) = - 1 / x 2 ⇒ f ′′ ⁢ ( 1 ) = - 1 f ′′′ ⁢ ( x ) = 2 / x 3 ⇒ f ′′′ ⁢ ( 1 ) = 2 f ( 4 ) ⁢ ( x ) = - 6 / x 4 ⇒ f ( 4 ) ⁢ ( 1 ) = - 6 ⋮ ⋮ f ( n ) ⁢ ( x ) = ⇒ f ( n ) ⁢ ( 1 ) = ( - 1 ) n + 1 ⁢ ( n - 1 ) ! x n ( - 1 ) n + 1 ⁢ ( n - 1 ) ! Figure 9.9.6: Derivatives of ln ⁡ x evaluated at x = 1 .

 p 6 ⁢ ( x ) = ( x - 1 ) - 1 2 ⁢ ( x - 1 ) 2 + 1 3 ⁢ ( x - 1 ) 3 - 1 4 ⁢ ( x - 1 ) 4 + 1 5 ⁢ ( x - 1 ) 5 - 1 6 ⁢ ( x - 1 ) 6 .

Since p 6 ⁢ ( x ) approximates ln ⁡ x well near x = 1 , we approximate ln ⁡ 1.5 ≈ p 6 ⁢ ( 1.5 ) : † † margin:

y Figure 9.9.7: A plot of y = ln ⁡ x and its 6 th degree Taylor polynomial at x = 1 .

 p 6 ⁢ ( 1.5 ) = ( 1.5 - 1 ) - 1 2 ⁢ ( 1.5 - 1 ) 2 + 1 3 ⁢ ( 1.5 - 1 ) 3 - 1 4 ⁢ ( 1.5 - 1 ) 4 + 1 5 ⁢ ( 1.5 - 1 ) 5 - 1 6 ⁢ ( 1.5 - 1 ) 6 = 259 640 ≈ 0.404688 .

This is a good approximation as a calculator shows that ln ⁡ 1.5 ≈ 0.4055 . Figure 9.9.7 plots y = ln ⁡ x with y = p 6 ⁢ ( x ) . We can see that ln ⁡ 1.5 ≈ p 6 ⁢ ( 1.5 ) .

We approximate ln ⁡ 2 with p 6 ⁢ ( 2 ) :

 p 6 ⁢ ( 2 ) = ( 2 - 1 ) - 1 2 ⁢ ( 2 - 1 ) 2 + 1 3 ⁢ ( 2 - 1 ) 3 - 1 4 ⁢ ( 2 - 1 ) 4 + 1 5 ⁢ ( 2 - 1 ) 5 - 1 6 ⁢ ( 2 - 1 ) 6 = 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 = 37 60 ≈ 0.616667 .

This approximation is not terribly impressive: a hand held calculator shows that ln ⁡ 2 ≈ 0.693147 . The graph in Figure 9.9.7 shows that p 6 ⁢ ( x ) provides less accurate approximations of ln ⁡ x as x gets close to 0 or 2.

y Figure 9.9.8: A plot of y = ln ⁡ x and its 20 th degree Taylor polynomial at x = 1 .

Surprisingly enough, even the 20 th degree Taylor polynomial fails to approximate ln ⁡ x for x > 2 , as shown in Figure 9.9.8 . We’ll soon discuss why this is.

Taylor polynomials are used to approximate functions f ⁢ ( x ) in mainly two situations:

When f ⁢ ( x ) is known, but perhaps “hard” to compute directly. For instance, we can define y = cos ⁡ x as either the ratio of sides of a right triangle (“adjacent over hypotenuse”) or with the unit circle. However, neither of these provides a convenient way of computing cos ⁡ 2 . A Taylor polynomial of sufficiently high degree can provide a reasonable method of computing such values using only operations usually hard-wired into a computer ( + , - , × and ÷ ).

When f ⁢ ( x ) is not known, but information about its derivatives is known. This occurs more often than one might think, especially in the study of differential equations.

In both situations, a critical piece of information to have is “How good is my approximation?” If we use a Taylor polynomial to compute cos ⁡ 2 , how do we know how accurate the approximation is?

We had the same problem when studying Numerical Integration. Theorem 8.7.1 provided bounds on the error when using, say, Simpson’s Rule to approximate a definite integral. These bounds allowed us to determine that, for example, using 10 subintervals provided an approximation within ± .01 of the exact value. The following theorem gives similar bounds for Taylor (and hence Maclaurin) polynomials.

###### Theorem 9.9.1 Taylor’s Theorem

Let f be a function whose ( n + 1 ) th derivative exists on an open interval I and let c be in I . Then, for each x in I , there exists z x between x and c such that

 f ⁢ ( x ) = ∑ k = 0 n f ( k ) ⁢ ( c ) k ! ⁢ ( x - c ) k + R n ⁢ ( x ) ,

where R n ⁢ ( x ) = f ( n + 1 ) ⁢ ( z x ) ( n + 1 ) ! ⁢ ( x - c ) n + 1 .

| R n ⁢ ( x ) | ≤ max z ⁡ | f ( n + 1 ) ⁢ ( z ) | ( n + 1 ) ! ⁢ | x - c | n + 1 , where z is between x and c .

The first part of Taylor’s Theorem states that f ⁢ ( x ) = p n ⁢ ( x ) + R n ⁢ ( x ) , where p n ⁢ ( x ) is the n th order Taylor polynomial and R n ⁢ ( x ) is the remainder, or error, in the Taylor approximation. The second part gives bounds on how big that error can be. If the ( n + 1 ) th derivative is large, the error may be large if x is far from c , the error may also be large. However, the ( n + 1 ) ! term in the denominator tends to ensure that the error gets smaller as n increases.

The following example computes error estimates for the approximations of ln ⁡ 1.5 and ln ⁡ 2 made in Example 9.9.2 .

###### Example 9.9.3 Finding error bounds of a Taylor polynomial

Use Theorem 9.9.1 to find error bounds when approximating ln ⁡ 1.5 and ln ⁡ 2 with p 6 ⁢ ( x ) , the Taylor polynomial of degree 6 of f ⁢ ( x ) = ln ⁡ x at x = 1 , as calculated in Example 9.9.2 .

We start with the approximation of ln ⁡ 1.5 with p 6 ⁢ ( 1.5 ) . The theorem references max ⁡ | f ( n + 1 ) ⁢ ( z ) | . In our situation, this is asking “How big can the 7 th derivative of y = ln ⁡ x be on the interval [ 1 , 1.5 ] ?” The seventh derivative is y = - 6 ! / x 7 . The largest absolute value it attains on I is 720. Thus we can bound the error as:

 | R 6 ⁢ ( 1.5 ) | ≤ max ⁡ | f ( 7 ) ⁢ ( z ) | 7 ! ⁢ | 1.5 - 1 | 7 ≤ 720 5040 ⋅ 1 2 7 ≈ 0.001 .

We computed p 6 ⁢ ( 1.5 ) = 0.404688 using a calculator, we find ln ⁡ 1.5 ≈ 0.405465 , so the actual error is about 0.000778 (or 0.2 % ), which is less than our bound of 0.001 . This affirms Taylor’s Theorem the theorem states that our approximation would be within about one thousandth of the actual value, whereas the approximation was actually closer.

The maximum value of the seventh derivative of f on [ 1 , 2 ] again 720 (as the largest values come at x = 1 ). Thus

 | R 6 ⁢ ( 2 ) | ≤ max ⁡ | f ( 7 ) ⁢ ( z ) | 7 ! ⁢ | 2 - 1 | 7 ≤ 720 5040 ⋅ 1 7 ≈ 0.28 .

This bound is not as nearly as good as before. Using the degree 6 Taylor polynomial at x = 1 will bring us within 0.3 of the correct answer. As p 6 ⁢ ( 2 ) ≈ 0.61667 , our error estimate guarantees that the actual value of ln ⁡ 2 is somewhere between 0.33667 and 0.89667 . These bounds are not particularly useful.

In reality, our approximation was only off by about 0.07 (or 11 % ). However, we are approximating ostensibly because we do not know the real answer. In order to be assured that we have a good approximation, we would have to resort to using a polynomial of higher degree.

We practice again. This time, we use Taylor’s theorem to find n that guarantees our approximation is within a certain amount.

###### Example 9.9.4 Finding sufficiently accurate Taylor polynomials

Find n such that the n th Taylor polynomial of f ⁢ ( x ) = cos ⁡ x at x = 0 approximates cos ⁡ 2 to within 0.001 of the actual answer. What is p n ⁢ ( 2 ) ?

Solution Following Taylor’s theorem, we need bounds on the size of the derivatives of f ⁢ ( x ) = cos ⁡ x . In the case of this trigonometric function, this is easy. All derivatives of cosine are ± sin ⁡ x or ± cos ⁡ x . In all cases, these functions are never greater than 1 in absolute value. We want the error to be less than 0.001 . To find the appropriate n , consider the following inequalities:

 max ⁡ | f ( n + 1 ) ⁢ ( z ) | ( n + 1 ) ! ⁢ | 2 - 0 | n + 1 ≤ 0.001 1 ( n + 1 ) ! ⋅ 2 n + 1 ≤ 0.001

We find an n that satisfies this last inequality with trial-and-error. When n = 8 , we have 2 8 + 1 ( 8 + 1 ) ! ≈ 0.0014 when n = 9 , we have 2 9 + 1 ( 9 + 1 ) ! ≈ 0.000282 < 0.001 . Thus we want to approximate cos ⁡ 2 with p 9 ⁢ ( 2 ) .

We now set out to compute p 9 ⁢ ( x ) . We again need a table of the derivatives of f ⁢ ( x ) = cos ⁡ x evaluated at x = 0 . A table of these values is given in Figure 9.9.9 . Notice how the derivatives, evaluated at x = 0 , follow a certain pattern. All the odd powers of x in the Taylor polynomial will disappear as their coefficient is 0. While our error bounds state that we need p 9 ⁢ ( x ) , our work shows that this will be the same as p 8 ⁢ ( x ) . † † margin: f ⁢ ( x ) = cos ⁡ x ⇒ f ⁢ ( 0 ) = 1 f ′ ⁢ ( x ) = - sin ⁡ x ⇒ f ′ ⁢ ( 0 ) = 0 f ′′ ⁢ ( x ) = - cos ⁡ x ⇒ f ′′ ⁢ ( 0 ) = - 1 f ′′′ ⁢ ( x ) = sin ⁡ x ⇒ f ′′′ ⁢ ( 0 ) = 0 f ( 4 ) ⁢ ( x ) = cos ⁡ x ⇒ f ( 4 ) ⁢ ( 0 ) = 1 f ( 5 ) ⁢ ( x ) = - sin ⁡ x ⇒ f ( 5 ) ⁢ ( 0 ) = 0 f ( 6 ) ⁢ ( x ) = - cos ⁡ x ⇒ f ( 6 ) ⁢ ( 0 ) = - 1 f ( 7 ) ⁢ ( x ) = sin ⁡ x ⇒ f ( 7 ) ⁢ ( 0 ) = 0 f ( 8 ) ⁢ ( x ) = cos ⁡ x ⇒ f ( 8 ) ⁢ ( 0 ) = 1 f ( 9 ) ⁢ ( x ) = - sin ⁡ x ⇒ f ( 9 ) ⁢ ( 0 ) = 0 Figure 9.9.9: A table of the derivatives of f ⁢ ( x ) = cos ⁡ x evaluated at x = 0 .

Since we are forming our polynomial at x = 0 , we are creating a Maclaurin polynomial, and: