calc2 study guide (section 2)

8/11/2019 Calc2 Study Guide (Section 2)

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Calculus 2Midterm 2Study Guide

By Christopher M. Brown

(Good Luck Guys!)

Section 8.1Sequences

The idea behind a sequence is that you have a domain of positive integers:

{1, 2, 3, 4}

And the sequence is a function that represents the variance or pattern of difference

between each index in this domain. For the above, simple sequence, an = n.

But if we consider another domain:

{2, , , ,an}From here we can find the function to be an=

+ .

Now we introduce the concept of convergence and divergence.

The idea for converge or divergence has to do with taking the limit of the sequence. Lets

take a look at our above function an=+ .

The definition of a Limit of a Sequence is:

Lim an= LIf we find that Lim anexists, meaning its not infinity, we say it converges. Otherwise,we say that the sequence is divergent.

So for an=+ we take the limit as Lim + = . We get thisfrom observing that the

numerator will always be significantly bigger than the denominator. As we have n1/2on

the bottom and n1on the top. Take into consideration the BOB0, BOTNO, BOSCO rule.

BOB0 (Bigger On Bottom Zero):If the highest exponent of the denominator is larger than the highest

exponent of the numerator, then the limit approaches 0.

BOTNO (Bigger On Top None):


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If the highest exponent of the numerator is larger than the highest

exponent of the denominator, then the limit does not exist meaning it

approaches .

BOSCO (Both Same Coefficients):

If the highest exponent of the numerator is the same as the largest

exponent of the denominator, then the limit approaches the coefficients of

the variables with the largest in the numerator and denominator. Note the

coefficients will still be in fraction form.

This summarizes the general idea behind sequences.

Now Lets take a look at the arithmetictheorems associated with sequences.

Theorem 1: As long as n is some positive integer, and anis some function.

Lim an= L. Move on. Theorem 2: Suppose Lim an= L and Lim n= M

1. Lim can= cL. Just shows that constants in front of the function can betaken out in front of the limit equation and multiplied to the end limit.

2. Lim

an bn) = L M. Shows if you have two functions being addedtogether, you can break them up into their own limits.

3. Lim anbn= LM. Shows if you have two functions being multiplied,then you can multiply their limits as well.

4. Lim an / bn= . Same as above but with division.5. Liman)p= Lp, if p > 0 and an> 0. Exponents can also be distributed to

the limit later.

Theorem 3: Squeeze theorem can sometimes help in determining if a more

complicated sequence converges or not by taking a sequence that is larger than an

at all points and a sequence that is smaller than it at all points, but approaches the

same limit as the larger one. Such that

an bn cnand Lim n= Lim cn= L, then Lim bn= L.


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Theorem 4: Has to do with absolute convergence.

If Lim |an|= 0, then Lim an= 0.Theorem 5: Has to do with the limit of a function within a function, showing that

you can pull limits inside the function.

If Lim an= L and the function f is continuous at L, thenLim fan)= f(Lim an) = f(L).

Another way to prove if a sequence is convergent is to see if it is both bounded and

monotonic. Being monotonic purely means that if the values of the domain of a sequence

are changing, that is to say they are decreasing or increasing, then it is monotonic.

a1< a2< a3< < an < an+1< or a1> a2> a3> > an > an+1>

Being bounded is to say that there exists a value M such that all values of the domain of

the sequence are less than or greater than M for all n 1.

Theorem 6: Every bounded, monotonic sequence is convergent.

Now lets show how we can apply these ideas to solving problems:

1.

Evaluate the limit: Lim

2. Evaluate the limit: Lim

.

3. Determine whether the sequence converges or diverges. If it converges, find its limit.

an=+



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an = tan


an =

+

6. Determine whether the sequence is monotonic and/or bounded.

an=3

7.

Determine whether the sequence is monotonic and/or bounded.

an =+

+

8. If a principal of P dollars is invested in an account earning interest at the rate of r per

year compounded monthly, then the accumulated amount A, at the end of n months is

An = P(1 +

)n

a. Write the first six terms of the sequence {An} if P = 10,000 and r = 0.105.

Interpret your results.

b. Does the sequence {An} converge or diverge?

9. An annuity is a sequence of payments made at regular intervals. Suppose that a sum

of $200 is deposited at the end of each month into an account earning interest at the

rate of 12% per year compounded monthly. Then the amount of deposit (called the

future value of annuity) at the end of the nth month is f(n) = 20,000[(1.01)n1].

Consider the sequence {an} defined by an = f(n).

a. Find the 24thterm of the sequence {an}, and interpret your result.

b. Evaluate Lim an, and interpret your result.

10.Use the Squeeze Theorem for Sequences to prove that Lim = 1 for a > 0.


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If you need further help with sequences look to these videos, great guides:

Sequence BasicsConvergence/Divergence with Sequences

Intro to Arithmetic Sequences

A Guide to Finding the n-th Term of a Arithmetic Sequence

Intro to Geometric Sequences

A Guide to Finding the n-th Term of a Geometric Sequence

Recursive Sequences

Monotonic and Bounded Sequences

The Squeeze Theorem and Absolute Value Theorem


Section 8.2Series

A series is most simply defined as summation of all of the terms in the domain of a

sequence. Or rather, an infinite sum/infinite series/series. Just like with sequences, theidea of the series is to get some function to represent the outcome, which for series would

be to represent the sum of the domain for the function an.

Given an infinite series:

= n= a1+ a2+ a3+ + an+ The nth partial sum of the series is

Sn = = k = a1+ a2+ a3+ + anIf the sequence of partial sums {Sn} converges to the number S, that is, if Lim Sn= S,then the series nconverges and has sum S, written

= n =a1+ a2+ a3+ + an + = SIf {Sn} diverges, then n diverges.Now lets cover the established types of series wewill see.
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In some cases you will be given a series that consists of unique values with varying signs,

and coincidentally, some of these values will cancel out with the values of the term

before it or after it. This is known as a telescoping series.

For example, look at the series

= .

an = = += +When we begin plugging in values for n, we notice that some of the values begin

to cancel out.

Sn = = = += =(

+= 2 - +

As we can see the values in red cancel out with values in the terms before and

after them.

From this we take the limit of the result, Lim Sn= Lim2 += 2, and weconclude from this that the given series is convergent and has sum 2; meaning

that,

= 2

Another type of series you will encounter often is the geometric series, which takes the

form of:

0

=

Where a represents the first value of the series, and r represents the common ratio

between the values.

Luckily for us, the geometric series has a nice, clean summation theorem for us to

follow.

Theorem 1: The Geometric Series

If |r| < 1, then the geometric series converges, and its sum is equal to the

value

If |r| 1, the geometric series diverges entirely.


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The next series is the harmonic serieswhich is relatively simple, in that it is just known

to be divergent.

1 = Theres no real reason to concern yourself with why this series is divergent, but if

you are, you can find the answer on page 747 of the textbook.

Now we will introduce the first of many tests that use when working with series to

determine convergence.

Theorem 2:

If the infinite sum = n converges, then lim n = 0.(Proof found on pg. 748) ((Note this is not the Divergence Test))

Theorem 3: The Divergence Test

From theorem 2, we get the Divergence Test which states that

if lim n does not exist or lim n 0, then = n diverges.(Proof found on pg. 748)

Theorem 4: Properties of Convergent Series

If = n = A and = n = B are convergent and c is any real number,then = n and = n bn) are also convergent, anda. = n = c = n = cAb. = n bn = = n = n = A B

The Divergence Test should be the first test you consider when determining

whether a series is convergent, as it is the easiest to perform.

So once again, lets practice these ideas with problems:

1. If the series

+

=

converges, find its sum.

2. Determine whether the series 2 5+= ) converges or diverges. If itconverges, find its sum.


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3. Determine if the series 8= converges or diverges. If it converges, find its sum.

4.

Determine whether the given series converges or diverges:

+

= . If itconverges find its sum.

5. Determine whether the given series converges or diverges:

[cos cos +]= . If it converges, find its sum.

6.

Determine whether the given series converges or diverges: ln += . If itconverges, find its sum.

7. Determine whether the given series converges or diverges: [

+= ].If it converges, find its sum.

8. Sum of Areas of Nested Triangles and Circles. An infinite sequence of nested

equilateral triangles and circles is constructed as follows: Beginning with an

equilateral triangle, followed by a circle, and so on, ad infinitum. Find the total area

of shaded regions.

9. Prove that += converges by showing that {Sn} is increasing and boundedabove, where Sn is the nth partial sum of the series.


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10.Prove or disprove: If nand n are both divergent, then n + bn) is divergent.

More instructional videos on series by PatrickJMT:

What is a Series

Telescoping Series

Geometric Series and the Test for Divergence

Test for Divergence for Series

Calculus 2

Midterm 2

Study Guide

Section 8.3The Integral Test

The integral test is simply a method to prove the convergence or divergence of an infinite

series by re-writing = n to where f(n) = an.Theorem 1: The Integral Test

Suppose that f is a continuous, positive, and decreasing function on [1, ).

If f(n) = anfor n 1, then

= n and Either both converge or both diver.

(Proof of this theorem can be found on page 753 of the textbook.)

Next we look at the p-Series. Which takes on a rather simple form, which happens to

include the previously seen harmonic series.

Ap-Series is identified by the form: = . For the harmonic series, p = 1.The theorem for the convergence of thep-Series is based off of the number 1.

Theorem 2: Convergence of thep-Series

If the value of p > 1 the p-Series = converges, if p 1.(Proof of this theorem can be found on page 755 of the textbook.)
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8. Use the Integral Test to show that []= converges if p > 1 anddiverges if p 1.

9. Find the value(s) of a for which the series [ + +]= converges. Justify youranswer.

10.Consider the series a. Evaluate , and deduce from the Integral Test that the given series is

convergent.

b. Show that the given series is a geometric series, and find its sum.

c.

Conclude that although the convergence of dx implies convergence ofthe infinite series, its value does not given the sum of the infinite series.Additional videos for further explanations:

Basics of Integral Test

Using the Integral Test for Series

P-series


Section 8.4The Comparison Test

The comparison test is used when a series proves to be too complex to solve with other

tests. By taking a series that is similar to the complex series, but is easier to test for

converge using one of the tests, we can determine based on its relationship to the

complex series.

Theorem 1: The Comparison Test

Suppose that n and n are series with positive terms.a.

If n is convergent and an bnfor all n, then nis also convergent.b. If n is divergent and an bnfor all n, then nis also divergent.(Proof can be found on page 759 of the textbook.)

In cases where the series found to be similar to the complex series, but does not satisfy

either of the arguments of the comparison test, we then look to the limit comparison test.
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Theorem 2: The Limit Comparison Test

Suppose that nand n are series with positive terms and

lim n/bn = LWhere L is a positive number. Then either both series converge or both diverge.

(Proof can be found on page 761 of the textbook.)

This basically means that if you have two series, and the limit of their quotient is

greater than 0, they both converge or both diverge. So if you use a test on either

one of them to prove if one of them converges, you have proved that the other one

converges as well. Same for divergence. So you can take a function similar to the

one youre working on, but is easier to prove is convergent, and use that in

relation to the original function to test for its convergence.

Problem time:

1. Determine whether the series is convergent or divergent. +++=

2. Determine whether the series is convergent or divergent. ++= .

3.

Which of the following is a valid argument for the convergence or divergence of the

series = ?(a)

> for n 3 so = diverges.

(b)

< for n 3 so = converges.(c)

> for n 3 so = converges.

(d) lim 0so = converges.(e)

0

satisfies the conditions

1. An+1 anfor all n (The series is decreasing.)

2. lim n = 0 (The limit of the series excluding the alternation is zero.)then the series converges.


Theorem 2: Error Estimate in Approximating an Alternating Series

Suppose

1

=

n is an alternating series satisfying

1. 0 an+1 anfor all n

(Every next term is less than the previous, but still greater than or

equal to zero.)

2. lim n = 0(The limit of the series without the alternating sequence is equal to

zero.)

If S is the sum of the series, then

|Rn| = |SSn| an+1

(Rnis the representation of the remainder after n terms of the

series. So when you take the sum, youre getting as close as

possible to the sum, S, by taking to n terms to get Sn, and Rnis the

difference between the actual sum and the sum you got from taking

it to n terms.


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In other words, the absolute value of the error incurred in approximating S

by Snis no larger than an+1, the first term omitted.


I feel like this one wasnt covered much in class, so lets go over quick

example.

Lets look at the series:

1!

=

So we test for convergence.

an+1= +! !+< ! nfor all n and

lim n= lim ! 0Which satisfies the Alternating Series Test, so we can conclude that it

converges.

To see how many terms of the series are needed to ensure the specified

accuracy of the approximation, we turn to Theorem 2. It tells us that

|Rn| = |SSn| an+1 =

+!

We require that |Rn| < 0.0005 or (n+1)! >

.= 2000(This is just a general assumption of required decimals to make the

accuracy for this theorem.)

So we begin plugging in values of n until we get a value that is less than

our required value of |Rn|, exclude that value and all of the ones after it.

S S5 = ! ! ! ! ! !(exclude !and all values beyond.)=1 (The next value is which is lessthan our required value, so we it does not make it into our

approximated sum.)

0.368


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For more explanations, check out these video tutorials:

Alternating SeriesAlternating Series Examples

Alternating Series - Error Estimation

Alternating Series - Error Estimation #2

Now that we understand that, lets work on more problems.

1. Determine the number of terms sufficient to obtain the sum of the

series accurate to three decimal places.

2+ 1!

=

2. Determine whether or not the series converges or diverges.

1 ln

=

3. Which of the following alternating series is convergent?

(a) +=

(b)

= (c) = (d) = (e) =

4. Determine whether the series converges or diverges.

1+

3

=

5. Determine whether the series converges or diverges.

16 ln

=
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6. Find an approximation of the sum of the series accurate to three

decimal places.

1 2

=

7. Determine the smallest number of terms to obtain the sum of the series

accurate to four decimal places.

2+ 1!

=

8.

(a) Show that

++= converges.(b) Find the sum of the series of part (a).

True or False:

9. If the alternating series 1= n, where an> 0, is divergent,then the series = n is also divergent.

10.

If the alternating series 1= n, where an> 0, converges, thenboth the series 1= 2n-1 and 1= 2nconverge.


Section 8.6Absolute Convergence; the Ratio and Root Tests

This section introduces the cases where a series cannot be defined as either

positive or alternating. Such an example would be the series = , wherethe first term is positive, but the following two terms are negative.

So by taking the absolute values of the series, we can then test it as a purelypositive series, and by confirming it as convergent in its absolute value form, we

confirm that the original series is absolutely convergent.

A series nis absolutely convergent if the series |n| is convergent.Sometimes we have the opposite case, where a series contains positive and

negative terms, and is convergent; but the absolute value of the series is divergent.


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If a series n is conditionally convergent if it is convergent but not absolutelyconvergent.

Theorem 1:

If a series

n is absolutely convergent, then it is convergent.

(Proof is on page 772 of the textbook.)

The ratio test is a test used only to test if an absolute series is convergent. The

idea is to take one term as the denominator, and the term after as the numerator,

and then take them as absolute. If their limit is less than 1, it converges. If their

limit is greater than 1, it diverges. If their limit is equal to 1, the test is

inconclusive.

Theorem 2: The Ratio Test

Let

nbe a series with nonzero terms.

(a) If lim = L < 1, then = n converges absolutely.(b)If lim = L > 1, or lim = , then = n diverges.(c) If lim = L = 1, test is inconclusive, and another test

should be used.

(Proof is on page 773 of the textbook.)

The next test, the Root test, comes in use when you have a series with the power

n. It works by removing the n power, and leaving just the absolute series without

the n power. Then the limit is taken to cover similar results to the ratio test above.

Theorem 3: The Root Test

Let = nbe a series.a. If lim|| = L < 1, then = n converges absolutely.b. If lim|| = L > 1 or lim|| = , then = ndiverges.

c. If lim|| = 1, the test is inconclusive, and another testshould be used.Guess what. Thats the last test.


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Now lets do some problems.

Determine whether these series are convergent, absolutely convergent,conditionally convergent, or divergent.

1. += 2. = 3. != 4. 1sin= 5. !!=

6. The series = is(a) absolutely convergent by the root test

(b) divergent by the ratio test

(c) divergent by the divergence test

(d) conditionally convergent by the alternating series test

(e) divergent by the integral test

7. Determine whether the series ( 1)= converges or diverges. Clearlyidentify any test(s) you are using and label all relevant data and/or properties.

8. Determine whether the series +

= converges absolutely, convergesconditionally, or diverges. Clearly identify any test(s) you are using and label all

relevant data and/or properties.

9. For the series = , which of the following statements are true?10. Determine whether the series is absolutely convergent, conditionally

convergent, or divergent.

1ln5

=


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For some additional help on this section, look to these videos:

Ratio Test

Ratio Test 2Ratio Test 3

Root Test

Root Test Example


Section 8.7Power Series

The power seriesis the introduction of series with non-constant terms due to containingpolynomials.

Power series are shown in form as:

=

Where ans are constants and are considered to be the coefficients of the series.

We can view a power series as a functionf defined by the rule

=

The domain of f is the set of all x values for which the power series converge, and the

rangeis the sums of the series obtained.

The value c is considered to be the center of the power series, giving us our first theorem.

Theorem 1: Convergence of a Power Series

Given a power series = , exactly one of the following is true:a. The series converges only at x = c.

b. The series converges for all x.

c. There is a positive number R such that the series converges for

|xc| < R and diverges for |xc| > R.
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This value R is known as the radius of convergence of the power series. For case

(a) of the theorem R = 0, and for case (b) R = .

The set for all the values of x for which the power series converges is called the

interval of convergenceof the power series. For case (a) the interval is the single

point c, for (b) the interval is infinite: (-,), for (c) the interval is (cR, c + R).Case (c) has a special problem in that it does not tell us whether or not the

endpoints x = cR and x = c + R are included in the interval of convergence. To

find out, plug these values into the power series and use a convergence test on the

resultant series.

When solving for R, use the Ratio Test to find which values of x satisfy the

requirements to converge by the Ratio Test. Sometimes youll use the Root Test.

Next we take a look at differentiating and integrating the functionf, wheref is a power

series. The way we do this is to do so term by term.

Theorem 2: Differentiation and Integration of Power Series

Take = with a radius of convergence R > 0. = = a0+ a1(xc) + a2(xc)2a3(xc)3+

With an interval of convergence (cR, c + R). The function is found to be

differentiable and integrable on all points. The derivative and indefinite integral of

fare written as so:

a. 2 3 = b. += = C

Now lets do some problems:

Find the radius of convergence and the interval of convergence of the following power

series:

1. + +=

2.

+

= 3. =

4. Find the radius of convergence of the series = .5. Let = . Find. What are the intervals of convergence off and?


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6. Find the interval of convergence of the series +!= . Clearly identify any test(s)you are using and label all relevant data and/or properties.

7. Find the radius of convergence and the interval of convergence of the power series:

! +2!= 8. Find the radius of convergence and the interval of convergence of the power series:

1! 5

=

9. If the radius of convergence of the power series is R, what is the radius ofconvergence of the power series ?maximum distance up the incline given by

10. Find the radius of convergence and interval of convergence of the power series:

1 7ln

=

For additional help on this topic, check out these videos:

Power Series - Finding the Interval of Convergence

Radius of Convergence

Differentiating and Integrating Power Series

Integrating a Power Series

Power Series Involving Natural Log


Section 8.8Taylor and Maclaurin Series

The idea behind this section is to find power series that will represent the function we are

looking at, we do this by using these two series.

Starting with the Taylor Series:

For this series, we are given a functionffrom which we know can be represented

by a power series that is centered at c.

The proof done for this series (an=

! ) is that iffhas a power seriesrepresentation, then the series must have the form given in the following theorem.
https://www.youtube.com/watch?v=01LzAU__J-0https://www.youtube.com/watch?v=01LzAU__J-0https://www.youtube.com/watch?v=MM3BXtVu9eMhttps://www.youtube.com/watch?v=MM3BXtVu9eMhttps://www.youtube.com/watch?v=qPl9nr8my2Qhttps://www.youtube.com/watch?v=qPl9nr8my2Qhttps://www.youtube.com/watch?v=q12rXAMfEXwhttps://www.youtube.com/watch?v=q12rXAMfEXwhttps://www.youtube.com/watch?v=r4K4GOAmKa8https://www.youtube.com/watch?v=r4K4GOAmKa8https://www.youtube.com/watch?v=r4K4GOAmKa8https://www.youtube.com/watch?v=q12rXAMfEXwhttps://www.youtube.com/watch?v=qPl9nr8my2Qhttps://www.youtube.com/watch?v=MM3BXtVu9eMhttps://www.youtube.com/watch?v=01LzAU__J-0


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Theorem 1: Taylor Series offat c.

Iffhas a power series representation at c, that is, if

= |xc| < RThenf

(n)

(c) exists for every positive integer nand

! Thus,

! = The Taylor Series is acquired from a function by taking several derivatives

of the function, and creating a series that will represent the pattern of all these

derivatives to the nth derivative.

The Maclaurin seriescomes from the special case of the Taylor series where c=0

and the series becomes

0!

=

There is also a special case for a function in which we get the Binomial Series,

this comes from getting a function in the form:

1 When k is any real number and |x| < 1. The is written as:

=

It is known that if k is positive, then the series converges for all x. However if it is

not positive, then you must use the Ratio Test to find the interval of convergence.

Here are a few common functions and their power series representations from the

textbook you might want to remember:

Maclaurin Series Interval of Convergence1.

1 = (-1, 1)

2. 1 !

!

!

!= (-, )3.sin !

!

!

+! 1=

+! (-, )

4. cos 1 !

!

!

! 1

!= (-, )


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5. ln1

1

= (-1, 1]6. sin

!

!+ !

!+= [-1, 1]

7. tan

+ 1

+= [-1, 1]8. 1 ()= 1 ! ! (-1, 1)

Problem time..

1. Find the Taylor Series offat the given value of c. Then find the radius of

convergence of the series.

11 2 , 12. Find the Taylor Series offat the given value of c. Then find the radius of


, 03. Find the Taylor Series offat the given value of c. Then find the radius of


ln1 , 04. Find the Taylor Series offat the given value of c. Then find the radius of


cos

, 0

5. Use the binomial series to find the power series representation of the function.

1 /6. Use the definition of a Taylor Series offat c(Theorem 1) to find the Taylor

series of 3. Then find the interval of convergence for theseries.

7. Use the definition of the Taylor Series to find the Taylor series of 3.8. Find the Maclaurin series for the function .9. Find the sum of the series != .10. Letfbe the function defined by

{ 00 0


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Show thatf cannot be represented by a Maclaurin series.

If you need additional instruction, be sure to check out these videos:

Taylor and Maclaurin Series - Example 1

Taylor and Maclaurin Series - Example 2Taylor / Maclaurin Series for Sin(x)

Maclaurin/Taylor Series in Approximating a Definite Integral

The Binomial Series


Section 8.9

Approximation by Taylor Polynomials

The difference between the power series and polynomials is that polynomials have a

finite number of terms, which is what we use to take the approximated sum of the power

series.

This section has to do with partial sums of the Taylor Series to approximate the real sum

of the series. The equation form looks like this:

Pn !=

! ! ! Which is called by the book the nth-degree Taylor polynomial of f at c. If c = 0, we

have the nth-degree Maclaurin polynomial of f.

The idea is to take the Taylor series to a certain degree of differentiation by increasing k

such that the series is as close as possible to that of the function.

So now we look to see how accurate we can be with the Taylors Forumla.

Theorem 1: Taylors Theorem

Iffhas derivatives up to order n + 1in an intervalI containing c,then for eachxinI, there exists a numberzbetweenx and csuch that

2!

! Where
https://www.youtube.com/watch?v=cjPoEZ0I5wQhttps://www.youtube.com/watch?v=cjPoEZ0I5wQhttps://www.youtube.com/watch?v=Os8OtXFBLkYhttps://www.youtube.com/watch?v=Os8OtXFBLkYhttps://www.youtube.com/watch?v=dp2ovDuWhrohttps://www.youtube.com/watch?v=dp2ovDuWhrohttps://www.youtube.com/watch?v=3ZOS69YTxQ8https://www.youtube.com/watch?v=3ZOS69YTxQ8https://www.youtube.com/watch?v=ocNLs4hmNtIhttps://www.youtube.com/watch?v=ocNLs4hmNtIhttps://www.youtube.com/watch?v=ocNLs4hmNtIhttps://www.youtube.com/watch?v=3ZOS69YTxQ8https://www.youtube.com/watch?v=dp2ovDuWhrohttps://www.youtube.com/watch?v=Os8OtXFBLkYhttps://www.youtube.com/watch?v=cjPoEZ0I5wQ


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+! +This is not that much of the test, so Im going to stop here at the first part.

Lets look at problems:

1. Find the 3rddegree Taylor polynomial for cos(x) generated at .

2. Find the Taylor polynomial and the Taylor remainder for the function andthe values of c and n.

2 3 1, 1, 43. Find the Taylor polynomial and the Taylor remainder for the function andthe values of c and n.

sin , 2 , 34. Find the Taylor polynomial and the Taylor remainder for the function andthe values of c and n.

tan , 4 , 25. Find the Taylor polynomial and the Taylor remainder for the function andthe values of c and n.

, 1, 3

For additional instruction on this section, check out these videos:

Taylor's Remainder Theorem

Taylor Polynomial to Approximate a Function

Additionally, for more practice, you can visitwww.Nixty.com,sign up for free and search for the

instructor PatrickJMT in the search engine, which should show you his sequence and series class.

Join the class, and he has some practice lessons that you can go through for additional review.
https://www.youtube.com/watch?v=yJWFAG1g4Jwhttps://www.youtube.com/watch?v=yJWFAG1g4Jwhttps://www.youtube.com/watch?v=WKvBJdl1qrMhttps://www.youtube.com/watch?v=WKvBJdl1qrMhttp://www.nixty.com/http://www.nixty.com/http://www.nixty.com/http://www.nixty.com/https://www.youtube.com/watch?v=WKvBJdl1qrMhttps://www.youtube.com/watch?v=yJWFAG1g4Jw

calc2 study guide (section 2)

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