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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=+
4. Determine whether the sequence converges or diverges. If it converges, find its limit.
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an = tan
5. Determine whether the sequence converges or diverges. If it converges, find its limit.
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
Calculus 2Midterm 2Study Guide
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
Calculus 2Midterm 2Study Guide
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.
(Proof can be found on page 766 of the textbook.)
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.
(Proof can be found on page 768 of the textbook.)
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.
Calculus 2Midterm 2Study Guide
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
Calculus 2Midterm 2Study Guide
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
Calculus 2Midterm 2Study Guide
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
convergence of the series.
, 03. Find the Taylor Series offat the given value of c. Then find the radius of
convergence of the series.
ln1 , 04. Find the Taylor Series offat the given value of c. Then find the radius of
convergence of the series.
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
Calculus 2Midterm 2Study Guide
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