1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
spayn [35]
3 years ago
15

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

a. (Enter your answers as a comma-separated list.) f(x) = 7 1 + x , a = 2
Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}

Step-by-step explanation:

The Taylor series of the function <em>f </em>at <em>a </em>(or about <em>a</em> or centered at <em>a</em>) is given by

f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k

To find the first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 you must:

In our case,

f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

  • f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}

Evaluate the function at the point: f\left(2\right)=\frac{7}{3}

  • f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}

  • f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}

  • f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}

  • f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}

Evaluate the function at the point: \left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}

Apply the Taylor series definition:

f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}

The first four nonzero terms of the Taylor series of \frac{7}{x + 1} around a=2 are:

f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}

You might be interested in
Need help as fast as possible will mark brainliest
MissTica

Answer:

Option C. is correct choice.

Step-by-step explanation:

20\pm 0.1\:^{\circ}\:C

Regards: Umer

6 0
3 years ago
5x + 2 (x+7) =14x - 7
qaws [65]
The answer is x=-23/21
5 0
3 years ago
Question 2 A birthday cake in the shape of the right-circular cylinder of radius 15cm and thickness 10cm is cut into smaller pie
finlep [7]

The volume of a piece of the cake will be 588.75 cm³.

The complete question is attached below.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

A birthday cake in the shape of the right-circular cylinder of radius 15 cm and thickness 10 cm is cut into smaller pieces.

One of the piece is shown.

Then the volume of piece of the cake will be

V = (30 / 360) x π x 15² × 10

V = 588.75 cm³

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

5 0
2 years ago
a dental company has rolls of floss that are 1872 in long they want to put the distance in the yard on their label what is the d
statuscvo [17]
Copy and paste ur answer onto internet it will tell u
4 0
3 years ago
HELP PLS ASAP!!
DedPeter [7]

Answer:

cos^(-1)(12/13) = 22.62°

Step-by-step explanation:

Inverse trig functions are used to find angles. I was taught SOH CAH TOA: sine = opposite/hypotenuse, cosine = adjacent / hypotenuse, tan = opposite / adjacent.

Here, we have an adjacent side length and the hypotenuse. So we use inverse cosine.

4 0
3 years ago
Other questions:
  • On a board measuring 1x100, each square is numbered from
    9·2 answers
  • Please help! ASAP plz&lt;3
    10·1 answer
  • A triangle has an area of 36 cm². The base and height are scaled by a factor of 5.
    10·1 answer
  • HELPWITH THIS QUESTION
    11·2 answers
  • Can someone help me?
    9·1 answer
  • What is 2/7 + 2/5 in simplest form
    15·1 answer
  • I need help <br><br> Solve the expression when X =5;3(X+2)=
    12·1 answer
  • What is the solution to the equation below? <br>-3 + sqrt 2x - 1 =8​
    11·1 answer
  • Julie has two bags of grapes. One weighs 5/8 pound and the ther weighs 1/4 pound. What is the difference in the two bags of grap
    9·2 answers
  • What is the volume of the right triangular prism in cubic meters?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!