Taylor Series question about error:

  • #1

Homework Statement


This is a three part question: It is based off the first two sections. I'm pretty sure the first two answers are correct, but I have no idea how to do the third question.

Write the First three nonzero terms and the general term of the Taylor series expansion about x=0 of f(x) = 5 sin (x/2)
5x/2 – 5x^3/48 + 5x^5/(5! * (2^5)….5[tex]\sum[/tex]^(2x + 1)/((2x + 1)!* (2^(2x + 1))

What is the interval of convergence for the series found in (a)?

I used the ratio test to prove the interval is all real numbers. Since this Taylor series uses the sin function, I'm sure it is all real numbers.

What is the minimum number of terms in (a) that are necessary to approximate f (x) on the interval with an error not exceeding .1? Show your method. My teacher says to use the interval found in B.



Homework Equations


Lagrange error
5x/2 – 5x^3/48 + 5x^5/(5! * (2^5)….5∑ X^(2x + 1)/((2x + 1)!* (2^(2x + 1))
f^(n+1) (x)/((n+1)!) * x^(n+1) ≤ .1





The Attempt at a Solution


What is the minimum number of terms in (a) that are necessary to approximate f (x) on the interval with an error not exceeding .1? Show your method. My teacher says to use the interval found in B.


This is the section that I need help on.

I used the lagrange error: f^(n+1) (x)/((n+1)!) * x^(n+1) ≤ .1

From here, I realize that this was too complicated, so I went through a trial and error and plug a number in to see what is accurate to the 1/10. I got 3 terms. I'm not sure how I am supposed to show a way without using a calculator or plug and chug. This is the 1976 BC7 Ap free response question.
 

Answers and Replies

  • #2
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6,394

Homework Statement


This is a three part question: It is based off the first two sections. I'm pretty sure the first two answers are correct, but I have no idea how to do the third question.

Write the First three nonzero terms and the general term of the Taylor series expansion about x=0 of f(x) = 5 sin (x/2)
5x/2 – 5x^3/48 + 5x^5/(5! * (2^5)….5[tex]\sum[/tex]^(2x + 1)/((2x + 1)!* (2^(2x + 1))
Your first three terms are right, but you general term isn't.

The Maclaurin series (Taylor series in powers of x) of sin(x) is
[tex]\sum_{n = 1}^{\infty} \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}[/tex]

Notice that general term has both x and n in it. Yours has only x.

The series for y = 5 sin(x/2) can be obtained pretty easily from the series for sin(x) by replacing x with x/2 and multiplying each term by 5.
What is the interval of convergence for the series found in (a)?

I used the ratio test to prove the interval is all real numbers. Since this Taylor series uses the sin function, I'm sure it is all real numbers.
Your work is suspect since you didn't have the right general term.
What is the minimum number of terms in (a) that are necessary to approximate f (x) on the interval with an error not exceeding .1? Show your method. My teacher says to use the interval found in B.



Homework Equations


Lagrange error
5x/2 – 5x^3/48 + 5x^5/(5! * (2^5)….5∑ X^(2x + 1)/((2x + 1)!* (2^(2x + 1))
f^(n+1) (x)/((n+1)!) * x^(n+1) ≤ .1
You have your first three terms + some gibberish.

The Attempt at a Solution


What is the minimum number of terms in (a) that are necessary to approximate f (x) on the interval with an error not exceeding .1? Show your method. My teacher says to use the interval found in B.


This is the section that I need help on.

I used the lagrange error: f^(n+1) (x)/((n+1)!) * x^(n+1) ≤ .1

From here, I realize that this was too complicated, so I went through a trial and error and plug a number in to see what is accurate to the 1/10. I got 3 terms. I'm not sure how I am supposed to show a way without using a calculator or plug and chug. This is the 1976 BC7 Ap free response question.
 

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