• kiwifruit
In summary, the question asks to find the quadratic MacLaurin polynomial for the function f(x) = x sin(x). The quadratic MacLaurin polynomial is obtained by setting a=0 in the Taylor polynomial formula and only considering the second degree term. This is because both the function and its first derivative are 0 at x=0, making the first non-zero coefficient the second degree term. Graphing the function and y=x^2 shows that they are similar close to the origin.
kiwifruit

Homework Statement

Given f(x) = x sin(x)

The Attempt at a Solution

i know that a maclaurin series is when a=0 in a taylor series.

i did the 1st-5th derivatives of f(x) and then used the formula for taylor polynomial and set a=0.

f'=sin(x) + x cos(x)
f''=2cos(x) -x sin(x)
f'''=-3sin(x) -xcos(x)
f''''=-4cos(x) +xsin(x)
f'''''=5sin(x) +xcos(x)

with the formula for taylor polynomial

Pn(x) = n
Ʃ [f^k (a) (x-a)^k] / k!
k=0

by setting a=0 i got Pn(x)=0+0+x^2 -0- (x^4)/6...however the answer seems to be P(x)=x^2
anyone able to explain to me how that is so?

i think i got it... is the quadratic maclaurin polynomial just the 4th order maclaurin polynomial??

No, the "quadratic" MacLaurin polynomial is NOT fourth degree. It is, as is any quadratic polynomial, second degree. You are correct that the function and first derivative are both 0 at x= 0 so the first non-zero coefficient is the second degree term. The "quadratic MacLaurin polynomial" for this function is just that second degree term, $x^2$.

Graph both y= x sin(x) and $y=x^2$ on the same coordinates. You should find that they remarkably similar close to the origin.

Attachments

• xsinx.bmp
23.8 KB · Views: 684
Last edited by a moderator:

What is a quadratic Maclaurin polynomial?

A quadratic Maclaurin polynomial is a polynomial function that approximates a given function using terms up to the quadratic power of the variable. It is centered at the origin and is a special case of the Maclaurin series, which is a representation of a function as an infinite sum of terms involving the derivatives of the function at the origin.

Why is finding a quadratic Maclaurin polynomial important?

Finding a quadratic Maclaurin polynomial is important because it allows us to approximate a complicated function with a simpler polynomial function, making it easier to perform calculations and analysis. It is also used in applications such as physics, engineering, and economics.

How do you find the quadratic Maclaurin polynomial for a given function?

To find the quadratic Maclaurin polynomial for a given function, you need to first find the first and second derivatives of the function at the origin. Then, you can use these derivatives to write out the Maclaurin series up to the quadratic term. Finally, you can simplify the series to get the quadratic Maclaurin polynomial.

What is the difference between a Maclaurin polynomial and a Taylor polynomial?

A Maclaurin polynomial is a special case of a Taylor polynomial, where the polynomial is centered at the origin (x=0). In contrast, a Taylor polynomial can be centered at any point on the function's domain. Both types of polynomials are used for approximating functions, but the Maclaurin polynomial is often used for functions that are symmetric about the origin.

How accurate is the quadratic Maclaurin polynomial approximation?

The accuracy of the quadratic Maclaurin polynomial approximation depends on the function being approximated and the number of terms used in the polynomial. Generally, the more terms used, the more accurate the approximation will be. However, for some functions, even a few terms can provide a good approximation.

• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus
Replies
19
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
36
Views
3K
• Calculus and Beyond Homework Help
Replies
2
Views
859