- #1
McAfee
- 96
- 1
Homework Statement
Homework Equations
The Attempt at a Solution
I keep getting the answer 1- 96/4!*x^4. Can't find where I'm going wrong.
Dick said:It's really hard to tell what you are doing from the blurry photograph. Can you explain? What are F(0), F'(0), F''(0) etc?
Dick said:Much clearer, thanks! Now I see where you are going wrong. f(0)=0, isn't it? It's f'(0) that is 1. And if you think about it, you didn't have to compute a lot of those terms involving high powers of x. But that's in retrospect.
McAfee said:I'm still not getting the correct answer.
I think f(0)=1. 0^4=0 0 times -4 equals 0 and then e to the power of 0 equals 1.
Dick said:Nooo. f(0) is the integral from 0 to 0 of e^(-4t^4). That's zero! f'(0) is exp(-4t^2) evaluated at t=0. Your derivative count is off by one!
McAfee said:So it should be x-(96/5!)*x^5 ?
Dick said:Yes, it should. But I don't like that '?' in the answer. You do see why, don't you?
McAfee said:Honestly, it is the right answer but I don't see why. Is there something special with finding a maclaurin polynomial of a function that includes e?
Dick said:No, nothing special about e. You don't agree f(0)=0? Suppose the function in your integral was just 1. What are f(0) and f'(0)?
McAfee said:f(0) would be 1 and f'(0) would be 0
Looking at it again is it because of the integral would be from 0 to 0
Dick said:If you are integrating 1 from 0 to x, then f(x)=x. I can say pretty definitely that f(0)=0. f'(0)=1.
McAfee said:Can it basically be a rule if you integrating from 0 to x, that the first term is 0 followed by the second term being x?
Dick said:You can't simplify it that much. Integrate 2 from 0 to x. The point of the example is just to show what you were doing wrong. f(0) is always zero, sure. f'(0) depends on the value of the integrand at 0. You just had your derivatives shifted by one. That's all.
A MacLaurin polynomial is a type of polynomial that is used to approximate a given function. It is centered at x = 0 (also known as the origin) and is a finite sum of terms that involve powers of x.
The degree of a MacLaurin polynomial is determined by the highest power of x in the polynomial. For example, a degree 5 MacLaurin polynomial will have terms that involve x^5, x^4, x^3, x^2, x, and a constant term.
The purpose of finding the MacLaurin polynomial of a function is to approximate the function using a simpler polynomial. This can be useful in situations where the function may be difficult to work with, but the polynomial can be easily manipulated and analyzed.
The MacLaurin polynomial of degree 5 is calculated by using the Taylor series expansion of the function and truncating it after the fifth degree term. This involves taking the derivatives of the function at x = 0 and plugging them into the formula for the Taylor series.
MacLaurin polynomials have various applications in mathematics and engineering, such as in numerical analysis, optimization, and approximation of real-world functions. They are also used in physics to approximate motion and in economics to model supply and demand curves.