Finding Taylor series for x^3 at a = -1

In summary: I appreciate your help!In summary, the Taylor series for f(x) centered at a=-1 is -1+3(x+1)-3(x+1)^2+(x+1)^3. This can be obtained by finding the derivatives of f(x)=x^3, substituting them into the formula for the Taylor series, and then simplifying by using the substitution y=x+1.
  • #1
mcdowellmg
55
0

Homework Statement



Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1

Homework Equations



f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution



I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0
I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?
 
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  • #2
mcdowellmg said:

Homework Statement



Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1


Homework Equations



f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution



I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0
I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?

You have listed f(x), f'(x), f''(x), f'''(x) but I don't see your correct values for

f(-1), f'(-1),f''(-1),f'''(-1)

which you should be using for coefficients.
 
  • #3
You can check your work by graphing the result & comparing the result with a graph of x3.
 
Last edited:
  • #4
mcdowellmg said:

Homework Statement



Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1


Homework Equations



f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution



I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0
So f'(-1)= 3, f''(-1)= 6(-1)= -6, f'''(-1)= 6. But did you cube f' and f''?

I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?
The simplest way to get this is to let y= x+ 1. Then x= y- 1 and [itex]x^3= (y- 1)^3[/itex]. Multiply that out and then replace y with x+ 1.
 
  • #5
Thanks! I did cube f' and f'' for some reason. I ended up with -1+3(x+1)-3(x+1)^2+(x+1)^3, which was correct.
 

What is a Taylor series?

A Taylor series is an infinite series representation of a function that is expressed as an infinite sum of terms. It is used in calculus to approximate functions as a polynomial series around a specific point (usually denoted as \(a\)). The Taylor series of a function \(f(x)\) at \(a\) is often denoted as \(T_a(x)\).

How do you find the Taylor series for \(x^3\) at \(a = -1\)?

To find the Taylor series for \(x^3\) at \(a = -1\), you can follow these steps:

  1. Find the derivatives of \(x^3\): Calculate the first few derivatives of \(x^3\) with respect to \(x\). In this case, you need the first few derivatives up to the desired degree of accuracy. For \(x^3\), the derivatives are:
    • \(f(x) = x^3\)
    • \(f'(x) = 3x^2\)
    • \(f''(x) = 6x\)
    • \(f'''(x) = 6\)
  2. Evaluate the derivatives at \(a = -1\): Substitute \(a = -1\) into each of the derivatives to find their values at \(x = -1\). This gives you the function values and derivatives at the expansion point:
    • \(f(-1) = (-1)^3 = -1\)
    • \(f'(-1) = 3(-1)^2 = 3\)
    • \(f''(-1) = 6(-1) = -6\)
    • \(f'''(-1) = 6\)
  3. Write down the Taylor series: Use the formula for the Taylor series expansion around \(a\): \[T_a(x) = f(a) + (x - a)f'(a) + \frac{{(x - a)^2}}{2!}f''(a) + \frac{{(x - a)^3}}{3!}f'''(a) + \ldots\] Substituting the values from step 2 and the derivatives from step 1: \[T_{-1}(x) = -1 + (x - (-1))(3) + \frac{{(x - (-1))^2}}{2!}(-6) + \frac{{(x - (-1))^3}}{3!}(6) + \ldots\]
  4. Simplify the series: Simplify the series by expanding the terms and combining like terms. In this case, the series for \(x^3\) at \(a = -1\) simplifies to: \[T_{-1}(x) = -1 + 3(x + 1) - 3(x + 1)^2 + 2(x + 1)^3 + \ldots\]

This is the Taylor series for \(x^3\) centered at \(a = -1\). You can continue the expansion to include higher-degree terms for more accuracy if needed.

What is the significance of finding a Taylor series?

Finding a Taylor series allows you to approximate a function with a polynomial, which can simplify calculations and provide insights into the behavior of the function around a specific point. Taylor series are widely used in calculus, physics, engineering, and various fields to analyze and approximate functions, especially in situations where the exact function is complex or not readily available.

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