10.1 Taylor Polynomials

Question 1

What is the purpose of Taylor polynomials?

Answer 1

To approximate a smooth function
𝑓(𝑥) near a specific point x=a.

Question 2

How do we create a Taylor polynomial and why do we create them?

Answer 2

It uses the function’s value and its derivatives at 𝑥=𝑎 to create a polynomial that approximates the function near that point.

Question 3

What is the formula for a Taylor polynomial of degree 𝑛 centred at x=a?

Answer 3

Tn(x)=f(a)+f ′(a)(x−a)+ f ′′(a) (x−a)^(2)/2!+⋯+ f’(n)(a)(x−a)^(n)/n!

Question 4

What does each term in a Taylor polynomial involve? (3)

Answer 4

Each term involves a derivative of the function at x=a, multiplied by (x−a) raised to a power, and divided by the factorial of the term’s degree.

Question 5

What is a Maclaurin polynomial?

Answer 5

A special case of the Taylor polynomial where a=0.

Question 6

What is the formula for a Maclaurin polynomial of degree n?

Answer 6

Tn(x)=f(0)+f ′(0)x+f ′′(0)x^(2)/2!+⋯+ f (n(0)x^(n)/n!

Question 7

What is the primary use of Taylor polynomials?

Answer 7

To provide a local approximation of a function near the point x=a.

Question 8

How does increasing the degree n of a Taylor polynomial affect the approximation?

Answer 8

It improves the approximation near x=a, assuming the function is smooth.

Question 9

Why are graphs important when studying Taylor polynomials?

Answer 9

They give a visual representation of how Taylor polynomials of increasing degrees better approximate the function near x=a.

Question 10

What is the interval of convergence mean in the context of Taylor polynomials?

Answer 10

The interval where the Taylor series (infinite version of Taylor polynomials) converges to the actual function.

Question 11

what are the steps to solve this problem; Find the Taylor polynomial of degree 2 for the function
𝑓(𝑥)=cos(𝑥) centred at x=0. (6)

Answer 11

1. Identify the function and point;

Function:
f(x)=cos(x)
Center: x=0

2. Find the first two derivatives of

f(x)=cos(x)
f′(x)=−sin(x)
f′(x)=−cos(x)

3. Evaluate the function and derivatives at the x value of interest, x=0:

𝑓(0)=cos(0)=1
f ′(0)=−sin(0)=0
f ′′(0)=−cos(0)=−1

4. Write the general Taylor polynomial formula:
   For degree 2:

𝑇2(𝑥)=𝑓(0)+𝑓′(0)𝑥+𝑓′′(0)x/2!

5. Substitute the values you found:

𝑇2(𝑥)==1+0⋅x+(-1/2)x^(2)

6. Simplify the polynomial:

𝑇2(𝑥)=1−1/2x^(2)