1. ## Taylor Polynomials

Please help! Don't worry too much about the plotting of the graph. I can work that out. Just on the actually math portion of the problem, I am not sure how to approach this question. How do you find all the various functions of T? What are non-zero terms (just any term that isn't lying on the x-axis?)? Thank you in advance.

2. Originally Posted by Acoustica
Please help! Don't worry too much about the plotting of the graph. I can work that out. Just on the actually math portion of the problem, I am not sure how to approach this question. How do you find all the various functions of T? Thank you in advance.
f(x) = sin(x) + cos(2x)
f'(x) = cos(x) - 2 sin(2x)
f''(x) = -sin(x) -4 cos(2x)
f'''(x) = -cos(x) +8 sin(2x)
f''''(x) = sin(x) + 16 cos(2x)

so if we are going to expand as a Taylor series about 0, we need:

f(0) = 1,
f'(0) = 1
f''(0) = -4
f'''(0) = -1
f''''(0) = 16

and so our series is:

f(x) = f(0) + f'(0) x + f''(0) x^2/2! + f'''(0) x^3/3! + f''''(0) x^4/4! + ..

.....= 1 + x - 2x^2 - x^3/6 + (2/3) x^4 + ...

(note this has gone too far as this takes us up to the 5th non-zero term)

RonL

3. Thanks! This is what I achieved for the graph and the functions I graphed.

Dotted line: sin(x) + cos(2x)
Green line: T1 = x+1
Blue line: T2 = (x-(x^3/6)) + (1-((2x)^2/2))
Red line: T3 = (x-(x^3/6)+(x^5/120)) + (1-((2x)^2/2)+((2x)^4/24))
Purple: T4 = (x-(x^3/6)+(x^5/120)-(x^7/5040)) + (1-((2x)^2/2)+((2x)^4/24)-((2x)^6/720))

The red and purple line overlap each other, so it is slightly hard to see.
Is what I got for T4 equivalent to what you wrote, "1 + x - 2x^2 - x^3/6"? I used the known series of sin(x) and cos(x) instead.
Also, the question asks to write down the polynomial T4, do I just write down the function I graphed? What does it mean when it says "Make sure that T4 has exactly four nonzero terms"?