2011-03-22, 04:31 AM
Well, the Taylor series approximation is a pretty simple concept, and I grasped it pretty easily at first, but I guess I lost it.
To create the series, you take the nth derivative at a and divide by n! starting at n=0 with f(a)/0! = f(a), so the series is f(a)+f'(a)(x-a)+(f''(a)(x-a)^2)/2+(f'''(a)(x-a)^3)/6 and so on. But first, it asks for an approximation that's accurate to like 4 decimal places with the least number of terms, so you use the remainder formula, the (n+1)th derivative at c, which since we're dealing with sin and cos would always be less than 1, divided by (n+1)! times (x-a)^(n+1) and that has to equal 10^-4 or less, which I think I figured out would only take two terms (when I did it the wrong way).
I think I've finally figured out what I did wrong. My problem was that I had no idea how to handle x-a and I was under the impression that x-pi/3 = pi/180 when really I should've done pi/180 - pi/3 = -59pi/180.
So then it's just guess and check with the remainder, I mean we didn't learn any guaranteed method to get the accuracy we wanted, but I guess n! might provide some clues.
R_7(x) = |f'''©|(-59pi/180)^8/8!, |f'''©| = |-cos c| < 1, so we have (-59pi/180)^8/8! which is 3.14 x 10^-5, which means the last term must be to the power of 7, which is 8 terms all together
So we need 7 terms of the Taylor series expansion so
f(pi/3) = sqrt(3)/2
f'(pi/3) = 1/2
f''(pi/3) = -sqrt(3)/2
f'''(pi/3) = -1/2
sqrt(3)/2 + (-59pi/180)/2 - (sqrt(3)(-59pi/180)^2)/4 - (-59pi/180)^3/12 + (sqrt(3)(-59pi/180)^4)/48 + (-59pi/180)^5/240 - (sqrt(3)(-59pi/180)^6)/1440 - (-59pi/180)^7/10080
Which, what do ya know, is 0.017427 and sin(pi/180) 0.017452
Too late.
This next part is somewhat unrelated.
Ohio State is also on Spring Break, and I'm unsure whether I should take 153 next quarter or not, with AP tests coming up and all. First, and this is the reason I didn't enjoy 161H (because of AP Calc AB), I've already been taught the material (whether I learned it is questionable, I mean, I feel I have, but, well, D+), and second, even if the D+ is forgiven, it'll still be there on my transcript for the rest of my life.
To create the series, you take the nth derivative at a and divide by n! starting at n=0 with f(a)/0! = f(a), so the series is f(a)+f'(a)(x-a)+(f''(a)(x-a)^2)/2+(f'''(a)(x-a)^3)/6 and so on. But first, it asks for an approximation that's accurate to like 4 decimal places with the least number of terms, so you use the remainder formula, the (n+1)th derivative at c, which since we're dealing with sin and cos would always be less than 1, divided by (n+1)! times (x-a)^(n+1) and that has to equal 10^-4 or less, which I think I figured out would only take two terms (when I did it the wrong way).
I think I've finally figured out what I did wrong. My problem was that I had no idea how to handle x-a and I was under the impression that x-pi/3 = pi/180 when really I should've done pi/180 - pi/3 = -59pi/180.
So then it's just guess and check with the remainder, I mean we didn't learn any guaranteed method to get the accuracy we wanted, but I guess n! might provide some clues.
R_7(x) = |f'''©|(-59pi/180)^8/8!, |f'''©| = |-cos c| < 1, so we have (-59pi/180)^8/8! which is 3.14 x 10^-5, which means the last term must be to the power of 7, which is 8 terms all together
So we need 7 terms of the Taylor series expansion so
f(pi/3) = sqrt(3)/2
f'(pi/3) = 1/2
f''(pi/3) = -sqrt(3)/2
f'''(pi/3) = -1/2
sqrt(3)/2 + (-59pi/180)/2 - (sqrt(3)(-59pi/180)^2)/4 - (-59pi/180)^3/12 + (sqrt(3)(-59pi/180)^4)/48 + (-59pi/180)^5/240 - (sqrt(3)(-59pi/180)^6)/1440 - (-59pi/180)^7/10080
Which, what do ya know, is 0.017427 and sin(pi/180) 0.017452
Too late.
This next part is somewhat unrelated.
Ohio State is also on Spring Break, and I'm unsure whether I should take 153 next quarter or not, with AP tests coming up and all. First, and this is the reason I didn't enjoy 161H (because of AP Calc AB), I've already been taught the material (whether I learned it is questionable, I mean, I feel I have, but, well, D+), and second, even if the D+ is forgiven, it'll still be there on my transcript for the rest of my life.