Question:
Help with Maclaurin Series sin^3 (x)?
jumpsteady88
2007-04-18 22:36:56 UTC
I need help finding the Maclaurin series of sin^3 (x). I have tried a few different approaches to come only to a standstill. When I asked my professor for assistance he said to first find the Maclaurin series of sin^2 (x) in the form of (1/2)[1-cos (2x)] and then multiply by the Maclaurin series of sin (x). I also found from my notes that the Maclaurin series of [sin (x)]*[cos (x)] = (1/2)[sin (2x)]. I am at a complete loss and this one problem may determine whether or not i graduate college.
Three answers:
Nishit V
2007-04-18 23:07:21 UTC
sin3 (x) = 3sinx - 4sin^3 x

so



sin^3 x = (3sinx - sin3x)/4



General maclaurin series is :

A Maclaurin series is a Taylor series expansion of a function about 0,

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

f(x) = (3sinx - sin3x)/4

f'(x) = 3cosx/4 - 3cos3x/4

f''(x) = -3sinx/4 +9sin3x/4

f'''(x) = -3cosx/4 + 27 cos3x/4

f''''(x) = 3sinx/4 -81sin3x/4

f'''''(x) = 3cosx/4 -243cos3x/4

f''''''(x) = -3sinx/4 +729sin3x/4

f'''''''(x) = -3cosx/4 +2187cos3x/4



f^n(x) = 0 if n is even

= (-1)^(n+1)/2[3^n -3]/4

f(0) = 0

f'(0) = 0

f''(0) = 0

f'''(0) = 6

f''''(0) = 0

f'''''(0) = -60

f'''''(0) = 0

f'''''(0) = 546



sin^3x=6/3!x^3 - 60/5!x^5...+ 546/7! x^7 + (f^((n))(0))/(n!)x^n+....



sin^3x=x^3 - x^5/2...+ 546/5040 x^7 + (-1)^(n+1)/2[3^n -3]/4 x^n/n!



For all n which is odd that is 1,3,5,7 ...

for even n the terms are 0
Nick Solly
2007-04-18 23:01:13 UTC
f(x) = sin^3 x f(0)=0

f ' (x) = 3*((sin^2 x)*(cos x)) f ' (0) = 0

f '' (x) = -3/4(sin x - 3 sin 3x) f '' (0) = 0

f ''' (x) = -3/4(cos x - 9 cos 3x) f''' (0) = 6

f '''' (x) = 3/4(sin x - 27 sin 3x) f'''' (0) = 0

f ''''' (x) = 3/4(cos x - 81*cos(3x)) f''''' (0) = -60

f '''''' (x) = -3/4(sin x - 243 sin 3x) f'''''' (0) = 0

f ''''''' (x) = -3/4(cos x - 729 cos 3x) f'''''' (0) = 546



That's probably enough so Maclaurin's Expansion of sin^3 (x):



sin^3 (x) = x^3 * (6/3!) + x^5 * (-60/5!) + x^7 * (546/7!) ....

= x^3 - (1/2)*x^5 + (13/120)*x^7 .....
novangelis
2007-04-18 22:43:39 UTC
sin³ x = (sin x - sin 3x)/4


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