For sum(Cn*x^n)
I'm supposed to enter in c3-c7
I already know that c4 and c6 are 0 because the derivative is something*sin(0)=0
but for the odd numbered c's I am having problems...
i know that the taylor series for sinx = sum( (-1^n) * x^(2n+1) )/ (2n+1)!
so i just substituted in 5x and multiplied by 5x^2 and got
5sum( (-1^n) * ( 5^(2n+1) * x^(2n+3) ) / (2n+1)!
so for c3 i got 5(-1^3)(5^7)/(7)! = -5^8/7!
but I am not getting the answer right for this. Can someone please explain what I am doing wrong.
2007-11-30 08:17:11 · 4 answers · asked by mula 2 in Science & Mathematics ➔ Mathematics
hi
wrong
c3 for n = 0
x^(2*0 + 3 ) = x^3 :)
5*[(-1)^0 * 5^(2*0 +1)/ (2*0 +1)! ] = 5*5 = 25
then
5 x^2 * ( 5x - ((5x)^3)/6 + ...) =
5 x^2 * ( 5 x - (125/6) x^3 + ...)
= 25 x^3 - (625/6) x^5 + ...
greetings
2007-11-30 08:28:24 · answer #1 · answered by railrule 7 · 0⤊ 0⤋
The 4th and 6th terms of the finished series are not 0.
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! + x^13/13! - x^14/14! + . . . . .
sin(5x) = 5x - (5x)^3/3! + (5x)^5/5! - (5x)^7/7! + (5x)^9/9! - (5x)^11/11! + (5x)^13/13! - (5x)^14/14! + . . . . .
5(x^2)sin(5x) = 5(x^2)[5x - (5x)^3/3! + (5x)^5/5! - (5x)^7/7! + (5x)^9/9! - (5x)^11/11! + (5x)^13/13! - (5x)^15/15! + . . . . . ]
If x is the 0th term,
c2 = + 5^6/5! ≈ 130.2083
c3 = - 5^8/7! ≈ 77.50496
c4 = + 5^10/9! ≈ 26.9114
c5 = - 5^12/11! ≈ 6.116237
c6 = + 5^14/13! ≈ 0.9801662
c7 = - 5^16/15! ≈ 0.116686
If you count x as the 1st term shift up 1
2007-11-30 12:12:47 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
It is [5x^2] * [ the Maclaurin expansion of sin(5x)]
Note 5x^2 is already a (finite) polynomial
2007-11-30 08:32:32 · answer #3 · answered by lienad14 6 · 0⤊ 0⤋
I am assuming there is no ! after sin(5x), am I correct ?
f(x) = 5(x^2)sin(5x) =
25*x^3-(625*x^5)/6+
(3125*x^7)/24-
(78125*x^9)/1008+
(1953125*x^11)/72576
2007-11-30 08:32:48 · answer #4 · answered by Anonymous · 0⤊ 0⤋