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OTTO Wrote: (01-22-2016, 09:19 PM)ICantGiveCredit Wrote: »ok ok is the answer actually 103. i want to know
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It's just four different ways of representing the sine function.
d/dx -cos(x) is
sin(x), from the derivatives of trigonometric functions
y"+y=0 is a second-order linear ordinary differential equation. The coefficient of y is 1, hence the solution is of the form y=c
1cos(x)+c
2sin(x). The conditions of y(0)=0 and y'(0)=1 mean that c
1=0 and c
2=1, so the solution is y=
sin(x)
(e
iz-e
-iz)/2i is the definition of the complex sine function,
sin(z)
And finally, the last expression is the Maclaurin Series expansion for
sin(x). It's calculated by f(x)=f(0)+f'(0)x+f"(0)x
2/2!+f‴(0)x
3/3!+... Since the n
th derivative of sin(x) at 0 is 1 for n=4k+1, -1 for n=4k+3, and 0 for n=2k, it's only the odd terms with alternating signs.
So I had sin(x), y=sin(x), sin(z), and sin(x). That sure is a lot of sins.
