Question

from this page,http://www.homeschoolmath.net/teaching/sine_calculator.php shows that how to calculate sine with +-*/. the

sin x = x - x^3/3! + x^5/5! - x^7/7! + ...,

what is the next sequence?
is it x^9/9! - x^11/11! ?
they always grow in the sequence of odds number?

Answer 1

Yes it is. This is the Taylor Series for sin(x).

You can find out more about this series at: http://en.wikipedia.org/wiki/Taylor_series

This site has some good graphs showing how each term in the series makes it fit sin(x) better and better: http://openstudy.com/study#/updates/518b81f0e4b0a13cf7422ffd

Answer 2

\[\sin ^{(4n+k)}(0) = \left\{ 0, when k =0; 1,when k = 1; 0 when k = 2; -1 when k =3 \right\}\]

I get that from wikipedia, http://en.wikipedia.org/wiki/Sine, the first equation in "series definition section". I don't understand why the sine has the index power of 4n+k and when k = 0,1,... then n = ???

Answer 3

\(\sin^{4n+k}(0)\) represents the value of the \((4n+k)^{th}\) derivative of the sine function at zero.

e.g.:
when n=0 and k=0, then 4n+k=0, so the value of the 0'th derivative of sin(0) is 0
when n=0 and k=1, then 4n+k=1, so the value of the 1'st derivative of sin(0) is 1
when n=0 and k=2, then 4n+k=2, so the value of the 2'nd derivative of sin(0) is 0
when n=0 and k=3, then 4n+k=3, so the value of the 3'rd derivative of sin(0) is -1
when n=1 and k=0, then 4n+k=4, so the value of the 4'th derivative of sin(0) is 0
etc

and we know that:
0'th derivative of sin(x) = sin(x), giving sin(0)=0
1'st derivative of sin(x) = cos(x), giving cos(0)=1
2'nd derivative of sin(x) = -sin(x), giving -sin(0)=0
3'rd derivative of sin(x) = -cos(x), giving -cos(0)=-1
4'th derivative of sin(x) = sin(x), giving sin(0)=0
etc