- #1
yungman
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This is not really a homework, I am trying to expand Si(x) into a series.The series expansion of Si(x) is given in articles:
[tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^{\infty}\frac {(-1)^k x^{2k+1}}{(2k+1)(2k+1)!}[/tex]
This is my work, I just cannot get the right answer:
[tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i[/tex]
[tex]\sin \theta=\sum_0^{\infty}\frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\Rightarrow\;\frac{\sin\theta}{\theta}=\sum_0^{\infty}\frac{(-1)^k \theta^{2k}}{(2k+1)!}[/tex]
[tex]\Delta \theta_i = \frac x m\;\hbox{ and }\; \theta_i= i\Delta \theta_i =i\frac x m[/tex]
[tex]\Rightarrow\;Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^k\left (i\frac x m\right )^{2k}}{(2k+1)!}\frac x m\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^ki^{2k} x^{2k+1}}{m^{2k+1}(2k+1)!} [/tex]
I don't know how to get the answer, please help.
[tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^{\infty}\frac {(-1)^k x^{2k+1}}{(2k+1)(2k+1)!}[/tex]
This is my work, I just cannot get the right answer:
[tex]Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i[/tex]
[tex]\sin \theta=\sum_0^{\infty}\frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\Rightarrow\;\frac{\sin\theta}{\theta}=\sum_0^{\infty}\frac{(-1)^k \theta^{2k}}{(2k+1)!}[/tex]
[tex]\Delta \theta_i = \frac x m\;\hbox{ and }\; \theta_i= i\Delta \theta_i =i\frac x m[/tex]
[tex]\Rightarrow\;Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^k\left (i\frac x m\right )^{2k}}{(2k+1)!}\frac x m\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^ki^{2k} x^{2k+1}}{m^{2k+1}(2k+1)!} [/tex]
I don't know how to get the answer, please help.
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