J jzellt Feb 2008 535 4 May 9, 2010 #1 Give the order of each of the zeros of the function: sinz / z Can someone please expalin or show the steps needed to do this? Thanks in advance...

Give the order of each of the zeros of the function: sinz / z Can someone please expalin or show the steps needed to do this? Thanks in advance...

chisigma MHF Hall of Honor Mar 2009 2,162 994 near Piacenza (Italy) May 9, 2010 #2 Considering the expansion of the function as 'infinite product'... \(\displaystyle \frac{\sin z}{z} = (1-\frac{z}{\pi}) (1+\frac{z}{\pi}) (1-\frac{z}{2 \pi}) (1+\frac{z}{2\pi}) \dots\) (1) ... You can easily verity that the zeroes are at \(\displaystyle z = k \pi\) with \(\displaystyle k \ne 0\) and each of them has order 1... Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\)

Considering the expansion of the function as 'infinite product'... \(\displaystyle \frac{\sin z}{z} = (1-\frac{z}{\pi}) (1+\frac{z}{\pi}) (1-\frac{z}{2 \pi}) (1+\frac{z}{2\pi}) \dots\) (1) ... You can easily verity that the zeroes are at \(\displaystyle z = k \pi\) with \(\displaystyle k \ne 0\) and each of them has order 1... Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\)

J jzellt Feb 2008 535 4 May 10, 2010 #3 Thanks for the quick reply. I guess I'm really behind here but how do you find that sinz/z = (1 - z/pi)(1 + z/pi)(1 - z/2pi)... ? Also, what tells you that the zeros are of order 1,2,3,...? Thanks Last edited: May 10, 2010

Thanks for the quick reply. I guess I'm really behind here but how do you find that sinz/z = (1 - z/pi)(1 + z/pi)(1 - z/2pi)... ? Also, what tells you that the zeros are of order 1,2,3,...? Thanks

H HallsofIvy MHF Helper Apr 2005 20,249 7,909 May 10, 2010 #4 His point was that if you write it as such an infinite product, you can see that each zero gives one factor and so has order one.

His point was that if you write it as such an infinite product, you can see that each zero gives one factor and so has order one.