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Shahram S.

asked • 08/03/17

Root existence for a function

I want to prove that the following f(x) always has a root. Due to periodicity, it would have many roots if there exist one. Can anybody help me with this? I tried to use intermediate value theorem but I could not come up with the proof.
 
f(x)=\sum{i=1 to K} (ni/di), where
ni= ai*sin(x+thetai)
di=2+cos(x-thetai)
and ai and thetai are given for every i. We also know that 0<=ai<=1 and -pi<=thetai<=pi .
 

1 Expert Answer

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Andy C. answered • 08/03/17

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Shahram S.

Thank you for the answer but please not that there is a summation over index i \in {1,2,3,...,K}. Actually, it is a summation over K terms and thetai and ai are given for every i. Therefore, we don't have just one theta but there are K of them within the summation. Domain of x is real numbers by the way.
 
 
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08/03/17

Andy C.

Yes, but the angles are restricted to the first and second quadrants.
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08/03/17

Shahram S.

I guess angles are not restricted to the the first and second because each one could be between -\pi and +\pi so they can be in any quadrant.
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08/04/17

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