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|dw:1332896219366:dw| You would need to finish the proof using mathematical induction. Show true for the first case...n=1|dw:1332896419560:dw| Assume true for n=k Prove true for n=k+1
Nothing wrong with his logic...he did not of course allude to a formal proof...as you stated....he indicated a more "logical acceptance" as shown with the first few cases I indicated....
I graphed \[1 - \cos (\Pi/x) - (\Pi/x)\] If what you are saying is correct, I will have to get a function which is always negative, right? then I saw that after 7 it becomes positive. You can graph it here http://www.coolmath.com/graphit/ and then you will get something like this: |dw:1332920038619:dw| then I tried this: \[1 - \sin (\Pi / 9) - (\Pi / 9) \] I got ~ 0.3089 so we should not be able to prove what you claimed. Let me know if I am wrong. I tried to prove it but I could not. :( then I found counter example.
I did not know how to edit my post, but thank you so much for the reply.
Hi alirez,...you must look at the interval...the domain from 0 to 1 and in the proof, where n is greater than or equal to 1.
Consider values in the first quadrant.
I found my mistake, I graphed 1−sin(Π/9)−(Π/9) and not 1−cos(Π/9)−(Π/9) so I tried again and you are right. I consider that x is greater than zero but I was graphing a wrong function. Now I will try to prove this: if n=1, 1−cos(Π) < (Π) then I have to prove n=k+1, 1 - cos(Π/(k+1)) < (Π/(k+1)) seems hard but give me some time to prove it. Thanks
these are ways I tried and I could not get answer 1) first step I will take this: 1−cos(Π/K) < (Π/K) now if I change K to K+1 cos(Π/K+1) is increasing, therefore 1- cos(Π/K+1) is decreasing. so I can conclude that: 1- cos(Π/K+1) < 1−cos(Π/K) < (Π/K) then: 1- cos(Π/K+1) < (Π/K) I stop here and I will start something else considering (Π/K), then we add a one to K. we will get (Π/K+1). (Π/K+1) is decreasing, therefore (Π/K+1) < (Π/K) So I have to prove (Π/K+1) > 1−cos(Π/K) which I tried few ways and I could not prove it. 2) tried all trigonometric formula, I had in mind and could get nothing 3) I tried to prove 1−cos(Π/k)<(Π/K) [1−cos(Π/x)] * [1+ cos(Π/x)] <(Π/x) * [1+ cos(Π/x)] 1- (cos(Π/x))^2 < (Π/x) + (Π/x) * cos(Π/x) (sin(Π/x))^2 < (Π/x) + (Π/x) * cos(Π/x) Once I read that sin (x), x approaches 0, then sin(x)=x (not sure where it is coming from) if I get that in mind I can prove this for when x is getting higher we know that sin (x) > (sin (x))^2 if sin (x)>0 and here sin (x) is greater than zero so we know that sin(Π/x) < (Π/x) and we can say (sin(Π/x))^2 < (Π/x) and then we know that (Π/x) * cos(Π/x) is a positive number so adding a positive value to greater side it will just make it greater. so, (sin(Π/x))^2 < (Π/x) + (Π/x) * cos(Π/x)
i dont agree to the above graph it will be ^^
Rohangrr what is that graph. why you don't agree with that?
@PROSS Could you help me to prove it using induction?
Hi Alireza, Start by looking at my first post. Show (1-cos(pi/1) is less than or equal to pi/1 is true (Note: The sum of the terms on the left will be less than or equal to the sum of the terms on the right.) Assume that (1-cos(pi/1))+(1-cos(pi/2))+...+(1-cos(pi/k)) is less than or equal to pi/1+pi/2+...pi/k is true. Prove that (1-cos(pi/1))+(1-cos(pi/2))+...+(1-cos(pi/k))+(1-cos(pi/(k+1)) is less than or equal to pi/1+pi/2+...pi/k+pi/(k+1) is true. then substitute... pi/1+pi/2+...+pi/k +(1-cos(pi/k+1) is less than or equal to pi/1+pi/2+...+pi/k+pi/(k+1). Then subtract equivalent values on both sides.... The leaves: (1-cos(pi/(k+1)) is less than or equal to pi/(k+1) Which is what were wanting to prove.
I have to digest what you just explained. Thanks for reply. :)