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# Pre-Calc Trigonometric Identities/Equations Proof

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I'm absolutely cluless on what to do and where to start this proof.

Recall that a function is periodic of period T if f(x+T) = f(x) for all x in the domain of f. The smallest positive value of T is called the fundamental period.

Given that f(x) = pi(x)^2-cos(x).

Show that f(x) is not a periodic function.

Hint: Assume f(x) is a periodic function of period T, where T > 0. Then, f(x+T) = f(x) for all x in the domain of f. Now, try to obtain a contradiction to the given assumption.

Here is the picture of the question: https://drive.google.com/file/d/1D37S5IxnONwUkm0SEFBmrTZ2h8drhB9M/view?usp=sharing

May 31, 2021

#1
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I'm going to try my best, but since I'm not too familiar with these questions, it might be wrong.

I'm going to follow the hint, and assume that f(x) is a periodic function of period T and try to prove that this equation is impossible.

f(x + T) = f(x)

pi*(x + T)^2 - cos(x + T) = pi*(x)^2 - cos(x)

pi*x^2 + 2*pi*xT + pi*T^2 - cos(x + T) = pi*(x)^2 - cos(x)

2*pi*xT + pi*T^2 - cos(x + T) = - cos(x)

2*pi*T(x + T) + cos(x) = cos(x + T)

2*pi*T(x + T) + cos(x) = cos(x) * cos(T) - sine(x) * sine(T)

And... now I'm stuck. :((

Maybe someone else can give help?

I'm really sorry that I wasn't able to help more.

=^._.^=

May 31, 2021
#2
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I think you might be correct but I'm not too sure. If we wanted to make the proof impossible, we need an expression such as a positive equals a negative and/or non matching sides. So I do believe that you can't go further for your answer besides simplying both sides if applicable.

The answer key given for this particular question is a dead end for me (can't see the equations), so I have to rely on others so I can understand this. There can be different solutions to this, so any general proof would work otherwise.

Thanks for the help though! I will wait until some other users answer this so I can get more feedback and compare.

stressedmathkid123  Jun 1, 2021
#3
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I actually figured it out. My phrasing is probably a bit off but it makes sense (to me at least)

We end up with 2pi(x)T + pi(T)^2 - cos(x+T) = -cos(x) after some simplifying, and distributing/expanding.

If we let x = 0

Then we end up with pi(T)^2 - cos(T) = -1. This makes our proof impossible. It's a contradiction because when we let x = 0, the function is not periodic.

UNLESS, if we let T = 0, then our proof is valid. We end up with -cos(x) = -cos(x) which then makes our function periodic. However, this isn't what we are looking for, as we aren't trying to find what makes the function periodic.

Jun 1, 2021
#4
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Omgosh, thank you.

Do you mind explaining how you got to 2pi(x)T + pi(T)^2 - cos(x+T) = -cos(x)?

Setting x as 0 is so smart. :))

=^._.^=

catmg  Jun 1, 2021
#5
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Yeah, sure!

1) Since we know that f(x+T) = f(x)

2) and f(x) = pi(x)^2 - cos (x), we can plug in f(x+T) for anywhere there is an f(x) for our manipulated side (left).

Since this is a proof, there are going to be two sides.

3) Then we get pi(x+T)^2 - cos(x+T) = pi(x)^2 - cos(x). Our left side going to be manipulated while our right side is going to be our "constant" with the original expression pi(x)^2 - cos(x).

4) We can expand the expression which gives us pi(x^2+2(x)T+T^2) - cos(x+T) = pi(x)^2 - cos(x)

5) Then we can foil the pi, which gives us pi(x)^2+ 2pi(x)T + pi(T)^2 - cos(x+T) = pi(x)^2 - cos(x)

6) We can then subtract pi(x)^2 from both sides which will leave us with 2pi(T)x + pi(T)^2 - cos(x+T) = -cos(x)

I might've switched the T and the x in some places (such as 2pi(x)T) compared from my previous answer (where it was 2pi(T)x), but it doesn't affect the actual proof. We can still test our proof just fine in this case. Where you arrange the x and T's is mostly whatever is easier to understand. It's multiplication, so it's fine to flip them generally. It's like 4x5 is the same as 5x4.

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I already explained this earlier but here's a more detailed explanation for the contradiction of the proof.

7) Testing the proof:

Now since we are given the hint that f(x) is a periodic function of period T when T > 0 for all x in the domain of f. We can test T = 0. Plug in 0 where ever there's a T and we get -cos(x) = -cos(x). This isn't what we are looking for because this shows that f(x) is a periodic function when T = 0 and is also a valid proof. Not a contradiction.

So since we need to show that f(x) is not a periodic function, we need to make the proof impossible. Since we tried T = 0, we can also test x = 0. Plug in 0 where ever there's an x, and we get pi(T)^2 - cos(T) = -1. This shows that f(x) is not a periodic function when x = 0 and is not a valid proof. It's a contradiction.

Solution: The function is not periodic when x = 0. Unless when T = 0, the function is periodic.

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I actually forgot to show the expanding and foiling part on my actual assignment since I did it on a separate piece of paper. So it looked like I skipped two steps lol.

stressedmathkid123  Jun 1, 2021