Proving cos2(x)/(n2 + 1) ≤ 1/(n2 + 1) - Proof and Reasoning

[Miike012]

I want to prove
cos²(x)/(n² + 1) ≤ 1/(n² + 1)

I know this is an obvious inequality but I want to know if my reasoning is correct.

For the expression cos²(x)/(n² + 1) to be as large as possible the numerator must → ∞ but cos²(x) is bounded above by 1.

cos²(x) = 1 for x = 2∏k where k ≥1 is an integer.

cos²(2∏k)/((2∏k)² + 1) = 1/ ((2∏k)² + 1) for integers k ≥ 1.

Now I want to compare n² + 1 to (2∏k)² + 1:

n² ≤ (2∏k)² where n and k are consecutive integers from 1 to ∞.

n² + 1 ≤ (2∏k)² + 1
and
1/(n² + 1) ≥ 1/ ((2∏k)² + 1 )

so cos²(x)/(n² + 1) ≤ 1/(n² + 1)

 

[Poopsilon]

This is unnecessarily complicated, if you can assume the identity (cos(x))^2 ≤ 1, then just divide both sides by (n^2 + 1)

 

[Miike012]

Yes I know that, I just wanted to see if I was correct in what I was saying.

 

[Mark44]

I want to prove
cos²(x)/(n² + 1) ≤ 1/(n² + 1)

I know this is an obvious inequality but I want to know if my reasoning is correct.

For the expression cos²(x)/(n² + 1) to be as large as possible the numerator must → ∞ but cos²(x) is bounded above by 1.

The numerator can't approach infinity

cos²(x) = 1 for x = 2∏k where k ≥1 is an integer.

cos²(2∏k)/((2∏k)² + 1) = 1/ ((2∏k)² + 1) for integers k ≥ 1.

Now I want to compare n² + 1 to (2∏k)² + 1:

n² ≤ (2∏k)² where n and k are consecutive integers from 1 to ∞.

Do you mean that if n = 2, k = 3, and if n = 3, k = 4? I get the idea that what you said isn't what you meant.

n² + 1 ≤ (2∏k)² + 1
and
1/(n² + 1) ≥ 1/ ((2∏k)² + 1 )

so cos²(x)/(n² + 1) ≤ 1/(n² + 1)

As already noted by poopsilon, what you have is much more complicated than what is needed, not to mention unclear. For any integer n (and for that matter any real number), -1 ≤ cos²(n) ≤ 1. This idea and the "squeeze" theorem are all you need to establish the inequality you started with.

 

[Miike012]

Mark44

"The numerator can't approach infinity"


Miike012
I know this that is why I said...
" but cos^2(x) is bounded above by 1."

Mark44

"Do you mean that if n = 2, k = 3, and if n = 3, k = 4? I get the idea that what you said isn't what you meant."


Miike012
"where n and k are consecutive integers from 1 to ∞."

So if I have the following inequality
n^2 ≤ (2∏k)^2

Then if n and k are consecutive integers from 1 to ∞ what I am meaning to say is...

when n is 1 k is 1
when n is 2 k is 2
n = 3, k = 3...

so,
(1)^2 ≤ (2∏(1))^2

(2)^2 ≤ (2∏(2))^2

(3)^2 ≤ (2∏(3))^2

And so on...

 

[Mark44]

Miike012
"where n and k are consecutive integers from 1 to ∞."

So if I have the following inequality
n^2 ≤ (2∏k)^2

Then if n and k are consecutive integers from 1 to ∞ what I am meaning to say is...

when n is 1 k is 1
when n is 2 k is 2
n = 3, k = 3...

Since n and k are equal at each step, there's no need for two variables. What you wrote was very confusing. You could have said
n² ≤ (2$\pi n)^2$, for n = 1, 2, 3, ...