Please solve this problem

Otis

Elite Member
Joined
Apr 22, 2015
Messages
3,682
But cos (pi^2) not equal ( cos pi )^2
That's true, and your exercise uses the latter. (It's called 'function notation'.)

\(\displaystyle \frac{1}{\cos(m\pi)^2} = \frac{1}{\cos(m\pi) \cdot \cos(m\pi)}\)



Evaluate factors [imath]\frac{1}{\cos(m\pi)^2}[/imath] for m=1, m=2, m=3 and m=4.

\(\displaystyle \frac{1}{\cos(\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2} \times \frac{1}{\cos(4\pi)^2}\)

How does the aggregate product change, as m increases? Using that pattern, extrapolate to m=1000.

😎
 

darongkun

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Joined
Oct 18, 2021
Messages
12
This is my understanding.
 

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Dr.Peterson

Elite Member
Joined
Nov 12, 2017
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12,598
This is my understanding.
Is that just your uninformed opinion which you are setting against those of more experienced people, or were you told this?

The way I read it, [imath]\cos(\pi x)[/imath] is function notation, and must be evaluated before applying the exponent. So [imath]\cos(\pi x)^2[/imath] means [imath][\cos(\pi x)]^2[/imath]. It would have been better if they had written the latter explicitly, but I think anyone with experience will take it this way.

In addition, if you interpret it this way, the problem becomes easily solvable! Otherwise it is not. Sometimes what you need to do is to try one interpretation (especially if someone with knowledge tells you to) and see what happens, then try another to see how it compares. Have you at least tried doing what you've been told? If not, why did you bother to ask here?
 
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