products-to-sum-question (1 Viewer)

[kooltrainer]

heres the question
i dunno why i cant get 0 if m don't equal to n

 

[ssglain]

Where's this question from?

I think either way the integral is 0.

Let f(x) = sin(mx)
f(-x) = sin(-mx) = -sin(mx)
f(-x) = -f(x)
.: odd function

Let g(x) = cos(nx)
g(-x)=cos(-nx)=cos(nx)
g(-x)=g(x)
.: even function

We know that odd function x even function = odd function. We also know that the integral of any odd function from -a to a is 0. So the given integral has to be 0.

Simply integrating the given function will confirm this. After letting m = n, remember to rewrite sin(mx)cos(mx) = (1/2)*sin(2mx) using the double angle result for sine. The answer's 0, not pi.

 

[kony]

We know that odd function x even function = odd function.

1.

(a) By considering an odd f(x) and an even g(x), or otherwise, prove this result [1]
(b) Hence show that x²arcsinx is odd. [2]
(b) hence, deduce the Reimann Hypothesis. [117]

hey how's it going.

 

[kooltrainer]

1.

(a) By considering an odd f(x) and an even g(x), or otherwise, prove this result [1]
(b) Hence show that x²arcsinx is odd. [2]
(b) hence, deduce the Reimann Hypothesis. [117]

hey how's it going.

wot daa, dunno wots reimann hypothesis or arcsinx

Where's this question from?

I think either way the integral is 0.

Let f(x) = sin(mx)
f(-x) = sin(-mx) = -sin(mx)
f(-x) = -f(x)
.: odd function

Let g(x) = cos(nx)
g(-x)=cos(-nx)=cos(nx)
g(-x)=g(x)
.: even function

We know that odd function x even function = odd function. We also know that the integral of any odd function from -a to a is 0. So the given integral has to be 0.

Simply integrating the given function will confirm this. After letting m = n, remember to rewrite sin(mx)cos(mx) = (1/2)*sin(2mx) using the double angle result for sine. The answer's 0, not pi.

this is from the products-to-sum chapter in the cambridge book ... u need to erm.. change sinmxcosnx to 0.5 sin(mx+nx)+cos(mx-nx)
after some manipulation u'll get pi .. but i reckon your method is also correct because i got 0 as well .. its suppose to be pi though =l

 

[ssglain]

(a) By considering an odd f(x) and an even g(x), or otherwise, prove this result [1]
(b) Hence show that x²arcsinx is odd. [2]
(b) hence, deduce the Reimann Hypothesis. [117]

Man, you crack me up.


@ kooltrainer:
Anyway, if he insists on rewriting the given integral using products-to-sums:
sin(mx)cos(nx) = 0.5{sin[(m + n)x] + sin[(m - n)x]}

It's still an odd function made up of odd bits, regardless of whether or not m=n or how you manipulate it.

Perhaps you'd like to give Mr Pender a buzz.