Question

Verify the Cauchy-Schwarz Inequality f(x)=x g(x)=cos(pi*x) Inner product integral is inside Verify the Cauchy-Schwarz Inequality f(x)=x g(x)=cos(pi*x) Inner product integral is inside @Mathematics

Answer 1

\[<f,g>=\int\limits_{-\pi}^{\pi}f(x)g(x)dx\]

Answer 2

I have a feeling that it might be a step in which it has to do with odd and even functions that i have forgotten

Answer 3

so <f,g> =\[\frac{x}{\pi}\sin(\pi^2)+\frac{\cos(\pi x)}{\pi^2}\] evaluated at -pi to pi

Answer 4

however is there anyway to simplify it

Answer 5

the right hand side of your equation is equal to 0 as
x*cos(pi*x) is an odd function, and the limits of the integral are from -pi to pi,

hope it helps

Answer 6

how it asks for cauchy schwarz... and that is not whatsoever cauchy

Answer 7

\[||<f,g>||\le||f|| *||g||\]

Answer 8

i don't believe that the left side is negative but i'm looking to simplify it

Answer 9

the left hand side in that cauchy equation will become zero, upon putting those functions, and the rhs however will not be zero because of the modulous function,
hence cauchy schwarz is verified

Answer 10

um how?

Answer 11

are just put the values
fx =x
gx= cospix
in that equation

Answer 12

alright and what do you get after that

Answer 13

we get 0 on the left hand side of the equation

Answer 14

i meant what do you get aftr integration as how do i know whether your integration is correct

Answer 15

\[x\cos(\pi x)\] is an odd function

Answer 16

u see the right hand side has modulous function, that means that the graph of the fx and the gx has to remain above the x axis, and hence when are under the graph is qualitatively analised it will come out to be a positive value, which is in any case greater zero

Answer 17

got it outcast????

Answer 18

\[2\sin(\pi^2)+\frac{\cos(\pi^2))}{\pi^2}-\frac{\cos(-\pi^2)}{\pi^2}\]

Answer 19

as it's going to be on a test, i can't just say that this is what happens?

Answer 20

for any a

\[\int\limits_{-a}^{a}x\cos(\pi x)dx=0\]

Answer 21

don't work so hard...use the fact that \[x\cos(\pi x)\] is odd

Answer 22

i need to show the work but i figured out what it was

Answer 23

\[\sin(\pi^2)+\sin(-\pi^2)=\sin(\pi^2)-\sin(\pi^2)=0\]

however what about getting rid of the cosine function... isn't cosine even?

Answer 24

o wait

\[\frac{\cos(\pi^2))}{\pi^2}-\cos(\pi^2)/\pi^2\]

Answer 25

i can't just say it's is odd i'd get marked down by my teacher he's very picky on showing work even if it's considered odd

Answer 26

been a while since i've used the odd and even function properties

Answer 27

see outcast , all u need to do do in this question is to verify whether the combination of the function is odd or even.
and let me tell u , u need to work hard on that part first an then solve these problems which involve multiple concepts.

so just grab any good calculus book, and solve problems until u develop some confidence in that particular area,

thats my word of advice

Answer 28

I have taken calculus i'm in calculus 3 right now, and I hardly ever see problems such as these and when we do them, It usually is a full through work out. We were never given the rule about odd function from -a to a simply on the idea (i'm guessing) that you can figure it out with analysis instead reference to the function and integral