does it defines inner product.. - Electronic Circuits Projects Diagrams Free

transgalactic · 1st October 2009, 01:23 PM

C[0,2]\\
f:[0,2]->c\\
$<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}$

i dont know how to solve the complement.
i tried to prove there easier version without the complement first.

<f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\
A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g( 1)+f(2)g(2)=<f,g>
B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+x f(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg>

C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y( 2))g(2)=<x,g>+<y,g>

step d:
<x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2)

i dont know how to prove that
<x,x> greater or equal 0
i dont know the values of x
??

skyhawk · 2nd October 2009, 03:04 AM

If x(0) is real, then x(0)x(0) > 0 because the square of any real number is positive, unless the number is zero.

Likewise x(1)y(1) > 0 and x(2)x(2) > 0 so <x,x> > 0 unless
x(0) = x(1) = x(2) = 0 in which case <x,x> = 0.

transgalactic · 3rd October 2009, 04:42 PM

how do you know that x(0) is real
i only know that x is continues [0,2]
?

skyhawk · 3rd October 2009, 05:21 PM

If x isn't real then you need the form of inner product that uses the complement (complex conjugate). Any number times its complex conjugate is greater than or equal to zero.

transgalactic · 3rd October 2009, 05:26 PM

ok i tried to make the original question into easier one but i made it harder

actually the original question is:
$<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}$

how to tackle the original one
?

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1st October 2009, 01:23 PM	#1
transgalactic	does it defines inner product..(analisys) C[0,2]\\ f:[0,2]->c\\ $<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}$ i dont know how to solve the complement. i tried to prove there easier version without the complement first. <f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\ A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g( 1)+f(2)g(2)=<f,g> B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+x f(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg> C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y( 2))g(2)=<x,g>+<y,g> step d: <x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2) i dont know how to prove that <x,x> greater or equal 0 i dont know the values of x ?? Last edited by transgalactic; 3rd October 2009 at 05:24 PM.

2nd October 2009, 03:04 AM	#2
skyhawk	If x(0) is real, then x(0)x(0) > 0 because the square of any real number is positive, unless the number is zero. Likewise x(1)y(1) > 0 and x(2)x(2) > 0 so <x,x> > 0 unless x(0) = x(1) = x(2) = 0 in which case <x,x> = 0. Last edited by skyhawk; 2nd October 2009 at 03:05 AM.

3rd October 2009, 04:42 PM	#3
transgalactic	how do you know that x(0) is real i only know that x is continues [0,2] ? Last edited by transgalactic; 3rd October 2009 at 04:53 PM.

3rd October 2009, 05:21 PM	#4
skyhawk	If x isn't real then you need the form of inner product that uses the complement (complex conjugate). Any number times its complex conjugate is greater than or equal to zero.

3rd October 2009, 05:26 PM	#5
transgalactic	ok i tried to make the original question into easier one but i made it harder actually the original question is: $<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}$ how to tackle the original one ? Last edited by transgalactic; 3rd October 2009 at 05:28 PM.