question to complex conjugation of a function which is complex for some values in the domain.
$begingroup$
I have some trouble understanding complex conjugation in this specific case. First I will define the function $f_n(x) = x^{n+frac{1}{2}}$ where $n in mathbb{N}$. Then i define a inner product $langle f_n, f_m rangle = int_{-1}^1
(f_n(x))^* f_m(x) dx$ where $a^*$ is the complex conjugated of some complex number a.
So my problem is that for some values of $x$ the function $f_n(x)$ takes complex values, specifically when $x in [-1,0)$. So in order to solve the problem i need to split up the integral and conjugate for only the values where $f_n(x)$ takes complex values?
This is the way I think the innerproduct is solved
$$ langle f_n,f_m rangle = int_0^1 x^{n+frac{1}{2}} x^{m+frac{1}{2}} dx + int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx$$
The first integral is $frac{1}{n+m+2}$. Solving the second integral using the substitution $x=-y$ we get
$$ int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx = int_{1}^0 left((-y)^{n+frac{1}{2}}right)^* (-y)^{m+frac{1}{2}}(-1)dy = int_0^1 left((-1)^{n+1/2} right)^* (-1)^{m+1/2} y^{m+n+1} dy = int_0^1 left(i (-1)^n right)^* i(-1)^{m} y^{m+n+1} dy = (-1)^{n+m} (-i)i frac{1}{n+m+2} = frac{(-1)^{n+m}}{n+m+2} $$
From this we get
$$ langle f_n,f_m rangle = frac{1+(-1)^{n+m}}{n+m+2} $$
Is this correct?
complex-numbers
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
$endgroup$
add a comment |
$begingroup$
I have some trouble understanding complex conjugation in this specific case. First I will define the function $f_n(x) = x^{n+frac{1}{2}}$ where $n in mathbb{N}$. Then i define a inner product $langle f_n, f_m rangle = int_{-1}^1
(f_n(x))^* f_m(x) dx$ where $a^*$ is the complex conjugated of some complex number a.
So my problem is that for some values of $x$ the function $f_n(x)$ takes complex values, specifically when $x in [-1,0)$. So in order to solve the problem i need to split up the integral and conjugate for only the values where $f_n(x)$ takes complex values?
This is the way I think the innerproduct is solved
$$ langle f_n,f_m rangle = int_0^1 x^{n+frac{1}{2}} x^{m+frac{1}{2}} dx + int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx$$
The first integral is $frac{1}{n+m+2}$. Solving the second integral using the substitution $x=-y$ we get
$$ int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx = int_{1}^0 left((-y)^{n+frac{1}{2}}right)^* (-y)^{m+frac{1}{2}}(-1)dy = int_0^1 left((-1)^{n+1/2} right)^* (-1)^{m+1/2} y^{m+n+1} dy = int_0^1 left(i (-1)^n right)^* i(-1)^{m} y^{m+n+1} dy = (-1)^{n+m} (-i)i frac{1}{n+m+2} = frac{(-1)^{n+m}}{n+m+2} $$
From this we get
$$ langle f_n,f_m rangle = frac{1+(-1)^{n+m}}{n+m+2} $$
Is this correct?
complex-numbers
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
$endgroup$
$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29
add a comment |
$begingroup$
I have some trouble understanding complex conjugation in this specific case. First I will define the function $f_n(x) = x^{n+frac{1}{2}}$ where $n in mathbb{N}$. Then i define a inner product $langle f_n, f_m rangle = int_{-1}^1
(f_n(x))^* f_m(x) dx$ where $a^*$ is the complex conjugated of some complex number a.
So my problem is that for some values of $x$ the function $f_n(x)$ takes complex values, specifically when $x in [-1,0)$. So in order to solve the problem i need to split up the integral and conjugate for only the values where $f_n(x)$ takes complex values?
This is the way I think the innerproduct is solved
$$ langle f_n,f_m rangle = int_0^1 x^{n+frac{1}{2}} x^{m+frac{1}{2}} dx + int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx$$
The first integral is $frac{1}{n+m+2}$. Solving the second integral using the substitution $x=-y$ we get
$$ int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx = int_{1}^0 left((-y)^{n+frac{1}{2}}right)^* (-y)^{m+frac{1}{2}}(-1)dy = int_0^1 left((-1)^{n+1/2} right)^* (-1)^{m+1/2} y^{m+n+1} dy = int_0^1 left(i (-1)^n right)^* i(-1)^{m} y^{m+n+1} dy = (-1)^{n+m} (-i)i frac{1}{n+m+2} = frac{(-1)^{n+m}}{n+m+2} $$
From this we get
$$ langle f_n,f_m rangle = frac{1+(-1)^{n+m}}{n+m+2} $$
Is this correct?
complex-numbers
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
$endgroup$
I have some trouble understanding complex conjugation in this specific case. First I will define the function $f_n(x) = x^{n+frac{1}{2}}$ where $n in mathbb{N}$. Then i define a inner product $langle f_n, f_m rangle = int_{-1}^1
(f_n(x))^* f_m(x) dx$ where $a^*$ is the complex conjugated of some complex number a.
So my problem is that for some values of $x$ the function $f_n(x)$ takes complex values, specifically when $x in [-1,0)$. So in order to solve the problem i need to split up the integral and conjugate for only the values where $f_n(x)$ takes complex values?
This is the way I think the innerproduct is solved
$$ langle f_n,f_m rangle = int_0^1 x^{n+frac{1}{2}} x^{m+frac{1}{2}} dx + int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx$$
The first integral is $frac{1}{n+m+2}$. Solving the second integral using the substitution $x=-y$ we get
$$ int_{-1}^0 left(x^{n+frac{1}{2}}right)^* x^{m+frac{1}{2}}dx = int_{1}^0 left((-y)^{n+frac{1}{2}}right)^* (-y)^{m+frac{1}{2}}(-1)dy = int_0^1 left((-1)^{n+1/2} right)^* (-1)^{m+1/2} y^{m+n+1} dy = int_0^1 left(i (-1)^n right)^* i(-1)^{m} y^{m+n+1} dy = (-1)^{n+m} (-i)i frac{1}{n+m+2} = frac{(-1)^{n+m}}{n+m+2} $$
From this we get
$$ langle f_n,f_m rangle = frac{1+(-1)^{n+m}}{n+m+2} $$
Is this correct?
complex-numbers
complex-numbers
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
asked Jan 15 at 21:58
Marcus Kristian NielsenMarcus Kristian Nielsen
1
1
$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29
add a comment |
$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29
$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29
$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29
add a comment |
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$begingroup$
Negative numbers have two square roots, so your inner product is not well defined.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:29