Mixing
integrating an expression with alomst quadratic denominator including Chebyshev polynomials of the first kind
$begingroup$
As a part of my research (in array processing - the specifics are not too exciting :-)), I cant resolve one specific integral.
Assume $rinleft[0 ,1right]$, $Ninmathbb{N}$.
The basic form is
$$
int_{cos{(x)}=-1}^{cos{(x)}=1}{
frac{
1-cos{left(Nxright)}
}{
N^{2}left(1-cos{(x)}right)
+r^{2}left(1-cos{left(Nxright)}right)
-Nrleft(1-cos{(x)}-cos{left(Nxright)}+cos{left(left(N-1right)xright)}right)
}
dx
}.
$$
Searching for hints, I came across the very informative "How to expand $cos{(nx)}$ with $cos{(x)}$?" discussion which lead me to investigate the expression using Chebyshev polynomials of the first kind, in light of the fact that $cos(nu)=T_n(cos(u)),$ but still no luck.
Anyhow, as can be seen in the wikipedia page (Chebyshev polynomials ),
$$
T_n(x)= sum_{k=0}^{lfloor n/2rfloor} binom{n}{2k} (x^2-1)^k x^{n-2k},
$$
and plugging it into the basic integral results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nrleft(1-x-T_N(x)+T_{N-1}(x)right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
}.
$$
Considering also the Chebyshev polynomials of the second kind,
$$
U_{N}left(xright)=frac{left(x+sqrt{x^{2}-1}right)^{N+1}-left(x-sqrt{x^{2}-1}right)^{N+1}}{2sqrt{x^{2}-1}},
$$
and the equality
$$
T_{N+1}(x) = xT_{N}(x)-left(1-x^{2}right)U_{N-1}(x)
$$
results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nr
left(
1-x+sqrt{1-x^{2}}
left(left(x+sqrt{x^{2}-1}right)^{N}-left(x-sqrt{x^{2}-1}right)^{N}right)
right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
},
$$
which has (as the headline states) "almost" quadratic form of the denominator.
This is where I am stuck and if someone has an idea for solving it I will be thrilled.
Thank you.
integration polynomials quadratic-forms
asked Jan 31 at 18:27
ItayyekaItayyeka
165
$endgroup$
add a comment |
$begingroup$
As a part of my research (in array processing - the specifics are not too exciting :-)), I cant resolve one specific integral.
Assume $rinleft[0 ,1right]$, $Ninmathbb{N}$.
The basic form is
$$
int_{cos{(x)}=-1}^{cos{(x)}=1}{
frac{
1-cos{left(Nxright)}
}{
N^{2}left(1-cos{(x)}right)
+r^{2}left(1-cos{left(Nxright)}right)
-Nrleft(1-cos{(x)}-cos{left(Nxright)}+cos{left(left(N-1right)xright)}right)
}
dx
}.
$$
Searching for hints, I came across the very informative "How to expand $cos{(nx)}$ with $cos{(x)}$?" discussion which lead me to investigate the expression using Chebyshev polynomials of the first kind, in light of the fact that $cos(nu)=T_n(cos(u)),$ but still no luck.
Anyhow, as can be seen in the wikipedia page (Chebyshev polynomials ),
$$
T_n(x)= sum_{k=0}^{lfloor n/2rfloor} binom{n}{2k} (x^2-1)^k x^{n-2k},
$$
and plugging it into the basic integral results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nrleft(1-x-T_N(x)+T_{N-1}(x)right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
}.
$$
Considering also the Chebyshev polynomials of the second kind,
$$
U_{N}left(xright)=frac{left(x+sqrt{x^{2}-1}right)^{N+1}-left(x-sqrt{x^{2}-1}right)^{N+1}}{2sqrt{x^{2}-1}},
$$
and the equality
$$
T_{N+1}(x) = xT_{N}(x)-left(1-x^{2}right)U_{N-1}(x)
$$
results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nr
left(
1-x+sqrt{1-x^{2}}
left(left(x+sqrt{x^{2}-1}right)^{N}-left(x-sqrt{x^{2}-1}right)^{N}right)
right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
},
$$
which has (as the headline states) "almost" quadratic form of the denominator.
This is where I am stuck and if someone has an idea for solving it I will be thrilled.
Thank you.
integration polynomials quadratic-forms
asked Jan 31 at 18:27
ItayyekaItayyeka
165
$endgroup$
add a comment |
$begingroup$
As a part of my research (in array processing - the specifics are not too exciting :-)), I cant resolve one specific integral.
Assume $rinleft[0 ,1right]$, $Ninmathbb{N}$.
The basic form is
$$
int_{cos{(x)}=-1}^{cos{(x)}=1}{
frac{
1-cos{left(Nxright)}
}{
N^{2}left(1-cos{(x)}right)
+r^{2}left(1-cos{left(Nxright)}right)
-Nrleft(1-cos{(x)}-cos{left(Nxright)}+cos{left(left(N-1right)xright)}right)
}
dx
}.
$$
Searching for hints, I came across the very informative "How to expand $cos{(nx)}$ with $cos{(x)}$?" discussion which lead me to investigate the expression using Chebyshev polynomials of the first kind, in light of the fact that $cos(nu)=T_n(cos(u)),$ but still no luck.
Anyhow, as can be seen in the wikipedia page (Chebyshev polynomials ),
$$
T_n(x)= sum_{k=0}^{lfloor n/2rfloor} binom{n}{2k} (x^2-1)^k x^{n-2k},
$$
and plugging it into the basic integral results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nrleft(1-x-T_N(x)+T_{N-1}(x)right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
}.
$$
Considering also the Chebyshev polynomials of the second kind,
$$
U_{N}left(xright)=frac{left(x+sqrt{x^{2}-1}right)^{N+1}-left(x-sqrt{x^{2}-1}right)^{N+1}}{2sqrt{x^{2}-1}},
$$
and the equality
$$
T_{N+1}(x) = xT_{N}(x)-left(1-x^{2}right)U_{N-1}(x)
$$
results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nr
left(
1-x+sqrt{1-x^{2}}
left(left(x+sqrt{x^{2}-1}right)^{N}-left(x-sqrt{x^{2}-1}right)^{N}right)
right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
},
$$
which has (as the headline states) "almost" quadratic form of the denominator.
This is where I am stuck and if someone has an idea for solving it I will be thrilled.
Thank you.
integration polynomials quadratic-forms
asked Jan 31 at 18:27
ItayyekaItayyeka
165
$endgroup$
As a part of my research (in array processing - the specifics are not too exciting :-)), I cant resolve one specific integral.
Assume $rinleft[0 ,1right]$, $Ninmathbb{N}$.
The basic form is
$$
int_{cos{(x)}=-1}^{cos{(x)}=1}{
frac{
1-cos{left(Nxright)}
}{
N^{2}left(1-cos{(x)}right)
+r^{2}left(1-cos{left(Nxright)}right)
-Nrleft(1-cos{(x)}-cos{left(Nxright)}+cos{left(left(N-1right)xright)}right)
}
dx
}.
$$
Searching for hints, I came across the very informative "How to expand $cos{(nx)}$ with $cos{(x)}$?" discussion which lead me to investigate the expression using Chebyshev polynomials of the first kind, in light of the fact that $cos(nu)=T_n(cos(u)),$ but still no luck.
Anyhow, as can be seen in the wikipedia page (Chebyshev polynomials ),
$$
T_n(x)= sum_{k=0}^{lfloor n/2rfloor} binom{n}{2k} (x^2-1)^k x^{n-2k},
$$
and plugging it into the basic integral results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nrleft(1-x-T_N(x)+T_{N-1}(x)right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
}.
$$
Considering also the Chebyshev polynomials of the second kind,
$$
U_{N}left(xright)=frac{left(x+sqrt{x^{2}-1}right)^{N+1}-left(x-sqrt{x^{2}-1}right)^{N+1}}{2sqrt{x^{2}-1}},
$$
and the equality
$$
T_{N+1}(x) = xT_{N}(x)-left(1-x^{2}right)U_{N-1}(x)
$$
results in
$$
int_{-1}^{1}{
frac{
1-T_N(x)
}{
N^{2}left(1-xright)
+r^{2}left(1-T_N(x)right)
-Nr
left(
1-x+sqrt{1-x^{2}}
left(left(x+sqrt{x^{2}-1}right)^{N}-left(x-sqrt{x^{2}-1}right)^{N}right)
right)
}
frac{-1}{sqrt{1-x^{2}}}
dx
},
$$
which has (as the headline states) "almost" quadratic form of the denominator.
This is where I am stuck and if someone has an idea for solving it I will be thrilled.
Thank you.
integration polynomials quadratic-forms
integration polynomials quadratic-forms
asked Jan 31 at 18:27
ItayyekaItayyeka
165
asked Jan 31 at 18:27
ItayyekaItayyeka
165
edited Jan 31 at 20:58
Itayyeka
asked Jan 31 at 18:27
ItayyekaItayyeka
165
asked Jan 31 at 18:27
ItayyekaItayyeka
165
asked Jan 31 at 18:27
ItayyekaItayyeka
165
165
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
A numerically achieved analytical answer:
sweeping many $[N,r]$ sets in MATHEMATICA, revealed that the finite integral (i.e. integration from $x=0$ to $x=2pi$) can be expressed as $$ int_{0}^{2pi}(...) = frac{2pi}{r^{2}-left(N-1right)r+N} $$ which is very nice and perfectly fits the fact that for $r=0$ it should be equal to $2pi/N$.
Now that I have the answer, should someone have an idea for analytical way of solving the integral, I will be very glad to hear about it.
Thanks
Itay.
answered Feb 6 at 20:07
ItayyekaItayyeka
165
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A numerically achieved analytical answer:
sweeping many $[N,r]$ sets in MATHEMATICA, revealed that the finite integral (i.e. integration from $x=0$ to $x=2pi$) can be expressed as $$ int_{0}^{2pi}(...) = frac{2pi}{r^{2}-left(N-1right)r+N} $$ which is very nice and perfectly fits the fact that for $r=0$ it should be equal to $2pi/N$.
Now that I have the answer, should someone have an idea for analytical way of solving the integral, I will be very glad to hear about it.
Thanks
Itay.
answered Feb 6 at 20:07
ItayyekaItayyeka
165
$endgroup$
add a comment |
$begingroup$
A numerically achieved analytical answer:
sweeping many $[N,r]$ sets in MATHEMATICA, revealed that the finite integral (i.e. integration from $x=0$ to $x=2pi$) can be expressed as $$ int_{0}^{2pi}(...) = frac{2pi}{r^{2}-left(N-1right)r+N} $$ which is very nice and perfectly fits the fact that for $r=0$ it should be equal to $2pi/N$.
Now that I have the answer, should someone have an idea for analytical way of solving the integral, I will be very glad to hear about it.
Thanks
Itay.
answered Feb 6 at 20:07
ItayyekaItayyeka
165
$endgroup$
add a comment |
$begingroup$
A numerically achieved analytical answer:
sweeping many $[N,r]$ sets in MATHEMATICA, revealed that the finite integral (i.e. integration from $x=0$ to $x=2pi$) can be expressed as $$ int_{0}^{2pi}(...) = frac{2pi}{r^{2}-left(N-1right)r+N} $$ which is very nice and perfectly fits the fact that for $r=0$ it should be equal to $2pi/N$.
Now that I have the answer, should someone have an idea for analytical way of solving the integral, I will be very glad to hear about it.
Thanks
Itay.
answered Feb 6 at 20:07
ItayyekaItayyeka
165
$endgroup$
A numerically achieved analytical answer:
sweeping many $[N,r]$ sets in MATHEMATICA, revealed that the finite integral (i.e. integration from $x=0$ to $x=2pi$) can be expressed as $$ int_{0}^{2pi}(...) = frac{2pi}{r^{2}-left(N-1right)r+N} $$ which is very nice and perfectly fits the fact that for $r=0$ it should be equal to $2pi/N$.
Now that I have the answer, should someone have an idea for analytical way of solving the integral, I will be very glad to hear about it.
Thanks
Itay.
answered Feb 6 at 20:07
ItayyekaItayyeka
165
edited Feb 7 at 6:54
answered Feb 6 at 20:07
ItayyekaItayyeka
165
answered Feb 6 at 20:07
ItayyekaItayyeka
165
answered Feb 6 at 20:07
ItayyekaItayyeka
165
165
add a comment |
add a comment |
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