Mixing
The Integral $int frac {dx}{(x^2-2ax+b)^n}$
$begingroup$
Recently I came across this general integral,
$$int frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±sqrt {a^2-b} = a±sqrt {∆}$$
Hence the integrand can be written as,
$$
frac {1}{(x^2-2ax+b)^n}
=
frac {1}{(x-a-sqrt ∆)^n(x-a+sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
frac {1}{(x^2-2ax+b)^n}
=
sum frac {A_r}{(x-a-sqrt ∆)^r} + sum frac {B_r}{(x-a+sqrt ∆)^r}
$$
Putting $-frac {1}{2sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$kappa log left( frac {x-a-sqrt ∆}{x-a+sqrt ∆}right) + frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} binom {2n-r-1}{n-1} D^{2n-r}$$
calculus indefinite-integrals partial-fractions
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
$endgroup$
|
show 1 more comment
$begingroup$
Recently I came across this general integral,
$$int frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±sqrt {a^2-b} = a±sqrt {∆}$$
Hence the integrand can be written as,
$$
frac {1}{(x^2-2ax+b)^n}
=
frac {1}{(x-a-sqrt ∆)^n(x-a+sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
frac {1}{(x^2-2ax+b)^n}
=
sum frac {A_r}{(x-a-sqrt ∆)^r} + sum frac {B_r}{(x-a+sqrt ∆)^r}
$$
Putting $-frac {1}{2sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$kappa log left( frac {x-a-sqrt ∆}{x-a+sqrt ∆}right) + frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} binom {2n-r-1}{n-1} D^{2n-r}$$
calculus indefinite-integrals partial-fractions
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
$endgroup$
$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
2
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46
|
show 1 more comment
$begingroup$
Recently I came across this general integral,
$$int frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±sqrt {a^2-b} = a±sqrt {∆}$$
Hence the integrand can be written as,
$$
frac {1}{(x^2-2ax+b)^n}
=
frac {1}{(x-a-sqrt ∆)^n(x-a+sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
frac {1}{(x^2-2ax+b)^n}
=
sum frac {A_r}{(x-a-sqrt ∆)^r} + sum frac {B_r}{(x-a+sqrt ∆)^r}
$$
Putting $-frac {1}{2sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$kappa log left( frac {x-a-sqrt ∆}{x-a+sqrt ∆}right) + frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} binom {2n-r-1}{n-1} D^{2n-r}$$
calculus indefinite-integrals partial-fractions
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
$endgroup$
Recently I came across this general integral,
$$int frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±sqrt {a^2-b} = a±sqrt {∆}$$
Hence the integrand can be written as,
$$
frac {1}{(x^2-2ax+b)^n}
=
frac {1}{(x-a-sqrt ∆)^n(x-a+sqrt ∆)^n}
$$
Resolving into partial fractions we have,
$$
frac {1}{(x^2-2ax+b)^n}
=
sum frac {A_r}{(x-a-sqrt ∆)^r} + sum frac {B_r}{(x-a+sqrt ∆)^r}
$$
Putting $-frac {1}{2sqrt ∆} = D$ , I could produce a table of the coefficients $A$ and $B$ for different $n$.
par
For $n=1$,
$$A_1=-D , B_1=D$$
For $n=2$,
$$A_1=2D^3 , B_1=-2D^3$$
$$A_2=D^2 , B_2 = D^2$$
For $n=3$,
$$A_1=-6D^5 , B_1=6D^5$$
$$A_2=-3D^4 , B_2 = -3D^4$$
$$A_3=-D^3, B_3=D^3$$
For $n=4$,
$$A_1=20D^7, B_1=-20D^7$$
$$A_2=10D^6 , B_2 = 10D^6$$
$$A_3=4D^5, B_3=-4D^5$$
$$A_4=D^4, B_4=D^4$$
For $n=5$,
$$A_1=-70D^9, B_1=70D^9$$
$$A_2=-35D^8, B_2 = -35D^8$$
$$A_3=-15D^7, B_3=15D^7$$
$$A_4=-5D^6, B_4=-5D^6$$
$$A_5=-D^5, B_4=D^5$$
Yet I am unable to deduce a general formula for the coefficients. If I have the coefficients, the integral is almost solved , for then I shall have a logarithmic term and a rational function in $x$. More directly, I seek a result of the form,
$$kappa log left( frac {x-a-sqrt ∆}{x-a+sqrt ∆}right) + frac {P(x)}{Q(x)}$$
Any help would be greatly appreciated.
Conjecture 1(Proved below)
$$A(n,r)= (-1)^n binom {2n-r-1}{n-1} D^{2n-r}$$
$$B(n,r)= (-1)^{n-r} binom {2n-r-1}{n-1} D^{2n-r}$$
calculus indefinite-integrals partial-fractions
calculus indefinite-integrals partial-fractions
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
edited Jan 26 at 4:00
Awe Kumar Jha
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
asked Jan 25 at 8:39
Awe Kumar JhaAwe Kumar Jha
516113
516113
$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
2
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46
|
show 1 more comment
$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
2
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46
$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
2
2
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-sqrt{Delta}$ and later defining $D=frac1{2sqrt{Delta}}$, we get
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{z^n(z+2sqrt{Delta})^n}=frac1{z^n}cdotfrac{(2sqrt{Delta})^{-n}}{(1+frac{z}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{(-D)^n}{z^n(1-Dz)^n} = frac{(-D)^n}{z^n}sum_{j=0}^{infty} binom{n+j-1}{j}D^jz^j\
&=(-1)^nsum_{j=0}^{infty}binom{n+j-1}{j}D^{n+j}z^{j-n}end{align*}
We claim that the coefficients $(-1)^nbinom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $frac1{z^{n-j}}$ in the partial fractions expansion of $frac1{z^n(z+2sqrt{delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^nbinom{2n-k-1}{n-k}D^{2n-k}=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$frac1{z^n(z+2sqrt{Delta})^n}=sum_{k=1}^n frac{A(n,k)}{z^k} +sum_{k=1}^n frac{B(n,k)}{(z+2sqrt{Delta})^k}=sum_{k=1}^n frac{A(n,k)}{(x-a-sqrt{Delta})^k} +sum_{k=1}^n frac{B(n,k)}{(x-a+sqrt{Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $frac1{z-apmsqrt{Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{(w-2sqrt{Delta})^nw^n}=frac1{w^n}cdotfrac{(-2sqrt{Delta})^{-n}}{(1-frac{w}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{D^n}{w^n(1+Dw)^n} = frac{D^n}{w^n}sum_{j=0}^{infty} binom{n+j-1}{j}(-D)^jw^j\
&=sum_{j=0}^{infty}(-1)^jbinom{n+j-1}{j}D^{n+j}w^{j-n}end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
$endgroup$
add a comment |
$begingroup$
Let $bneq a^2$,
$$S(n)=int frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-left( 2 A n-C-3 Aright) , {{x}^{2}}-left( left( 2 B-2 A aright) n+left( 2 C+4 Aright) a-2 Bright) x\+2 B a n-left( -C-Aright) b-2 B a$$
$$A=frac{1}{2 left( b-{{a}^{2}}right) , left( n-1right) },B=-frac{a}{2 left( b-{{a}^{2}}right) , left( n-1right) },\C=frac{2 n-3}{2 left( b-{{a}^{2}}right) , left( n-1right) }$$
Then
$$S(n)=frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
frac{2n-3}{2(n-1)(b-a^2)}S(n-1), ; n>1$$
$$S(1)=int frac {dx}{x^2-2ax+b}$$
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
$endgroup$
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
add a comment |
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2 Answers
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2 Answers
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$begingroup$
All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-sqrt{Delta}$ and later defining $D=frac1{2sqrt{Delta}}$, we get
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{z^n(z+2sqrt{Delta})^n}=frac1{z^n}cdotfrac{(2sqrt{Delta})^{-n}}{(1+frac{z}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{(-D)^n}{z^n(1-Dz)^n} = frac{(-D)^n}{z^n}sum_{j=0}^{infty} binom{n+j-1}{j}D^jz^j\
&=(-1)^nsum_{j=0}^{infty}binom{n+j-1}{j}D^{n+j}z^{j-n}end{align*}
We claim that the coefficients $(-1)^nbinom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $frac1{z^{n-j}}$ in the partial fractions expansion of $frac1{z^n(z+2sqrt{delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^nbinom{2n-k-1}{n-k}D^{2n-k}=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$frac1{z^n(z+2sqrt{Delta})^n}=sum_{k=1}^n frac{A(n,k)}{z^k} +sum_{k=1}^n frac{B(n,k)}{(z+2sqrt{Delta})^k}=sum_{k=1}^n frac{A(n,k)}{(x-a-sqrt{Delta})^k} +sum_{k=1}^n frac{B(n,k)}{(x-a+sqrt{Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $frac1{z-apmsqrt{Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{(w-2sqrt{Delta})^nw^n}=frac1{w^n}cdotfrac{(-2sqrt{Delta})^{-n}}{(1-frac{w}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{D^n}{w^n(1+Dw)^n} = frac{D^n}{w^n}sum_{j=0}^{infty} binom{n+j-1}{j}(-D)^jw^j\
&=sum_{j=0}^{infty}(-1)^jbinom{n+j-1}{j}D^{n+j}w^{j-n}end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
$endgroup$
add a comment |
$begingroup$
All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-sqrt{Delta}$ and later defining $D=frac1{2sqrt{Delta}}$, we get
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{z^n(z+2sqrt{Delta})^n}=frac1{z^n}cdotfrac{(2sqrt{Delta})^{-n}}{(1+frac{z}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{(-D)^n}{z^n(1-Dz)^n} = frac{(-D)^n}{z^n}sum_{j=0}^{infty} binom{n+j-1}{j}D^jz^j\
&=(-1)^nsum_{j=0}^{infty}binom{n+j-1}{j}D^{n+j}z^{j-n}end{align*}
We claim that the coefficients $(-1)^nbinom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $frac1{z^{n-j}}$ in the partial fractions expansion of $frac1{z^n(z+2sqrt{delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^nbinom{2n-k-1}{n-k}D^{2n-k}=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$frac1{z^n(z+2sqrt{Delta})^n}=sum_{k=1}^n frac{A(n,k)}{z^k} +sum_{k=1}^n frac{B(n,k)}{(z+2sqrt{Delta})^k}=sum_{k=1}^n frac{A(n,k)}{(x-a-sqrt{Delta})^k} +sum_{k=1}^n frac{B(n,k)}{(x-a+sqrt{Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $frac1{z-apmsqrt{Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{(w-2sqrt{Delta})^nw^n}=frac1{w^n}cdotfrac{(-2sqrt{Delta})^{-n}}{(1-frac{w}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{D^n}{w^n(1+Dw)^n} = frac{D^n}{w^n}sum_{j=0}^{infty} binom{n+j-1}{j}(-D)^jw^j\
&=sum_{j=0}^{infty}(-1)^jbinom{n+j-1}{j}D^{n+j}w^{j-n}end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
$endgroup$
add a comment |
$begingroup$
All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-sqrt{Delta}$ and later defining $D=frac1{2sqrt{Delta}}$, we get
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{z^n(z+2sqrt{Delta})^n}=frac1{z^n}cdotfrac{(2sqrt{Delta})^{-n}}{(1+frac{z}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{(-D)^n}{z^n(1-Dz)^n} = frac{(-D)^n}{z^n}sum_{j=0}^{infty} binom{n+j-1}{j}D^jz^j\
&=(-1)^nsum_{j=0}^{infty}binom{n+j-1}{j}D^{n+j}z^{j-n}end{align*}
We claim that the coefficients $(-1)^nbinom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $frac1{z^{n-j}}$ in the partial fractions expansion of $frac1{z^n(z+2sqrt{delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^nbinom{2n-k-1}{n-k}D^{2n-k}=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$frac1{z^n(z+2sqrt{Delta})^n}=sum_{k=1}^n frac{A(n,k)}{z^k} +sum_{k=1}^n frac{B(n,k)}{(z+2sqrt{Delta})^k}=sum_{k=1}^n frac{A(n,k)}{(x-a-sqrt{Delta})^k} +sum_{k=1}^n frac{B(n,k)}{(x-a+sqrt{Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $frac1{z-apmsqrt{Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{(w-2sqrt{Delta})^nw^n}=frac1{w^n}cdotfrac{(-2sqrt{Delta})^{-n}}{(1-frac{w}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{D^n}{w^n(1+Dw)^n} = frac{D^n}{w^n}sum_{j=0}^{infty} binom{n+j-1}{j}(-D)^jw^j\
&=sum_{j=0}^{infty}(-1)^jbinom{n+j-1}{j}D^{n+j}w^{j-n}end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
$endgroup$
All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-sqrt{Delta}$ and later defining $D=frac1{2sqrt{Delta}}$, we get
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{z^n(z+2sqrt{Delta})^n}=frac1{z^n}cdotfrac{(2sqrt{Delta})^{-n}}{(1+frac{z}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{(-D)^n}{z^n(1-Dz)^n} = frac{(-D)^n}{z^n}sum_{j=0}^{infty} binom{n+j-1}{j}D^jz^j\
&=(-1)^nsum_{j=0}^{infty}binom{n+j-1}{j}D^{n+j}z^{j-n}end{align*}
We claim that the coefficients $(-1)^nbinom{n+j-1}{j}D^{n+j}$ for $j<n$ are precisely the coefficients of $frac1{z^{n-j}}$ in the partial fractions expansion of $frac1{z^n(z+2sqrt{delta})^n}$. Why? Subtract the negative-exponent terms of the Laurent series from the partial fractions expansion. The difference is locally bounded, with a nice power series. But then, the only terms in the partial fractions expansion that aren't locally bounded are the $frac1{z^k}$ terms - so their coefficients all have to match with the terms from the Laurent series.
Let $k=n-j$, and we get $A(n,k)=(-1)^nbinom{2n-k-1}{n-k}D^{2n-k}=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ in the partial fractions expansion
$$frac1{z^n(z+2sqrt{Delta})^n}=sum_{k=1}^n frac{A(n,k)}{z^k} +sum_{k=1}^n frac{B(n,k)}{(z+2sqrt{Delta})^k}=sum_{k=1}^n frac{A(n,k)}{(x-a-sqrt{Delta})^k} +sum_{k=1}^n frac{B(n,k)}{(x-a+sqrt{Delta})^k}$$
Oh, yes - in my comment, I didn't actually define my notation, and the update to the question imported that without defining it. The purpose is clear; we're just putting both parameters in the notation instead of just the power $k$ of $frac1{z-apmsqrt{Delta}}$. Formally, the definition is the line just above.
That's half of the conjecture. For the other half, we expand around the other root.
begin{align*}frac1{(x^2-2ax+b)^n} &= frac1{(x-a-sqrt{Delta})^n(x-a+sqrt{Delta})^n}=frac1{(w-2sqrt{Delta})^nw^n}=frac1{w^n}cdotfrac{(-2sqrt{Delta})^{-n}}{(1-frac{w}{2sqrt{Delta}})^n}\
frac1{(x^2-2ax+b)^n} &= frac{D^n}{w^n(1+Dw)^n} = frac{D^n}{w^n}sum_{j=0}^{infty} binom{n+j-1}{j}(-D)^jw^j\
&=sum_{j=0}^{infty}(-1)^jbinom{n+j-1}{j}D^{n+j}w^{j-n}end{align*}
Again, extract the negative-exponent terms to get $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-k}D^{2n-k} =(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. The conjecture is confirmed, and we have our general formula.
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
answered Jan 25 at 23:34
jmerryjmerry
14.9k1632
14.9k1632
add a comment |
add a comment |
$begingroup$
Let $bneq a^2$,
$$S(n)=int frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-left( 2 A n-C-3 Aright) , {{x}^{2}}-left( left( 2 B-2 A aright) n+left( 2 C+4 Aright) a-2 Bright) x\+2 B a n-left( -C-Aright) b-2 B a$$
$$A=frac{1}{2 left( b-{{a}^{2}}right) , left( n-1right) },B=-frac{a}{2 left( b-{{a}^{2}}right) , left( n-1right) },\C=frac{2 n-3}{2 left( b-{{a}^{2}}right) , left( n-1right) }$$
Then
$$S(n)=frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
frac{2n-3}{2(n-1)(b-a^2)}S(n-1), ; n>1$$
$$S(1)=int frac {dx}{x^2-2ax+b}$$
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
$endgroup$
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
add a comment |
$begingroup$
Let $bneq a^2$,
$$S(n)=int frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-left( 2 A n-C-3 Aright) , {{x}^{2}}-left( left( 2 B-2 A aright) n+left( 2 C+4 Aright) a-2 Bright) x\+2 B a n-left( -C-Aright) b-2 B a$$
$$A=frac{1}{2 left( b-{{a}^{2}}right) , left( n-1right) },B=-frac{a}{2 left( b-{{a}^{2}}right) , left( n-1right) },\C=frac{2 n-3}{2 left( b-{{a}^{2}}right) , left( n-1right) }$$
Then
$$S(n)=frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
frac{2n-3}{2(n-1)(b-a^2)}S(n-1), ; n>1$$
$$S(1)=int frac {dx}{x^2-2ax+b}$$
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
$endgroup$
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
add a comment |
$begingroup$
Let $bneq a^2$,
$$S(n)=int frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-left( 2 A n-C-3 Aright) , {{x}^{2}}-left( left( 2 B-2 A aright) n+left( 2 C+4 Aright) a-2 Bright) x\+2 B a n-left( -C-Aright) b-2 B a$$
$$A=frac{1}{2 left( b-{{a}^{2}}right) , left( n-1right) },B=-frac{a}{2 left( b-{{a}^{2}}right) , left( n-1right) },\C=frac{2 n-3}{2 left( b-{{a}^{2}}right) , left( n-1right) }$$
Then
$$S(n)=frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
frac{2n-3}{2(n-1)(b-a^2)}S(n-1), ; n>1$$
$$S(1)=int frac {dx}{x^2-2ax+b}$$
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
$endgroup$
Let $bneq a^2$,
$$S(n)=int frac {dx}{(x^2-2ax+b)^n}$$
With method of undetermined coefficients we find formula
$$S(n)=frac{Ax+B}{(x^2-2ax+b)^{n-1}}+CS(n-1)$$
We get
$$1=-left( 2 A n-C-3 Aright) , {{x}^{2}}-left( left( 2 B-2 A aright) n+left( 2 C+4 Aright) a-2 Bright) x\+2 B a n-left( -C-Aright) b-2 B a$$
$$A=frac{1}{2 left( b-{{a}^{2}}right) , left( n-1right) },B=-frac{a}{2 left( b-{{a}^{2}}right) , left( n-1right) },\C=frac{2 n-3}{2 left( b-{{a}^{2}}right) , left( n-1right) }$$
Then
$$S(n)=frac{x-a}{2(n-1)(b-a^2)(x^2-2ax+b)^{n-1}}+
frac{2n-3}{2(n-1)(b-a^2)}S(n-1), ; n>1$$
$$S(1)=int frac {dx}{x^2-2ax+b}$$
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
edited Jan 25 at 13:17
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
answered Jan 25 at 13:05
Aleksas DomarkasAleksas Domarkas
1,53816
1,53816
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
add a comment |
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
1
1
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
Moreover, this recurrence can be solved in closed form, giving $$int frac {dx} {(x - a)^n (x - b)^n} = frac {a + b - 2 x} {2 (2 n - 1) (x - a)^n (x - b)^n} ,{_2hspace{-1px}F_1} {left( 1, n; n + frac 1 2; -frac {(a - b)^2} {4 (x - a) (x - b)} right)}.$$
$endgroup$
– Maxim
Jan 25 at 14:56
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
$begingroup$
@Aleksas Somaras, though I did not mark your answer ,yet I appreciate your approach of using reduction formula and therefore it is my humble request that please don't remove your answer.
$endgroup$
– Awe Kumar Jha
Jan 26 at 4:12
add a comment |
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$begingroup$
The notation $;Delta;$ is usually connected to the discriminant of a quadratic equation. In your case, it'd be $;x_{1,2}=apmsqrtDelta;$...which leads to the second comment: what if $;Delta<0;$ ? Then you have two complex non-real roots...
$endgroup$
– DonAntonio
Jan 25 at 8:46
$begingroup$
The decomposition still makes sense for negative discriminant - it'll just lead to an answer in complex form.
$endgroup$
– jmerry
Jan 25 at 8:49
$begingroup$
@DonAntonio, As for the notation I have amended it, as for the sign of the discriminant , in case it is negative the logarithmic term with imaginary argument can be written as inverse sine function.
$endgroup$
– Awe Kumar Jha
Jan 25 at 9:15
$begingroup$
$frac{A_k}{B_k} = (-1)^k$ for $k = 1,...,n$ except when n = 3. Are the signs right in that case for k = 2, 3?
$endgroup$
– Paul
Jan 25 at 9:18
2
$begingroup$
A conjecture: $A(n,k)=(-1)^nbinom{2n-k-1}{n-1}D^{2n-k}$ and $B(n,k)=(-1)^{n-k}binom{2n-k-1}{n-1}D^{2n-k}$. It fits the data given (with the corrected signs). No full solution yet, and I feel like moving on to something else for a while.
$endgroup$
– jmerry
Jan 25 at 9:46