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How to quickly solve partial fractions equation?
The 2019 Stack Overflow Developer Survey Results Are InHow can this indefinite integral be solved without partial fractions?Separation of variables and substituion; first integral from the Euler-Differential Equation for the minimal surface problemHow to set up partial fractions?How to solve $int frac,dx(x^3 + x + 1)^3$?How to solve this integral by parts?Integration of rational functions by partial fractionsCompute $int _0 ^infty fracx^alpha1+x^2, mathrm d x$ for $-1<alpha<1$How can $int fracdx(x+a)^2(x+b)^2$ be found?Solve $int frac1cos^2(x)+cos(x)+1dx$How do I solve this trivial complex integral $int^w_0 fracbz(z-a)(z+a)textrmdz$?
$begingroup$
Often I am dealing with an integral of let's say:
$$intfracdt(t-2)(t+3)$$
or
$$int fracdtt(t-4)$$
or to make this a more general case in which I am interested the most:
$$int fracdt(t+alpha)(t+beta) quad quad alpha, beta in mathbbR$$
Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.
What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):
$$frac1(t+alpha)(t+beta) = fracAt+alpha + fracBt+beta$$
After solving this I end up with some $A, B$ coefficients and I can solve the integral.
Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.
(bonus question: what if there were more variables, like 3 variables?)
I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.
calculus integration indefinite-integrals quadratics partial-fractions
$endgroup$
add a comment |
$begingroup$
Often I am dealing with an integral of let's say:
$$intfracdt(t-2)(t+3)$$
or
$$int fracdtt(t-4)$$
or to make this a more general case in which I am interested the most:
$$int fracdt(t+alpha)(t+beta) quad quad alpha, beta in mathbbR$$
Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.
What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):
$$frac1(t+alpha)(t+beta) = fracAt+alpha + fracBt+beta$$
After solving this I end up with some $A, B$ coefficients and I can solve the integral.
Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.
(bonus question: what if there were more variables, like 3 variables?)
I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.
calculus integration indefinite-integrals quadratics partial-fractions
$endgroup$
1
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago
add a comment |
$begingroup$
Often I am dealing with an integral of let's say:
$$intfracdt(t-2)(t+3)$$
or
$$int fracdtt(t-4)$$
or to make this a more general case in which I am interested the most:
$$int fracdt(t+alpha)(t+beta) quad quad alpha, beta in mathbbR$$
Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.
What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):
$$frac1(t+alpha)(t+beta) = fracAt+alpha + fracBt+beta$$
After solving this I end up with some $A, B$ coefficients and I can solve the integral.
Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.
(bonus question: what if there were more variables, like 3 variables?)
I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.
calculus integration indefinite-integrals quadratics partial-fractions
$endgroup$
Often I am dealing with an integral of let's say:
$$intfracdt(t-2)(t+3)$$
or
$$int fracdtt(t-4)$$
or to make this a more general case in which I am interested the most:
$$int fracdt(t+alpha)(t+beta) quad quad alpha, beta in mathbbR$$
Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.
What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):
$$frac1(t+alpha)(t+beta) = fracAt+alpha + fracBt+beta$$
After solving this I end up with some $A, B$ coefficients and I can solve the integral.
Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.
(bonus question: what if there were more variables, like 3 variables?)
I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.
calculus integration indefinite-integrals quadratics partial-fractions
calculus integration indefinite-integrals quadratics partial-fractions
edited 3 hours ago
weno
asked 3 hours ago
wenoweno
42311
42311
1
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago
add a comment |
1
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago
1
1
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac1alpha - beta$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac1beta - alpha$$
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
add a comment |
$begingroup$
If your fraction is in form of $$ frac 1(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.
Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$
$$A_3=frac 1(-4-1)(-4-3)=frac135$$
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.
Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne k^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.
For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_2-a_1
$.
For $n=3$,
$b_1
=dfrac1(a_1-a_2)(a_1-a_3)
$,
$b_2
=dfrac1(a_2-a_1)(a_2-a_3)
$,
$b_3
=dfrac1(a_3-a_1)(a_3-a_2)
$.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac1alpha - beta$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac1beta - alpha$$
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
add a comment |
$begingroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac1alpha - beta$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac1beta - alpha$$
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
add a comment |
$begingroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac1alpha - beta$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac1beta - alpha$$
$endgroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac1alpha - beta$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac1beta - alpha$$
answered 3 hours ago
DairDair
1,96711124
1,96711124
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
add a comment |
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
$begingroup$
Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $frac1(t+alpha)(t+beta)(t+gamma)$ and $alpha neq beta neq gamma$
$endgroup$
– weno
3 hours ago
add a comment |
$begingroup$
If your fraction is in form of $$ frac 1(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.
Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$
$$A_3=frac 1(-4-1)(-4-3)=frac135$$
$endgroup$
add a comment |
$begingroup$
If your fraction is in form of $$ frac 1(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.
Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$
$$A_3=frac 1(-4-1)(-4-3)=frac135$$
$endgroup$
add a comment |
$begingroup$
If your fraction is in form of $$ frac 1(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.
Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$
$$A_3=frac 1(-4-1)(-4-3)=frac135$$
$endgroup$
If your fraction is in form of $$ frac 1(t-alpha_1)(t-alpha_2)(t-alpha_3)...(t-alpha_k)$$ where the $alpha s$ are different, there is an easy way to find the partial fraction decomposition.
Let $t=alpha_i$ and cover $(t-alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$
$$A_3=frac 1(-4-1)(-4-3)=frac135$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.
Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne k^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.
For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_2-a_1
$.
For $n=3$,
$b_1
=dfrac1(a_1-a_2)(a_1-a_3)
$,
$b_2
=dfrac1(a_2-a_1)(a_2-a_3)
$,
$b_3
=dfrac1(a_3-a_1)(a_3-a_2)
$.
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.
Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne k^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.
For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_2-a_1
$.
For $n=3$,
$b_1
=dfrac1(a_1-a_2)(a_1-a_3)
$,
$b_2
=dfrac1(a_2-a_1)(a_2-a_3)
$,
$b_3
=dfrac1(a_3-a_1)(a_3-a_2)
$.
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.
Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne k^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.
For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_2-a_1
$.
For $n=3$,
$b_1
=dfrac1(a_1-a_2)(a_1-a_3)
$,
$b_2
=dfrac1(a_2-a_1)(a_2-a_3)
$,
$b_3
=dfrac1(a_3-a_1)(a_3-a_2)
$.
$endgroup$
Here's your answer
for general $n$.
$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.
Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne k^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.
For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_2-a_1
$.
For $n=3$,
$b_1
=dfrac1(a_1-a_2)(a_1-a_3)
$,
$b_2
=dfrac1(a_2-a_1)(a_2-a_3)
$,
$b_3
=dfrac1(a_3-a_1)(a_3-a_2)
$.
answered 13 mins ago
marty cohenmarty cohen
75.3k549130
75.3k549130
add a comment |
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1
$begingroup$
Don't you mean find $A,B$? $alpha,beta$ are usually given.
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
Yes. Fixed. Thank you!
$endgroup$
– weno
3 hours ago
$begingroup$
Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $alpha ne beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though.
$endgroup$
– Eevee Trainer
3 hours ago