How to quickly solve partial fractions equation?
$begingroup$
Often I am dealing with an integral of let's say:
$$intfrac{dt}{(t-2)(t+3)}$$
or
$$int frac{dt}{t(t-4)}$$
or to make this a more general case in which I am interested the most:
$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$
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):
$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$
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
asked 3 hours ago
wenoweno
42311
$endgroup$
add a comment |
$begingroup$
Often I am dealing with an integral of let's say:
$$intfrac{dt}{(t-2)(t+3)}$$
or
$$int frac{dt}{t(t-4)}$$
or to make this a more general case in which I am interested the most:
$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$
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):
$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$
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
asked 3 hours ago
wenoweno
42311
$endgroup$
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:
$$intfrac{dt}{(t-2)(t+3)}$$
or
$$int frac{dt}{t(t-4)}$$
or to make this a more general case in which I am interested the most:
$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$
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):
$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$
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
asked 3 hours ago
wenoweno
42311
$endgroup$
Often I am dealing with an integral of let's say:
$$intfrac{dt}{(t-2)(t+3)}$$
or
$$int frac{dt}{t(t-4)}$$
or to make this a more general case in which I am interested the most:
$$int frac{dt}{(t+alpha)(t+beta)} quad quad alpha, beta in mathbb{R}$$
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):
$$frac{1}{(t+alpha)(t+beta)} = frac{A}{t+alpha} + frac{B}{t+beta}$$
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
asked 3 hours ago
wenoweno
42311
asked 3 hours ago
wenoweno
42311
edited 3 hours ago
weno
asked 3 hours ago
wenoweno
42311
asked 3 hours ago
wenoweno
42311
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 $beta neq alpha$):
$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac{1}{alpha - beta}$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac{1}{beta - alpha}$$
answered 3 hours ago
DairDair
1,96711124
$endgroup$
$begingroup$
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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$
$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1{prod_{k=1}^n (x-a_k)}
=sum_{k=1}^n dfrac{b_k}{x-a_k}
$.
Therefore
$1
=sum_{k=1}^n dfrac{b_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
=dfrac1{prod_{j=1, jne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=dfrac1{a_1-a_2}
$,
$b_2
=dfrac1{a_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 15 mins ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
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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 $beta neq alpha$):
$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac{1}{alpha - beta}$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac{1}{beta - alpha}$$
answered 3 hours ago
DairDair
1,96711124
$endgroup$
$begingroup$
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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $beta neq alpha$):
$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac{1}{alpha - beta}$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac{1}{beta - alpha}$$
answered 3 hours ago
DairDair
1,96711124
$endgroup$
$begingroup$
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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $beta neq alpha$):
$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac{1}{alpha - beta}$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac{1}{beta - alpha}$$
answered 3 hours ago
DairDair
1,96711124
$endgroup$
Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):
$$frac{1}{(t + alpha)(t + beta)} = frac{A}{t + alpha} + frac{B}{t + beta}$$
$$1 = A(t + beta) + B(t + alpha)$$
Evaluating $beta$ for $t$:
$$1 = B(alpha - beta)$$
$$B = frac{1}{alpha - beta}$$
Similarly, for $A$, sub in $-alpha$:
$$1 = A(beta - alpha)$$
$$A = frac{1}{beta - alpha}$$
answered 3 hours ago
DairDair
1,96711124
answered 3 hours ago
DairDair
1,96711124
answered 3 hours ago
DairDair
1,96711124
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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $frac{1}{(t+alpha)(t+beta)(t+gamma)}$ 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 $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$
$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
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 $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$
$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
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 $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$
$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
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 $$ frac{1}{(t-1)(t-3)(t+4)} = frac {A_1}{(t-1)} + frac{ A_2}{(t-3)} +frac { A_3}{(t+4)}$$
Where $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$
$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered 2 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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 $n$.
$dfrac1{prod_{k=1}^n (x-a_k)}
=sum_{k=1}^n dfrac{b_k}{x-a_k}
$.
Therefore
$1
=sum_{k=1}^n dfrac{b_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
=dfrac1{prod_{j=1, jne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=dfrac1{a_1-a_2}
$,
$b_2
=dfrac1{a_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 15 mins ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1{prod_{k=1}^n (x-a_k)}
=sum_{k=1}^n dfrac{b_k}{x-a_k}
$.
Therefore
$1
=sum_{k=1}^n dfrac{b_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
=dfrac1{prod_{j=1, jne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=dfrac1{a_1-a_2}
$,
$b_2
=dfrac1{a_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 15 mins ago
marty cohenmarty cohen
75.3k549130
$endgroup$
add a comment |
$begingroup$
Here's your answer
for general $n$.
$dfrac1{prod_{k=1}^n (x-a_k)}
=sum_{k=1}^n dfrac{b_k}{x-a_k}
$.
Therefore
$1
=sum_{k=1}^n dfrac{b_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
=dfrac1{prod_{j=1, jne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=dfrac1{a_1-a_2}
$,
$b_2
=dfrac1{a_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 15 mins ago
marty cohenmarty cohen
75.3k549130
$endgroup$
Here's your answer
for general $n$.
$dfrac1{prod_{k=1}^n (x-a_k)}
=sum_{k=1}^n dfrac{b_k}{x-a_k}
$.
Therefore
$1
=sum_{k=1}^n dfrac{b_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
=dfrac1{prod_{j=1, jne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=dfrac1{a_1-a_2}
$,
$b_2
=dfrac1{a_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 15 mins ago
marty cohenmarty cohen
75.3k549130
answered 15 mins ago
marty cohenmarty cohen
75.3k549130
answered 15 mins ago
marty cohenmarty cohen
75.3k549130
answered 15 mins ago
marty cohenmarty cohen
75.3k549130
75.3k549130
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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