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
1,95711124
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 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 1 min 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
1,95711124
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 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
1,95711124
$endgroup$
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 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
1,95711124
$endgroup$
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
1,95711124
answered 3 hours ago
DairDair
1,95711124
answered 3 hours ago
DairDair
1,95711124
answered 3 hours ago
DairDair
1,95711124
1,95711124
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 hours ago
add a comment |
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 hours ago
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 hours ago
$begingroup$
I'll be coming back to this post. This is what I was looking for.
$endgroup$
– weno
2 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
2 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
2 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 1 min 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 1 min 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 1 min 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 1 min ago
marty cohenmarty cohen
75.3k549130
answered 1 min ago
marty cohenmarty cohen
75.3k549130
answered 1 min ago
marty cohenmarty cohen
75.3k549130
answered 1 min 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