How to quickly solve partial fractions equation?












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










3 Answers
3






active

oldest

votes


















3












$begingroup$

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 $-alpha$:



$$1 = A(beta - alpha)$$



$$A = frac{1}{beta - alpha}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















1












$begingroup$

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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



$$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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)}
    $    .






    share|cite









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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 $-alpha$:



      $$1 = A(beta - alpha)$$



      $$A = frac{1}{beta - alpha}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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


















      3












      $begingroup$

      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 $-alpha$:



      $$1 = A(beta - alpha)$$



      $$A = frac{1}{beta - alpha}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        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
















      3












      3








      3





      $begingroup$

      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 $-alpha$:



      $$1 = A(beta - alpha)$$



      $$A = frac{1}{beta - alpha}$$






      share|cite|improve this answer









      $endgroup$



      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 $-alpha$:



      $$1 = A(beta - alpha)$$



      $$A = frac{1}{beta - alpha}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      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




















      • $begingroup$
        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













      1












      $begingroup$

      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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
      $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



      $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
        $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



        $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
          $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



          $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$






          share|cite|improve this answer









          $endgroup$



          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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
          $$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$



          $$A_3=frac {1}{(-4-1)(-4-3)}=frac{1}{35}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Mohammad Riazi-KermaniMohammad Riazi-Kermani

          41.6k42061




          41.6k42061























              0












              $begingroup$

              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)}
              $              .






              share|cite









              $endgroup$


















                0












                $begingroup$

                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)}
                $                .






                share|cite









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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)}
                  $                  .






                  share|cite









                  $endgroup$



                  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)}
                  $                  .







                  share|cite












                  share|cite



                  share|cite










                  answered 9 mins ago









                  marty cohenmarty cohen

                  75.3k549130




                  75.3k549130






























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