Knowing $prod_{i=1}^n (c_1 - a_i) = d_1$ can we determine $prod_{i=1}^n (c_2 - a_i)$?












1














For known set of real values ${a_i}$ and $c_1$, if we know



$$prod_{i=1}^n (c_1 - a_i) = d_1$$



is there a way to determine



$$prod_{i=1}^n (c_2 - a_i)$$



without having to redo the multiplications?



(Forgot to include additional relevant assumptions):
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $max {a_i}$. The product is strictly positive. Thank you.






algebra-precalculus







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  • No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
    – T. Bongers
    Dec 9 at 0:30










  • Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
    – sheppa28
    Dec 9 at 0:33
















1














For known set of real values ${a_i}$ and $c_1$, if we know



$$prod_{i=1}^n (c_1 - a_i) = d_1$$



is there a way to determine



$$prod_{i=1}^n (c_2 - a_i)$$



without having to redo the multiplications?



(Forgot to include additional relevant assumptions):
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $max {a_i}$. The product is strictly positive. Thank you.






algebra-precalculus







share|cite|improve this question
























  • No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
    – T. Bongers
    Dec 9 at 0:30










  • Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
    – sheppa28
    Dec 9 at 0:33














1












1








1







For known set of real values ${a_i}$ and $c_1$, if we know



$$prod_{i=1}^n (c_1 - a_i) = d_1$$



is there a way to determine



$$prod_{i=1}^n (c_2 - a_i)$$



without having to redo the multiplications?



(Forgot to include additional relevant assumptions):
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $max {a_i}$. The product is strictly positive. Thank you.






algebra-precalculus







share|cite|improve this question















For known set of real values ${a_i}$ and $c_1$, if we know



$$prod_{i=1}^n (c_1 - a_i) = d_1$$



is there a way to determine



$$prod_{i=1}^n (c_2 - a_i)$$



without having to redo the multiplications?



(Forgot to include additional relevant assumptions):
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $max {a_i}$. The product is strictly positive. Thank you.






algebra-precalculus




algebra-precalculus






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share|cite|improve this question













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edited Dec 9 at 0:35

























asked Dec 9 at 0:29









sheppa28

318110




318110












  • No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
    – T. Bongers
    Dec 9 at 0:30










  • Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
    – sheppa28
    Dec 9 at 0:33


















  • No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
    – T. Bongers
    Dec 9 at 0:30










  • Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
    – sheppa28
    Dec 9 at 0:33
















No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
– T. Bongers
Dec 9 at 0:30




No: Consider the case that some $a_i$ coincides with $c_1$, giving a zero product. Then if $c_2 notin {a_i}$, the second product is certainly unrelated to $d_1$.
– T. Bongers
Dec 9 at 0:30












Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
– sheppa28
Dec 9 at 0:33




Ah, yes; I should have clarified; which I'll edit - no a_i is equal to c1 or c2; in fact both {c1,c2} are greater than the max {a_i}.
– sheppa28
Dec 9 at 0:33










2 Answers
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0














Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$



It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.






share|cite|improve this answer























  • I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
    – Martin Argerami
    Dec 9 at 1:11










  • @MartinArgerami you are right, I overlooked that.
    – dezdichado
    Dec 9 at 1:21



















1














Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.



Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
$$
p(x)=sum_{k=0}^n (-1)^{n}e_n(a_1,ldots,a_n),x^{n-k}.
$$
Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=ldots=a_n=0$. Then you are asking something like




If I now that $pi^n=d_1$, find $e^n$.




Which can be answer, but not in a "polynomial" way.






share|cite|improve this answer























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    2 Answers
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    active

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    2 Answers
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    active

    oldest

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    0














    Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$



    It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.






    share|cite|improve this answer























    • I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
      – Martin Argerami
      Dec 9 at 1:11










    • @MartinArgerami you are right, I overlooked that.
      – dezdichado
      Dec 9 at 1:21
















    0














    Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$



    It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.






    share|cite|improve this answer























    • I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
      – Martin Argerami
      Dec 9 at 1:11










    • @MartinArgerami you are right, I overlooked that.
      – dezdichado
      Dec 9 at 1:21














    0












    0








    0






    Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$



    It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.






    share|cite|improve this answer














    Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$



    It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 at 1:22

























    answered Dec 9 at 1:07









    dezdichado

    6,1781929




    6,1781929












    • I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
      – Martin Argerami
      Dec 9 at 1:11










    • @MartinArgerami you are right, I overlooked that.
      – dezdichado
      Dec 9 at 1:21


















    • I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
      – Martin Argerami
      Dec 9 at 1:11










    • @MartinArgerami you are right, I overlooked that.
      – dezdichado
      Dec 9 at 1:21
















    I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
    – Martin Argerami
    Dec 9 at 1:11




    I don't think polynomial interpolation is relevant. Interpolating a polynomial with a polynomial is not very useful, and here $P$ is explicit. Meanwhile, you do have the values at $n+1$ points, namely $c_1$ and $a_1,ldots,a_n$.
    – Martin Argerami
    Dec 9 at 1:11












    @MartinArgerami you are right, I overlooked that.
    – dezdichado
    Dec 9 at 1:21




    @MartinArgerami you are right, I overlooked that.
    – dezdichado
    Dec 9 at 1:21











    1














    Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.



    Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
    $$
    p(x)=sum_{k=0}^n (-1)^{n}e_n(a_1,ldots,a_n),x^{n-k}.
    $$
    Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=ldots=a_n=0$. Then you are asking something like




    If I now that $pi^n=d_1$, find $e^n$.




    Which can be answer, but not in a "polynomial" way.






    share|cite|improve this answer




























      1














      Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.



      Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
      $$
      p(x)=sum_{k=0}^n (-1)^{n}e_n(a_1,ldots,a_n),x^{n-k}.
      $$
      Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=ldots=a_n=0$. Then you are asking something like




      If I now that $pi^n=d_1$, find $e^n$.




      Which can be answer, but not in a "polynomial" way.






      share|cite|improve this answer


























        1












        1








        1






        Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.



        Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
        $$
        p(x)=sum_{k=0}^n (-1)^{n}e_n(a_1,ldots,a_n),x^{n-k}.
        $$
        Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=ldots=a_n=0$. Then you are asking something like




        If I now that $pi^n=d_1$, find $e^n$.




        Which can be answer, but not in a "polynomial" way.






        share|cite|improve this answer














        Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.



        Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
        $$
        p(x)=sum_{k=0}^n (-1)^{n}e_n(a_1,ldots,a_n),x^{n-k}.
        $$
        Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=ldots=a_n=0$. Then you are asking something like




        If I now that $pi^n=d_1$, find $e^n$.




        Which can be answer, but not in a "polynomial" way.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 at 1:13

























        answered Dec 9 at 1:07









        Martin Argerami

        124k1176174




        124k1176174






























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