Knowing $prod_{i=1}^n (c_1 - a_i) = d_1$ can we determine $prod_{i=1}^n (c_2 - a_i)$?
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
asked Dec 9 at 0:29
sheppa28
318110
add a comment |
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
asked Dec 9 at 0:29
sheppa28
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
add a comment |
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
asked Dec 9 at 0:29
sheppa28
318110
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
asked Dec 9 at 0:29
sheppa28
318110
asked Dec 9 at 0:29
sheppa28
318110
edited Dec 9 at 0:35
asked Dec 9 at 0:29
sheppa28
318110
asked Dec 9 at 0:29
sheppa28
318110
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
add a comment |
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
add a comment |
2 Answers
2
active
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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.
answered Dec 9 at 1:07
dezdichado
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
add a comment |
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.
answered Dec 9 at 1:07
Martin Argerami
124k1176174
add a comment |
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2 Answers
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2 Answers
2
active
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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.
answered Dec 9 at 1:07
dezdichado
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
add a comment |
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.
answered Dec 9 at 1:07
dezdichado
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
add a comment |
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.
answered Dec 9 at 1:07
dezdichado
6,1781929
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.
answered Dec 9 at 1:07
dezdichado
6,1781929
edited Dec 9 at 1:22
answered Dec 9 at 1:07
dezdichado
6,1781929
answered Dec 9 at 1:07
dezdichado
6,1781929
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
add a comment |
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
add a comment |
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.
answered Dec 9 at 1:07
Martin Argerami
124k1176174
add a comment |
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.
answered Dec 9 at 1:07
Martin Argerami
124k1176174
add a comment |
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.
answered Dec 9 at 1:07
Martin Argerami
124k1176174
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.
answered Dec 9 at 1:07
Martin Argerami
124k1176174
edited Dec 9 at 1:13
answered Dec 9 at 1:07
Martin Argerami
124k1176174
answered Dec 9 at 1:07
Martin Argerami
124k1176174
answered Dec 9 at 1:07
Martin Argerami
124k1176174
124k1176174
add a comment |
add a comment |
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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