Calculate $sum^{20}_{k=1}frac{1}{x_k-x_k^2}$ where $x_k$ are roots of $P(x)=x^{20}+x^{10}+x^5+2$
$begingroup$
We have the polynomial $P(x)=x^{20}+x^{10}+x^5+2$, which has roots $x_1,x_2,x_3,...,x_{20}$. Calculate the sum $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$
What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$
I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.
Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.
polynomials contest-math roots
edited Mar 23 at 16:13
Jean Marie
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
We have the polynomial $P(x)=x^{20}+x^{10}+x^5+2$, which has roots $x_1,x_2,x_3,...,x_{20}$. Calculate the sum $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$
What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$
I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.
Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.
polynomials contest-math roots
edited Mar 23 at 16:13
Jean Marie
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
6
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11
add a comment |
$begingroup$
We have the polynomial $P(x)=x^{20}+x^{10}+x^5+2$, which has roots $x_1,x_2,x_3,...,x_{20}$. Calculate the sum $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$
What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$
I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.
Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.
polynomials contest-math roots
edited Mar 23 at 16:13
Jean Marie
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We have the polynomial $P(x)=x^{20}+x^{10}+x^5+2$, which has roots $x_1,x_2,x_3,...,x_{20}$. Calculate the sum $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}$$
What I've noticed: $$sum^{20}_{k=1}frac{1}{x_k-x_k^2}=sum^{20}_{k=1}left(frac{1}{x_k}+frac{1}{1-x_k}right)$$
I know how to calculate the first sum: $sum^{20}_{k=1}frac{1}{x_k}$.
Please help me calculate the second one: $sum^{20}_{k=1}frac{1}{1-x_k}$.
polynomials contest-math roots
polynomials contest-math roots
edited Mar 23 at 16:13
Jean Marie
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 23 at 16:13
Jean Marie
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 23 at 16:13
Jean Marie
31k42255
edited Mar 23 at 16:13
Jean Marie
31k42255
edited Mar 23 at 16:13
Jean Marie
31k42255
31k42255
asked Mar 22 at 19:22
P. MillerP. Miller
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 22 at 19:22
P. MillerP. Miller
462
asked Mar 22 at 19:22
P. MillerP. Miller
462
462
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P. Miller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
6
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11
add a comment |
2
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
6
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11
2
2
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
6
6
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
and $P'(x)= 20x^{19}+10x^9+5x^4$
we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Hint:
Set $y=1-x$. If the $x_k$ satisfy the equation $;x^{20}+x^{10}+x^{5}+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$
Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?
answered Mar 22 at 19:44
BernardBernard
123k741117
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
and $P'(x)= 20x^{19}+10x^9+5x^4$
we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
and $P'(x)= 20x^{19}+10x^9+5x^4$
we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
and $P'(x)= 20x^{19}+10x^9+5x^4$
we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
Since $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
and $P'(x)= 20x^{19}+10x^9+5x^4$
we have $$sum_{k=1}^{20}frac{1}{1-x_k}=frac{P'(1)}{P(1)} = {35over 5}=7$$
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
answered Mar 22 at 19:50
Maria MazurMaria Mazur
48.5k1260121
48.5k1260121
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Why is $$frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
$begingroup$
Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva
$endgroup$
– Maria Mazur
2 days ago
add a comment |
$begingroup$
Hint:
Set $y=1-x$. If the $x_k$ satisfy the equation $;x^{20}+x^{10}+x^{5}+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$
Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?
answered Mar 22 at 19:44
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Set $y=1-x$. If the $x_k$ satisfy the equation $;x^{20}+x^{10}+x^{5}+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$
Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?
answered Mar 22 at 19:44
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Set $y=1-x$. If the $x_k$ satisfy the equation $;x^{20}+x^{10}+x^{5}+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$
Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?
answered Mar 22 at 19:44
BernardBernard
123k741117
$endgroup$
Hint:
Set $y=1-x$. If the $x_k$ satisfy the equation $;x^{20}+x^{10}+x^{5}+2=0$, the corresponding $:y_k$ satisfy the equation
$$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$
Can you find the constant term and the coefficient of $y$ in this equation, to use Vieta's relations?
answered Mar 22 at 19:44
BernardBernard
123k741117
answered Mar 22 at 19:44
BernardBernard
123k741117
answered Mar 22 at 19:44
BernardBernard
123k741117
answered Mar 22 at 19:44
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
P. Miller is a new contributor. Be nice, and check out our Code of Conduct.
P. Miller is a new contributor. Be nice, and check out our Code of Conduct.
P. Miller is a new contributor. Be nice, and check out our Code of Conduct.
P. Miller is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help?
$endgroup$
– Sil
Mar 22 at 19:38
6
$begingroup$
Hint: $frac{P'(x)}{P(x)} = sum_{k=1}^{20}frac{1}{x-x_k}$
$endgroup$
– achille hui
Mar 22 at 19:44
$begingroup$
Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags.
$endgroup$
– Jean Marie
Mar 23 at 16:11