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Find the value of $(1-2alpha_1)(1-2alpha_2)dots (1-2alpha_6)$
How to derive the share of a prize given scoresFind the minimum value.How to measure monotonicity of a list of valuesFind the value.Find the value belowProof by $PMI$ of a formula for the sum of the first $n$ terms of a given sequenceSum of sixth power of roots of $x^3-x-1=0$Conceptual problem in Theory Of EquationsFind surface area of 3D rectangle given one edge's dimensions and areaif 1, $alpha_1$, $alpha_2$, $alpha_3$, $ldots$, $alpha_n-1$ are nth roots of unity then…
$begingroup$
If $alpha_1,alpha_2,dots,alpha_6$ are roots of $x^6+x^2+1=0$, then find the value of $(1-2alpha_1)(1-2alpha_2)dots (1-2alpha_6)$.
My attempt:
From the given equation we have, $S_1=0,S_2=0,S_3=0,S_4=1,S_5=0,S_6=1$. ($S_k$ is the sum of the products of the elements of $alpha_1,alpha_2,dots,alpha_6 $ taken $k$ at a time).
$prod^6_i=1(y-a_i)=y^6-S_1'y^5+S_2'y^4-S_3'y^3+S_4'y^2-S_5'y+S_6'$. ($S_k$ is the sum of the products of the elements of $a_k:a_k=2alpha_k$ taken $k$ at a time.)
Putting $y=1$ in the above equation, we get, $prod^6_i=1(1-a_i)=1-S_1'+S_2'-S_3'+S_4'-S_5'+S_6'=1-2S_1+4S_2-8S_3+16S_4-32S_5+64=1-0+0-0+16-0+64=81$
Is my process correct? What are some other methods to solve this question?
algebra-precalculus
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
$endgroup$
add a comment |
$begingroup$
If $alpha_1,alpha_2,dots,alpha_6$ are roots of $x^6+x^2+1=0$, then find the value of $(1-2alpha_1)(1-2alpha_2)dots (1-2alpha_6)$.
My attempt:
From the given equation we have, $S_1=0,S_2=0,S_3=0,S_4=1,S_5=0,S_6=1$. ($S_k$ is the sum of the products of the elements of $alpha_1,alpha_2,dots,alpha_6 $ taken $k$ at a time).
$prod^6_i=1(y-a_i)=y^6-S_1'y^5+S_2'y^4-S_3'y^3+S_4'y^2-S_5'y+S_6'$. ($S_k$ is the sum of the products of the elements of $a_k:a_k=2alpha_k$ taken $k$ at a time.)
Putting $y=1$ in the above equation, we get, $prod^6_i=1(1-a_i)=1-S_1'+S_2'-S_3'+S_4'-S_5'+S_6'=1-2S_1+4S_2-8S_3+16S_4-32S_5+64=1-0+0-0+16-0+64=81$
Is my process correct? What are some other methods to solve this question?
algebra-precalculus
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
$endgroup$
add a comment |
$begingroup$
If $alpha_1,alpha_2,dots,alpha_6$ are roots of $x^6+x^2+1=0$, then find the value of $(1-2alpha_1)(1-2alpha_2)dots (1-2alpha_6)$.
My attempt:
From the given equation we have, $S_1=0,S_2=0,S_3=0,S_4=1,S_5=0,S_6=1$. ($S_k$ is the sum of the products of the elements of $alpha_1,alpha_2,dots,alpha_6 $ taken $k$ at a time).
$prod^6_i=1(y-a_i)=y^6-S_1'y^5+S_2'y^4-S_3'y^3+S_4'y^2-S_5'y+S_6'$. ($S_k$ is the sum of the products of the elements of $a_k:a_k=2alpha_k$ taken $k$ at a time.)
Putting $y=1$ in the above equation, we get, $prod^6_i=1(1-a_i)=1-S_1'+S_2'-S_3'+S_4'-S_5'+S_6'=1-2S_1+4S_2-8S_3+16S_4-32S_5+64=1-0+0-0+16-0+64=81$
Is my process correct? What are some other methods to solve this question?
algebra-precalculus
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
$endgroup$
If $alpha_1,alpha_2,dots,alpha_6$ are roots of $x^6+x^2+1=0$, then find the value of $(1-2alpha_1)(1-2alpha_2)dots (1-2alpha_6)$.
My attempt:
From the given equation we have, $S_1=0,S_2=0,S_3=0,S_4=1,S_5=0,S_6=1$. ($S_k$ is the sum of the products of the elements of $alpha_1,alpha_2,dots,alpha_6 $ taken $k$ at a time).
$prod^6_i=1(y-a_i)=y^6-S_1'y^5+S_2'y^4-S_3'y^3+S_4'y^2-S_5'y+S_6'$. ($S_k$ is the sum of the products of the elements of $a_k:a_k=2alpha_k$ taken $k$ at a time.)
Putting $y=1$ in the above equation, we get, $prod^6_i=1(1-a_i)=1-S_1'+S_2'-S_3'+S_4'-S_5'+S_6'=1-2S_1+4S_2-8S_3+16S_4-32S_5+64=1-0+0-0+16-0+64=81$
Is my process correct? What are some other methods to solve this question?
algebra-precalculus
algebra-precalculus
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
edited Mar 29 at 14:03
MrAP
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
asked Mar 29 at 10:24
MrAPMrAP
1,29121432
1,29121432
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$x^6+x^2+1=prod_i=1^6(x-alpha_i).$$
Put $x=0.5$ and multiply $2^6$,
$$prod_i=1^6(1-2alpha_i)=2^6prod_i=1^6(0.5-alpha_i)=2^6left(0.5^6+0.5^2+1right).$$
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
$endgroup$
add a comment |
$begingroup$
If the $alpha_k$ are roots of $x^6+x^2+1$, then the $1-2alpha_k$ are roots of
$$left(frac1-x2right)^6+left(frac1-x2right)^2+1.$$
The extreme terms of this polynomial are
$$fracx^62^6$$ and $$frac12^6+frac12^2+1$$ and by Vieta the product of the roots is the ratio of the coefficients, $81$.
For the sum of these numbers,
$$-frac-dfrac 62^6dfrac12^6=6,$$
which is simply $6cdot1-2cdot0$, where $0$ is the sum of the roots of the original polynomial.
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have
$$x^6+x^2+1=prod_i=1^6(x-alpha_i).$$
Put $x=0.5$ and multiply $2^6$,
$$prod_i=1^6(1-2alpha_i)=2^6prod_i=1^6(0.5-alpha_i)=2^6left(0.5^6+0.5^2+1right).$$
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
$endgroup$
add a comment |
$begingroup$
We have
$$x^6+x^2+1=prod_i=1^6(x-alpha_i).$$
Put $x=0.5$ and multiply $2^6$,
$$prod_i=1^6(1-2alpha_i)=2^6prod_i=1^6(0.5-alpha_i)=2^6left(0.5^6+0.5^2+1right).$$
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
$endgroup$
add a comment |
$begingroup$
We have
$$x^6+x^2+1=prod_i=1^6(x-alpha_i).$$
Put $x=0.5$ and multiply $2^6$,
$$prod_i=1^6(1-2alpha_i)=2^6prod_i=1^6(0.5-alpha_i)=2^6left(0.5^6+0.5^2+1right).$$
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
$endgroup$
We have
$$x^6+x^2+1=prod_i=1^6(x-alpha_i).$$
Put $x=0.5$ and multiply $2^6$,
$$prod_i=1^6(1-2alpha_i)=2^6prod_i=1^6(0.5-alpha_i)=2^6left(0.5^6+0.5^2+1right).$$
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
answered Mar 29 at 10:37
TianlaluTianlalu
3,28521238
3,28521238
add a comment |
add a comment |
$begingroup$
If the $alpha_k$ are roots of $x^6+x^2+1$, then the $1-2alpha_k$ are roots of
$$left(frac1-x2right)^6+left(frac1-x2right)^2+1.$$
The extreme terms of this polynomial are
$$fracx^62^6$$ and $$frac12^6+frac12^2+1$$ and by Vieta the product of the roots is the ratio of the coefficients, $81$.
For the sum of these numbers,
$$-frac-dfrac 62^6dfrac12^6=6,$$
which is simply $6cdot1-2cdot0$, where $0$ is the sum of the roots of the original polynomial.
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
If the $alpha_k$ are roots of $x^6+x^2+1$, then the $1-2alpha_k$ are roots of
$$left(frac1-x2right)^6+left(frac1-x2right)^2+1.$$
The extreme terms of this polynomial are
$$fracx^62^6$$ and $$frac12^6+frac12^2+1$$ and by Vieta the product of the roots is the ratio of the coefficients, $81$.
For the sum of these numbers,
$$-frac-dfrac 62^6dfrac12^6=6,$$
which is simply $6cdot1-2cdot0$, where $0$ is the sum of the roots of the original polynomial.
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
If the $alpha_k$ are roots of $x^6+x^2+1$, then the $1-2alpha_k$ are roots of
$$left(frac1-x2right)^6+left(frac1-x2right)^2+1.$$
The extreme terms of this polynomial are
$$fracx^62^6$$ and $$frac12^6+frac12^2+1$$ and by Vieta the product of the roots is the ratio of the coefficients, $81$.
For the sum of these numbers,
$$-frac-dfrac 62^6dfrac12^6=6,$$
which is simply $6cdot1-2cdot0$, where $0$ is the sum of the roots of the original polynomial.
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
$endgroup$
If the $alpha_k$ are roots of $x^6+x^2+1$, then the $1-2alpha_k$ are roots of
$$left(frac1-x2right)^6+left(frac1-x2right)^2+1.$$
The extreme terms of this polynomial are
$$fracx^62^6$$ and $$frac12^6+frac12^2+1$$ and by Vieta the product of the roots is the ratio of the coefficients, $81$.
For the sum of these numbers,
$$-frac-dfrac 62^6dfrac12^6=6,$$
which is simply $6cdot1-2cdot0$, where $0$ is the sum of the roots of the original polynomial.
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
edited Mar 29 at 10:52
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
answered Mar 29 at 10:38
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
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