Mixing
Properties of roots
Clash Royale CLAN TAG#URR8PPP
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I've been struggling to understand the procedure shown below. Somehow, by setting x=0 I managed to get to the equation (1.12). However, I think didn't get it really as I do not know how to compare the coefficients (which is apparently important to get to (1.13) ).
Kind regards
polynomials roots factoring
edited Aug 3 at 10:16
egreg
164k1180187
asked Aug 3 at 10:03
MrDerDart
92
add a comment |Â
up vote
0
down vote
favorite
I've been struggling to understand the procedure shown below. Somehow, by setting x=0 I managed to get to the equation (1.12). However, I think didn't get it really as I do not know how to compare the coefficients (which is apparently important to get to (1.13) ).
Kind regards
polynomials roots factoring
edited Aug 3 at 10:16
egreg
164k1180187
asked Aug 3 at 10:03
MrDerDart
92
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been struggling to understand the procedure shown below. Somehow, by setting x=0 I managed to get to the equation (1.12). However, I think didn't get it really as I do not know how to compare the coefficients (which is apparently important to get to (1.13) ).
Kind regards
polynomials roots factoring
edited Aug 3 at 10:16
egreg
164k1180187
asked Aug 3 at 10:03
MrDerDart
92
I've been struggling to understand the procedure shown below. Somehow, by setting x=0 I managed to get to the equation (1.12). However, I think didn't get it really as I do not know how to compare the coefficients (which is apparently important to get to (1.13) ).
Kind regards
polynomials roots factoring
edited Aug 3 at 10:16
egreg
164k1180187
asked Aug 3 at 10:03
MrDerDart
92
edited Aug 3 at 10:16
egreg
164k1180187
edited Aug 3 at 10:16
egreg
164k1180187
edited Aug 3 at 10:16
egreg
164k1180187
164k1180187
asked Aug 3 at 10:03
MrDerDart
92
asked Aug 3 at 10:03
MrDerDart
92
asked Aug 3 at 10:03
MrDerDart
92
92
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
I think didn't get it really as I do not know how to compare the coefficients
All it means is that $sum_i=0^n a_ix^i=sum_i=0^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $alpha_i$, the roots.
The author's point is that the $a_0$ and $a_n-1$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.
answered Aug 3 at 10:20
rschwieb
99.5k1190225
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
add a comment |Â
up vote
0
down vote
When you have a product of binomials like these
$$
alpha_n(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0
$$
Which for simplicity we are going to convert it to a monic polynomial:
$$(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0$$
You can express it also like this:
$$
(x-alpha_j)prod_ineq j^n(x-alpha_i)iff 1le j le n
$$
Now lets expand $prod_ineq j^n(x-alpha_i)$:
$$
(x-alpha_j)(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0) = x(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
-
alpha_j(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
$$
After expanding the equation from above we focus on the term $ (b_n-2-alpha_j)x^n-1 $.
Now we need to figure out what $b_n-2$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2:
$$
(x-alpha_1)(x-alpha_2) = x^2+(-alpha_1-alpha_2)x+alpha_1alpha_2
$$
For n = 3:
$$
(x-alpha_1)(x^2+(-alpha_2-alpha_3)x+alpha_2alpha_3) =
x^3+(-alpha_1-alpha_2-alpha_3)x^2 +...-alpha_1alpha_2alpha_3
$$
From this part it is easy to see that $b_n-2 = b_n-3-alpha_n$ where $b_0 = -alpha_2$.
That implies $a_n-1=b_n-2-alpha_1=-(alpha_1+alpha_2+alpha_3+..+alpha_n)=-sum_k=1^nalpha_k$
$$a_n-1=-sum_k=1^nalpha_kto -a_n-1=sum_k=1^nalpha_k$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$.
$$-fraca_n-1a_n=sum_k=1^nalpha_k$$
answered Aug 3 at 14:48
MarioE
486
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I think didn't get it really as I do not know how to compare the coefficients
All it means is that $sum_i=0^n a_ix^i=sum_i=0^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $alpha_i$, the roots.
The author's point is that the $a_0$ and $a_n-1$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.
answered Aug 3 at 10:20
rschwieb
99.5k1190225
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
add a comment |Â
up vote
1
down vote
I think didn't get it really as I do not know how to compare the coefficients
All it means is that $sum_i=0^n a_ix^i=sum_i=0^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $alpha_i$, the roots.
The author's point is that the $a_0$ and $a_n-1$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.
answered Aug 3 at 10:20
rschwieb
99.5k1190225
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think didn't get it really as I do not know how to compare the coefficients
All it means is that $sum_i=0^n a_ix^i=sum_i=0^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $alpha_i$, the roots.
The author's point is that the $a_0$ and $a_n-1$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.
answered Aug 3 at 10:20
rschwieb
99.5k1190225
I think didn't get it really as I do not know how to compare the coefficients
All it means is that $sum_i=0^n a_ix^i=sum_i=0^n b_ix^i$ if and only if $a_i=b_i$ for all $i$. "Setting $x=0$" is one way of looking at the constant term of a polynomial, but it's not really necessary if you follow the fact I just mentioned.
You're presented with three ways of writing polynomials, but if you multiply them out, their coefficients have to match. The first equation gives you labels for the coefficients ($a_i$) and the latter two, if multiplied out, will give you coefficients in terms of $alpha_i$, the roots.
The author's point is that the $a_0$ and $a_n-1$, are easy to compute in terms of the roots, but it is not very useful to compute the rest of the coefficients this way.
answered Aug 3 at 10:20
rschwieb
99.5k1190225
answered Aug 3 at 10:20
rschwieb
99.5k1190225
answered Aug 3 at 10:20
rschwieb
99.5k1190225
answered Aug 3 at 10:20
rschwieb
99.5k1190225
99.5k1190225
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
add a comment |Â
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
Thanks very much for your reply. Could you please elaborate in a step-by-step explanation how you get to the desired result? I might have missed the point by trying to multiply the third equation out (as n can be very large).
– MrDerDart
Aug 3 at 11:09
1
1
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
@MrDerDart No, I'm not going to do that for you: multiplying the third one out is the entire point. Just ask yourself "after this mess gets multiplied out, what are all the things that look like $lambda x^n-1$ for some scalar $lambda$? Gather those up, and you have your solution. Perhaps you should just start out with toy examples to convince yourself how it works: take $(x-1)(x-2)(x-3)$ and try that first. It's not that hard... you don't have to ask strangers to do it!
– rschwieb
Aug 3 at 13:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
After carefully multiplying out the third equation with n = 5, I finally noticed the pattern and hence, I got to the solution. Thank you!
– MrDerDart
Aug 3 at 14:28
add a comment |Â
up vote
0
down vote
When you have a product of binomials like these
$$
alpha_n(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0
$$
Which for simplicity we are going to convert it to a monic polynomial:
$$(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0$$
You can express it also like this:
$$
(x-alpha_j)prod_ineq j^n(x-alpha_i)iff 1le j le n
$$
Now lets expand $prod_ineq j^n(x-alpha_i)$:
$$
(x-alpha_j)(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0) = x(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
-
alpha_j(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
$$
After expanding the equation from above we focus on the term $ (b_n-2-alpha_j)x^n-1 $.
Now we need to figure out what $b_n-2$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2:
$$
(x-alpha_1)(x-alpha_2) = x^2+(-alpha_1-alpha_2)x+alpha_1alpha_2
$$
For n = 3:
$$
(x-alpha_1)(x^2+(-alpha_2-alpha_3)x+alpha_2alpha_3) =
x^3+(-alpha_1-alpha_2-alpha_3)x^2 +...-alpha_1alpha_2alpha_3
$$
From this part it is easy to see that $b_n-2 = b_n-3-alpha_n$ where $b_0 = -alpha_2$.
That implies $a_n-1=b_n-2-alpha_1=-(alpha_1+alpha_2+alpha_3+..+alpha_n)=-sum_k=1^nalpha_k$
$$a_n-1=-sum_k=1^nalpha_kto -a_n-1=sum_k=1^nalpha_k$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$.
$$-fraca_n-1a_n=sum_k=1^nalpha_k$$
answered Aug 3 at 14:48
MarioE
486
add a comment |Â
up vote
0
down vote
When you have a product of binomials like these
$$
alpha_n(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0
$$
Which for simplicity we are going to convert it to a monic polynomial:
$$(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0$$
You can express it also like this:
$$
(x-alpha_j)prod_ineq j^n(x-alpha_i)iff 1le j le n
$$
Now lets expand $prod_ineq j^n(x-alpha_i)$:
$$
(x-alpha_j)(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0) = x(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
-
alpha_j(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
$$
After expanding the equation from above we focus on the term $ (b_n-2-alpha_j)x^n-1 $.
Now we need to figure out what $b_n-2$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2:
$$
(x-alpha_1)(x-alpha_2) = x^2+(-alpha_1-alpha_2)x+alpha_1alpha_2
$$
For n = 3:
$$
(x-alpha_1)(x^2+(-alpha_2-alpha_3)x+alpha_2alpha_3) =
x^3+(-alpha_1-alpha_2-alpha_3)x^2 +...-alpha_1alpha_2alpha_3
$$
From this part it is easy to see that $b_n-2 = b_n-3-alpha_n$ where $b_0 = -alpha_2$.
That implies $a_n-1=b_n-2-alpha_1=-(alpha_1+alpha_2+alpha_3+..+alpha_n)=-sum_k=1^nalpha_k$
$$a_n-1=-sum_k=1^nalpha_kto -a_n-1=sum_k=1^nalpha_k$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$.
$$-fraca_n-1a_n=sum_k=1^nalpha_k$$
answered Aug 3 at 14:48
MarioE
486
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When you have a product of binomials like these
$$
alpha_n(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0
$$
Which for simplicity we are going to convert it to a monic polynomial:
$$(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0$$
You can express it also like this:
$$
(x-alpha_j)prod_ineq j^n(x-alpha_i)iff 1le j le n
$$
Now lets expand $prod_ineq j^n(x-alpha_i)$:
$$
(x-alpha_j)(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0) = x(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
-
alpha_j(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
$$
After expanding the equation from above we focus on the term $ (b_n-2-alpha_j)x^n-1 $.
Now we need to figure out what $b_n-2$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2:
$$
(x-alpha_1)(x-alpha_2) = x^2+(-alpha_1-alpha_2)x+alpha_1alpha_2
$$
For n = 3:
$$
(x-alpha_1)(x^2+(-alpha_2-alpha_3)x+alpha_2alpha_3) =
x^3+(-alpha_1-alpha_2-alpha_3)x^2 +...-alpha_1alpha_2alpha_3
$$
From this part it is easy to see that $b_n-2 = b_n-3-alpha_n$ where $b_0 = -alpha_2$.
That implies $a_n-1=b_n-2-alpha_1=-(alpha_1+alpha_2+alpha_3+..+alpha_n)=-sum_k=1^nalpha_k$
$$a_n-1=-sum_k=1^nalpha_kto -a_n-1=sum_k=1^nalpha_k$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$.
$$-fraca_n-1a_n=sum_k=1^nalpha_k$$
answered Aug 3 at 14:48
MarioE
486
When you have a product of binomials like these
$$
alpha_n(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0
$$
Which for simplicity we are going to convert it to a monic polynomial:
$$(x-alpha_1)(x-alpha_2)dotsm(x-alpha_n)=0$$
You can express it also like this:
$$
(x-alpha_j)prod_ineq j^n(x-alpha_i)iff 1le j le n
$$
Now lets expand $prod_ineq j^n(x-alpha_i)$:
$$
(x-alpha_j)(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0) = x(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
-
alpha_j(x^n-1 + b_n-2x^n-2 + b_n-3x^n-3 + dots + b_1x + b_0)
$$
After expanding the equation from above we focus on the term $ (b_n-2-alpha_j)x^n-1 $.
Now we need to figure out what $b_n-2$ is. Lets start with the first two cases.
I am going to omit non-relevant terms just for simplicity.
For n = 2:
$$
(x-alpha_1)(x-alpha_2) = x^2+(-alpha_1-alpha_2)x+alpha_1alpha_2
$$
For n = 3:
$$
(x-alpha_1)(x^2+(-alpha_2-alpha_3)x+alpha_2alpha_3) =
x^3+(-alpha_1-alpha_2-alpha_3)x^2 +...-alpha_1alpha_2alpha_3
$$
From this part it is easy to see that $b_n-2 = b_n-3-alpha_n$ where $b_0 = -alpha_2$.
That implies $a_n-1=b_n-2-alpha_1=-(alpha_1+alpha_2+alpha_3+..+alpha_n)=-sum_k=1^nalpha_k$
$$a_n-1=-sum_k=1^nalpha_kto -a_n-1=sum_k=1^nalpha_k$$
Now remember that the polynomial equation given in the picture is not necessarily monic so we fix that by dividing the left side of the equation by $a_n$.
$$-fraca_n-1a_n=sum_k=1^nalpha_k$$
answered Aug 3 at 14:48
MarioE
486
answered Aug 3 at 14:48
MarioE
486
answered Aug 3 at 14:48
MarioE
486
answered Aug 3 at 14:48
MarioE
486
486
add a comment |Â
add a comment |Â
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