Mixing
Why can one equate polynomials by “inside power”s, when solving through expansion?
$begingroup$
Why can one equate $x^3-x+epsilon=0$ by "inside power"s, when solving through expansion?
https://www.maths.nottingham.ac.uk/plp/pmzjb1/G13AMD/book2.pdf, p.1
The author expands $x^3-x+epsilon=0$ with $x=x_0+epsilon x_1 + epsilon^2 x_2 + O(epsilon^3)$, like
$$(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)^3-(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)+color{red}{epsilon}=0 space (eq.1)$$
(and further)
Then groups terms by $epsilon$ powers.
Then equates terms grouped by $epsilon$ powers with what's "inside powers" in $(eq.1)$.
So that e.g. for $O(epsilon)$ level:
$$(3x_0-1)x_1+1=color{red}{2x_1}+color{red}{1}=0$$
My question:
Why is one allowed to equate things that are "inside powers", like the first redded term?
polynomials
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
$endgroup$
add a comment |
$begingroup$
Why can one equate $x^3-x+epsilon=0$ by "inside power"s, when solving through expansion?
https://www.maths.nottingham.ac.uk/plp/pmzjb1/G13AMD/book2.pdf, p.1
The author expands $x^3-x+epsilon=0$ with $x=x_0+epsilon x_1 + epsilon^2 x_2 + O(epsilon^3)$, like
$$(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)^3-(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)+color{red}{epsilon}=0 space (eq.1)$$
(and further)
Then groups terms by $epsilon$ powers.
Then equates terms grouped by $epsilon$ powers with what's "inside powers" in $(eq.1)$.
So that e.g. for $O(epsilon)$ level:
$$(3x_0-1)x_1+1=color{red}{2x_1}+color{red}{1}=0$$
My question:
Why is one allowed to equate things that are "inside powers", like the first redded term?
polynomials
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
$endgroup$
add a comment |
$begingroup$
Why can one equate $x^3-x+epsilon=0$ by "inside power"s, when solving through expansion?
https://www.maths.nottingham.ac.uk/plp/pmzjb1/G13AMD/book2.pdf, p.1
The author expands $x^3-x+epsilon=0$ with $x=x_0+epsilon x_1 + epsilon^2 x_2 + O(epsilon^3)$, like
$$(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)^3-(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)+color{red}{epsilon}=0 space (eq.1)$$
(and further)
Then groups terms by $epsilon$ powers.
Then equates terms grouped by $epsilon$ powers with what's "inside powers" in $(eq.1)$.
So that e.g. for $O(epsilon)$ level:
$$(3x_0-1)x_1+1=color{red}{2x_1}+color{red}{1}=0$$
My question:
Why is one allowed to equate things that are "inside powers", like the first redded term?
polynomials
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
$endgroup$
Why can one equate $x^3-x+epsilon=0$ by "inside power"s, when solving through expansion?
https://www.maths.nottingham.ac.uk/plp/pmzjb1/G13AMD/book2.pdf, p.1
The author expands $x^3-x+epsilon=0$ with $x=x_0+epsilon x_1 + epsilon^2 x_2 + O(epsilon^3)$, like
$$(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)^3-(x_0+color{red}{epsilon x_1} + epsilon^2 x_2)+color{red}{epsilon}=0 space (eq.1)$$
(and further)
Then groups terms by $epsilon$ powers.
Then equates terms grouped by $epsilon$ powers with what's "inside powers" in $(eq.1)$.
So that e.g. for $O(epsilon)$ level:
$$(3x_0-1)x_1+1=color{red}{2x_1}+color{red}{1}=0$$
My question:
Why is one allowed to equate things that are "inside powers", like the first redded term?
polynomials
polynomials
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
asked Jan 13 at 13:02
mavaviljmavavilj
2,77011037
2,77011037
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Since when $epsilon=0$ the given equation has simple roots at $x=-1,0,1$ labeled by $x_1,x_2,x_3$, respectively, we can apply inverse function theorem near each $x_i$ to conclude that parametrized root $x_i(epsilon)$ is in fact an analytic function of $epsilonin mathbb{C}$. (with the domain being $|epsilon|<eta$ for sufficiently small $eta>0$.)
answered Jan 13 at 16:34
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
he is trying to find an approximate solution for small value of $epsilon$.SInce right hand side is zero, the author equates all the coefficients of $epsilon$ to Zero.
answered Jan 13 at 13:10
HarishHarish
567415
$endgroup$
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Since when $epsilon=0$ the given equation has simple roots at $x=-1,0,1$ labeled by $x_1,x_2,x_3$, respectively, we can apply inverse function theorem near each $x_i$ to conclude that parametrized root $x_i(epsilon)$ is in fact an analytic function of $epsilonin mathbb{C}$. (with the domain being $|epsilon|<eta$ for sufficiently small $eta>0$.)
answered Jan 13 at 16:34
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
Since when $epsilon=0$ the given equation has simple roots at $x=-1,0,1$ labeled by $x_1,x_2,x_3$, respectively, we can apply inverse function theorem near each $x_i$ to conclude that parametrized root $x_i(epsilon)$ is in fact an analytic function of $epsilonin mathbb{C}$. (with the domain being $|epsilon|<eta$ for sufficiently small $eta>0$.)
answered Jan 13 at 16:34
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
Since when $epsilon=0$ the given equation has simple roots at $x=-1,0,1$ labeled by $x_1,x_2,x_3$, respectively, we can apply inverse function theorem near each $x_i$ to conclude that parametrized root $x_i(epsilon)$ is in fact an analytic function of $epsilonin mathbb{C}$. (with the domain being $|epsilon|<eta$ for sufficiently small $eta>0$.)
answered Jan 13 at 16:34
SongSong
12.9k631
$endgroup$
Since when $epsilon=0$ the given equation has simple roots at $x=-1,0,1$ labeled by $x_1,x_2,x_3$, respectively, we can apply inverse function theorem near each $x_i$ to conclude that parametrized root $x_i(epsilon)$ is in fact an analytic function of $epsilonin mathbb{C}$. (with the domain being $|epsilon|<eta$ for sufficiently small $eta>0$.)
answered Jan 13 at 16:34
SongSong
12.9k631
edited Jan 13 at 16:41
answered Jan 13 at 16:34
SongSong
12.9k631
answered Jan 13 at 16:34
SongSong
12.9k631
answered Jan 13 at 16:34
SongSong
12.9k631
12.9k631
add a comment |
add a comment |
$begingroup$
he is trying to find an approximate solution for small value of $epsilon$.SInce right hand side is zero, the author equates all the coefficients of $epsilon$ to Zero.
answered Jan 13 at 13:10
HarishHarish
567415
$endgroup$
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
add a comment |
$begingroup$
he is trying to find an approximate solution for small value of $epsilon$.SInce right hand side is zero, the author equates all the coefficients of $epsilon$ to Zero.
answered Jan 13 at 13:10
HarishHarish
567415
$endgroup$
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
add a comment |
$begingroup$
he is trying to find an approximate solution for small value of $epsilon$.SInce right hand side is zero, the author equates all the coefficients of $epsilon$ to Zero.
answered Jan 13 at 13:10
HarishHarish
567415
$endgroup$
he is trying to find an approximate solution for small value of $epsilon$.SInce right hand side is zero, the author equates all the coefficients of $epsilon$ to Zero.
answered Jan 13 at 13:10
HarishHarish
567415
answered Jan 13 at 13:10
HarishHarish
567415
answered Jan 13 at 13:10
HarishHarish
567415
answered Jan 13 at 13:10
HarishHarish
567415
567415
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
add a comment |
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
So you mean that he assumes that in $(eq.1)$ each of the terms: $x^3, -x, epsilon$ must be zero (so everything inside them must as well) in order to have zero on R.H.S.?
$endgroup$
– mavavilj
Jan 13 at 13:24
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
$begingroup$
What if some of the terms had been multiplied by scalar? E.g. $4x^3-x+epsilon=0$. Would the $4$ be unreflected?
$endgroup$
– mavavilj
Jan 13 at 13:42
add a comment |
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