Prove uniqueness and existence of ANY polynomial points/pairs The Next CEO of Stack...
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Prove uniqueness and existence of ANY polynomial points/pairs
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In the book "A Programmer's introduction to mathematics" by J Kun., page 15, it states:
Theorem 2.2: For any integer $n>=0$ and any list of $n+1$ points $(x_0,y_0),(x_1,y_1)...(x_n,y_n)$ in $mathbb{R^2}$ with $x_0 < x_1 < ... < x_n$, there exists a unique degree $n$ polynomial $p(x)$ such that $p(x_i)=y_i$ for all $i$.
As I was reading this, I thought: does this mean that if I make up any pairs of points, where $x_1 < x_n $, there will be a function that will satisfy it?
So I made these up:
(2,10),(3,15),(4,23),(5,30)
Which of course don't have a solution but do satisfy the constraint imposed on $x$ (in the theorem there is no constraint imposed on $y$) although they do have an approximation .
So my question is this, and forgive my noobiness,
1) is the statement exact and I'm missing something
OR
2) is this just normal math lingo and everyone knows that of course there won't be a solution for any sequence of pairs but if such a sequence is the output of a polynomial function, then, it will be the only one, ie. there will be no other polynomial function that will output the same exact sequence of pairs?
I lean toward option 2) but want to be absolutely sure I'm not missing something.
polynomials proof-explanation
asked Mar 16 at 17:12
danoobis mdanoobis m
32
$endgroup$
add a comment |
$begingroup$
In the book "A Programmer's introduction to mathematics" by J Kun., page 15, it states:
Theorem 2.2: For any integer $n>=0$ and any list of $n+1$ points $(x_0,y_0),(x_1,y_1)...(x_n,y_n)$ in $mathbb{R^2}$ with $x_0 < x_1 < ... < x_n$, there exists a unique degree $n$ polynomial $p(x)$ such that $p(x_i)=y_i$ for all $i$.
As I was reading this, I thought: does this mean that if I make up any pairs of points, where $x_1 < x_n $, there will be a function that will satisfy it?
So I made these up:
(2,10),(3,15),(4,23),(5,30)
Which of course don't have a solution but do satisfy the constraint imposed on $x$ (in the theorem there is no constraint imposed on $y$) although they do have an approximation .
So my question is this, and forgive my noobiness,
1) is the statement exact and I'm missing something
OR
2) is this just normal math lingo and everyone knows that of course there won't be a solution for any sequence of pairs but if such a sequence is the output of a polynomial function, then, it will be the only one, ie. there will be no other polynomial function that will output the same exact sequence of pairs?
I lean toward option 2) but want to be absolutely sure I'm not missing something.
polynomials proof-explanation
asked Mar 16 at 17:12
danoobis mdanoobis m
32
$endgroup$
$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04
add a comment |
$begingroup$
In the book "A Programmer's introduction to mathematics" by J Kun., page 15, it states:
Theorem 2.2: For any integer $n>=0$ and any list of $n+1$ points $(x_0,y_0),(x_1,y_1)...(x_n,y_n)$ in $mathbb{R^2}$ with $x_0 < x_1 < ... < x_n$, there exists a unique degree $n$ polynomial $p(x)$ such that $p(x_i)=y_i$ for all $i$.
As I was reading this, I thought: does this mean that if I make up any pairs of points, where $x_1 < x_n $, there will be a function that will satisfy it?
So I made these up:
(2,10),(3,15),(4,23),(5,30)
Which of course don't have a solution but do satisfy the constraint imposed on $x$ (in the theorem there is no constraint imposed on $y$) although they do have an approximation .
So my question is this, and forgive my noobiness,
1) is the statement exact and I'm missing something
OR
2) is this just normal math lingo and everyone knows that of course there won't be a solution for any sequence of pairs but if such a sequence is the output of a polynomial function, then, it will be the only one, ie. there will be no other polynomial function that will output the same exact sequence of pairs?
I lean toward option 2) but want to be absolutely sure I'm not missing something.
polynomials proof-explanation
asked Mar 16 at 17:12
danoobis mdanoobis m
32
$endgroup$
In the book "A Programmer's introduction to mathematics" by J Kun., page 15, it states:
Theorem 2.2: For any integer $n>=0$ and any list of $n+1$ points $(x_0,y_0),(x_1,y_1)...(x_n,y_n)$ in $mathbb{R^2}$ with $x_0 < x_1 < ... < x_n$, there exists a unique degree $n$ polynomial $p(x)$ such that $p(x_i)=y_i$ for all $i$.
As I was reading this, I thought: does this mean that if I make up any pairs of points, where $x_1 < x_n $, there will be a function that will satisfy it?
So I made these up:
(2,10),(3,15),(4,23),(5,30)
Which of course don't have a solution but do satisfy the constraint imposed on $x$ (in the theorem there is no constraint imposed on $y$) although they do have an approximation .
So my question is this, and forgive my noobiness,
1) is the statement exact and I'm missing something
OR
2) is this just normal math lingo and everyone knows that of course there won't be a solution for any sequence of pairs but if such a sequence is the output of a polynomial function, then, it will be the only one, ie. there will be no other polynomial function that will output the same exact sequence of pairs?
I lean toward option 2) but want to be absolutely sure I'm not missing something.
polynomials proof-explanation
polynomials proof-explanation
asked Mar 16 at 17:12
danoobis mdanoobis m
32
asked Mar 16 at 17:12
danoobis mdanoobis m
32
asked Mar 16 at 17:12
danoobis mdanoobis m
32
asked Mar 16 at 17:12
danoobis mdanoobis m
32
asked Mar 16 at 17:12
danoobis mdanoobis m
32
32
$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04
add a comment |
$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04
$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Lagrange polynomials are constructed such that they pass through a set of points ${(x_i,y_i)}_{i=1}^n$, with $x_1<ldots<x_n$.
The link has the proof, but in general terms, the polynomials are a sum of terms that are zero at all the other points except $x_i$, where that term has the value $y_i$.
You may also want to take a look at some explanation of Lagrange polynomials.
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
$endgroup$
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Lagrange polynomials are constructed such that they pass through a set of points ${(x_i,y_i)}_{i=1}^n$, with $x_1<ldots<x_n$.
The link has the proof, but in general terms, the polynomials are a sum of terms that are zero at all the other points except $x_i$, where that term has the value $y_i$.
You may also want to take a look at some explanation of Lagrange polynomials.
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
$endgroup$
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
add a comment |
$begingroup$
Lagrange polynomials are constructed such that they pass through a set of points ${(x_i,y_i)}_{i=1}^n$, with $x_1<ldots<x_n$.
The link has the proof, but in general terms, the polynomials are a sum of terms that are zero at all the other points except $x_i$, where that term has the value $y_i$.
You may also want to take a look at some explanation of Lagrange polynomials.
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
$endgroup$
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
add a comment |
$begingroup$
Lagrange polynomials are constructed such that they pass through a set of points ${(x_i,y_i)}_{i=1}^n$, with $x_1<ldots<x_n$.
The link has the proof, but in general terms, the polynomials are a sum of terms that are zero at all the other points except $x_i$, where that term has the value $y_i$.
You may also want to take a look at some explanation of Lagrange polynomials.
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
$endgroup$
Lagrange polynomials are constructed such that they pass through a set of points ${(x_i,y_i)}_{i=1}^n$, with $x_1<ldots<x_n$.
The link has the proof, but in general terms, the polynomials are a sum of terms that are zero at all the other points except $x_i$, where that term has the value $y_i$.
You may also want to take a look at some explanation of Lagrange polynomials.
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
answered Mar 16 at 18:08
ErtxiemErtxiem
55811
55811
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
add a comment |
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
Thanks for that 2nd link, it is particularly helpful!
$endgroup$
– danoobis m
Mar 17 at 0:04
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
$begingroup$
You're welcome, @danoobis-m.
$endgroup$
– Ertxiem
Mar 17 at 1:06
add a comment |
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$begingroup$
1 is true and you are indeed missing something - note that the cubic here will have rational coefficients not necessarily integral
$endgroup$
– Conrad
Mar 16 at 21:37
$begingroup$
Thank you! Indeed, I hadn't realized it the coefficients could be rational! But still (and again forgive my noobiness) why can't an exact solution be found for such a 'simple' equation, ie one with four terms? It's sort of confusing that we can prove there is such a solution but we can't actually find it..
$endgroup$
– danoobis m
Mar 16 at 23:58
$begingroup$
We can find it by equating coefficients sure, just that computations can be complicated - here it's 4 linear equations with 4 unknowns but fairly high coefficients. The Lagrange interpolation formula does it also but it's fairly impractical as each term had a definite formula but you have to add 4 here and again computations explode fast - usually ad hoc computations should first be tried in specific examples
$endgroup$
– Conrad
Mar 17 at 2:04