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Questions on a self-made theorem about polynomials
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I recently came up with this theorem:
For any complex polynomial $P$ degree $n$:
$$ sumlimits_k=0^n+1(-1)^kbinomn+1kP(a+kb) = 0quad forall a,b inmathbbC$$
Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
- Has this been discovered? If yes, what's the formal name for this phenomenon?
- Is it significant/Are there important consequences of this being true?
- Can this be generalized to non-polynomials?
polynomials summation binomial-coefficients
asked 1 hour ago
Felix Fourcolor
937
add a comment |Â
up vote
3
down vote
favorite
I recently came up with this theorem:
For any complex polynomial $P$ degree $n$:
$$ sumlimits_k=0^n+1(-1)^kbinomn+1kP(a+kb) = 0quad forall a,b inmathbbC$$
Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
- Has this been discovered? If yes, what's the formal name for this phenomenon?
- Is it significant/Are there important consequences of this being true?
- Can this be generalized to non-polynomials?
polynomials summation binomial-coefficients
asked 1 hour ago
Felix Fourcolor
937
1
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
1
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I recently came up with this theorem:
For any complex polynomial $P$ degree $n$:
$$ sumlimits_k=0^n+1(-1)^kbinomn+1kP(a+kb) = 0quad forall a,b inmathbbC$$
Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
- Has this been discovered? If yes, what's the formal name for this phenomenon?
- Is it significant/Are there important consequences of this being true?
- Can this be generalized to non-polynomials?
polynomials summation binomial-coefficients
asked 1 hour ago
Felix Fourcolor
937
I recently came up with this theorem:
For any complex polynomial $P$ degree $n$:
$$ sumlimits_k=0^n+1(-1)^kbinomn+1kP(a+kb) = 0quad forall a,b inmathbbC$$
Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
- Has this been discovered? If yes, what's the formal name for this phenomenon?
- Is it significant/Are there important consequences of this being true?
- Can this be generalized to non-polynomials?
polynomials summation binomial-coefficients
polynomials summation binomial-coefficients
asked 1 hour ago
Felix Fourcolor
937
asked 1 hour ago
Felix Fourcolor
937
edited 1 hour ago
asked 1 hour ago
Felix Fourcolor
937
asked 1 hour ago
Felix Fourcolor
937
asked 1 hour ago
Felix Fourcolor
937
937
1
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
1
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago
add a comment |Â
1
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
1
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago
1
1
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
1
1
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago
add a comment |Â
1 Answer
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In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $Delta$: $Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $Delta_b$ for your operator.
The most important feature of the $Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $Delta_b P(x) = a_dleft((x+b)^d-x^dright)+Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+dchoose 1bx^d-1+ldots$, so $(x+b)^d-x^d=dchoose 1bx^d-1+ldots$ is a polynomial of degree at most $d-1$, and thus $Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_dleft((x+b)^d-x^dright)$ and $Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $Delta_bP(x)$ is exactly $d-1$ when $b neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $Delta_b$ reduces the degree by one, then $(Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(Delta_b)^d+1P(x)$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $Delta$ is very similar to a derivative in many ways, with the Newton polynomials $xchoose d=frac1d!x(x-1)(x-2)cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
edited 13 mins ago
darij grinberg
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $Delta$: $Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $Delta_b$ for your operator.
The most important feature of the $Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $Delta_b P(x) = a_dleft((x+b)^d-x^dright)+Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+dchoose 1bx^d-1+ldots$, so $(x+b)^d-x^d=dchoose 1bx^d-1+ldots$ is a polynomial of degree at most $d-1$, and thus $Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_dleft((x+b)^d-x^dright)$ and $Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $Delta_bP(x)$ is exactly $d-1$ when $b neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $Delta_b$ reduces the degree by one, then $(Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(Delta_b)^d+1P(x)$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $Delta$ is very similar to a derivative in many ways, with the Newton polynomials $xchoose d=frac1d!x(x-1)(x-2)cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
edited 13 mins ago
darij grinberg
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
add a comment |Â
up vote
5
down vote
In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $Delta$: $Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $Delta_b$ for your operator.
The most important feature of the $Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $Delta_b P(x) = a_dleft((x+b)^d-x^dright)+Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+dchoose 1bx^d-1+ldots$, so $(x+b)^d-x^d=dchoose 1bx^d-1+ldots$ is a polynomial of degree at most $d-1$, and thus $Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_dleft((x+b)^d-x^dright)$ and $Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $Delta_bP(x)$ is exactly $d-1$ when $b neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $Delta_b$ reduces the degree by one, then $(Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(Delta_b)^d+1P(x)$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $Delta$ is very similar to a derivative in many ways, with the Newton polynomials $xchoose d=frac1d!x(x-1)(x-2)cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
edited 13 mins ago
darij grinberg
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
add a comment |Â
up vote
5
down vote
up vote
5
down vote
In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $Delta$: $Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $Delta_b$ for your operator.
The most important feature of the $Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $Delta_b P(x) = a_dleft((x+b)^d-x^dright)+Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+dchoose 1bx^d-1+ldots$, so $(x+b)^d-x^d=dchoose 1bx^d-1+ldots$ is a polynomial of degree at most $d-1$, and thus $Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_dleft((x+b)^d-x^dright)$ and $Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $Delta_bP(x)$ is exactly $d-1$ when $b neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $Delta_b$ reduces the degree by one, then $(Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(Delta_b)^d+1P(x)$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $Delta$ is very similar to a derivative in many ways, with the Newton polynomials $xchoose d=frac1d!x(x-1)(x-2)cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
edited 13 mins ago
darij grinberg
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $Delta$: $Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $Delta_b$ for your operator.
The most important feature of the $Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $Delta_b P(x) = a_dleft((x+b)^d-x^dright)+Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+dchoose 1bx^d-1+ldots$, so $(x+b)^d-x^d=dchoose 1bx^d-1+ldots$ is a polynomial of degree at most $d-1$, and thus $Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_dleft((x+b)^d-x^dright)$ and $Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $Delta_bP(x)$ is exactly $d-1$ when $b neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $Delta_b$ reduces the degree by one, then $(Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(Delta_b)^d+1P(x)$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $Delta$ is very similar to a derivative in many ways, with the Newton polynomials $xchoose d=frac1d!x(x-1)(x-2)cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
edited 13 mins ago
darij grinberg
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
edited 13 mins ago
darij grinberg
9,47132960
edited 13 mins ago
darij grinberg
9,47132960
edited 13 mins ago
darij grinberg
9,47132960
9,47132960
answered 1 hour ago
Steven Stadnicki
40.3k765119
answered 1 hour ago
Steven Stadnicki
40.3k765119
answered 1 hour ago
Steven Stadnicki
40.3k765119
40.3k765119
add a comment |Â
add a comment |Â
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1
Tiny note: you use $n$ differently in two different contexts (as your difference and as your degree) and you should probably use different variables for those two things.
– Steven Stadnicki
1 hour ago
1
I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 leq m leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html
– Hw Chu
1 hour ago
For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Qleft(xright) = P left(a+xbright)$.
– darij grinberg
6 mins ago