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Say you have a polynomial function like $f(x) = 2x^3 - 6x^2 + 8$. This function won't be point symmetrical over $(0, 0)$, nor will it be symmetrical over the y-axis, but after plotting it, you can see that it's point symmetrical over the point $(1, 4)$.

For a polynomial function like $g(x) = (x+1)^3$, you can tell that it was shifted by one unit to the left on the x-axis, $h(x) = x^3$ is symmetrical over the point $(0, 0)$, so $g(x)$ will be symmetrical over the point $(-1, 0)$. I don't really see how you could use this approach for $f(x)$ yet though.

One approach I found that works for $f(x)$ is to take the second derivative and set it to $0$.

$$f'(x) = 6x^2 - 12x$$ $$f''(x) = 12x - 12$$ $$f''(x) = 0$$ $$12x = 12$$ $$\fbox{x = 1}$$

Now to find the corresponding y value, plug in $x = 1$ into $f(x)$.

This second derivative approach doesn't seem to work for all functions though, there could be a point where the graph changes concavity but still isn't point symmetrical over.

Now my question is, is there a way to determine over which point any function like this will be symmetrical over without plotting it?


In the comments, it was stated that for a point of symmetry to exist, all the derivatives of the function must be (anti-)symmetric with respect to a line x = a. How can you check if a function is (anti-)symmetric to this line algebraically?

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    $\begingroup$ $f(2a-x) = 2b-f(x)\implies f'(2a-x) =f'(x) \implies f^{(k)}(2a-x)= (-1)^{k-1}f^{(k)}(x)$. This means that for a point of symmetry to exist, all the derivatives of the function must be (anti-)symmetric with respect to a line $x=a$ $\endgroup$
    – Exodd
    Sep 25, 2022 at 12:46
  • $\begingroup$ For an odd-degree function of degree $d$, you can see the $d-1$ derivative is a line, that is anti-symmetric with respect to the line $x=a$ where $a$ is its root $\endgroup$
    – Exodd
    Sep 25, 2022 at 12:49
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    $\begingroup$ If $f(x)$ is symmetric wrt the point $(a,b)$ then $f(x)=y$ implies $f(2a-x) = 2b-y$ $\endgroup$
    – Exodd
    Sep 25, 2022 at 13:58
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    $\begingroup$ Here it seems we are not talking about functions in general, but about polynomials, so the word "polynomials" should appear, at least in the question text and probably in the title too. Also there should be a "polynomial" tag. $\endgroup$ Oct 8, 2022 at 10:01
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    $\begingroup$ Since $a=0$ in this case, $f'(2a-x) = f'(-x) = 3x^2 = f'(x)$ $\endgroup$
    – Exodd
    Oct 8, 2022 at 11:30

1 Answer 1

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If $f(x) = a_n x^n + a_{n-1} x^{n-1} +...$, the only possible abscissa for a symmetry point is the roots mean, i.e. $-\frac {a_{n-1}} {n \, a_n}$.

First proof
As was said in the comments, if $f$ is symmetric or anti-symmetric, all its derivatives are symmetric or anti-symmetric around the same abscissa. This alternates: even degree polynomials are symmetric (this is because of the behavior on $\infty$), odd degree polynomials are anti-symmetric.

So we can find this abscissa $a$ with the $n-1$th derivative: $f^{(n-1)}(x) = a_n \, n! \, x + (n-1)! \, a_{n-1} = 0$
$x=-\frac {a_{n-1}} {n \, a_n}$.

Tentative of a second proof
Another possible proof would proceed like this.
If there is a symmetry or anti-symmetry point $(a,b)$, we can vertically translate $f$ by subtracting $b$: $g(x)=f(x)-b$.
Now $g$ is (anti-)symmetric around $(a,0)$. So its real roots are symmetric around $a$.

If $g$ has all its $n$ roots real, because of the symmetry, the roots mean will be $a$, so $a=-\frac {a_{n-1}} {n \, a_n}$.

Now how to deal with the case where $g$ has some non-real complex roots?
One possible way would be to show that we can build a polynomial $h$ such that $h-g$ is a polynomial with degree at most $n-2$, and $h$ is (anti-)symmetric around a point with abscissa $a$ such as $g$. We would then be able to apply $a=-\frac {a_{n-1}} {n \, a_n}$: as we have only added an $n-2$ degree polynomial, $h$ has same first two coefficients $a_n$ and $a_{n-1}$ as $f$ and $g$.
Unfortunately I don't succeed in proving that.

Third proof, which gives the complete list of conditions
The method here is to directly expand the (anti-)symmetry equation.

If $n$ is odd, $f$ is anti-symmetric, so
$f(a+x)+f(a-x)=2b$
$\sum_{k=0}^n a_k(a+x)^k + \sum_{k=0}^n a_k(a-x)^k=2b$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} + \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=2b$
This can be split into separate equations, one for each power of $x$, i.e. one for each $j$. Equations for $j$ odd give $0=0$. Equations for $j$ even and $\ge 2$ give the following:
$\forall j$ even and $\ge 2$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} + \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
$\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$
For $j=n-1$, we get
$\sum_{k=n-1}^n a_k {k \choose {n-1}} a^{k-(n-1)} = 0$
$a_{n-1} a^0 + a_n \, n \, a^1 = 0$
$a = - \frac {a_{n-1}} {n \, a_n}$
The other conditions are obtained with the other even values for $j$. The value for $b$ can be computed from $j=0$ (but that's not much different from computing $b=f(a)$). Unfortunately, the conditions grow ugly when $j$ decreases; but it is the necessary and sufficient list and there does not seem to be a simpler way.
As example, the next condition, for $j=n-3$, expands into:
$3n^2a_n^2a_{n-3} - 3n(n-2)a_na_{n-1}a_{n-2}+(n-1)(n-2)a_{n-1}^3=0$

If $n$ is even, $f$ is symmetric, so
$f(a+x)-f(a-x)=0$
$\sum_{k=0}^n a_k(a+x)^k - \sum_{k=0}^n a_k(a-x)^k=0$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} - \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=0$
As above, this splits into separate equations, one for each $j$. Equations for $j$ even give $0=0$. Equations for $j$ odd give:
$\forall j$ odd and $\ge 1$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} - \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
so $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ as above.
For $j=n-1$, we get $a = - \frac {a_{n-1}} {n \, a_n}$ as before.

The two cases (n even; n odd) can be summarized in the following conditions:
$f(x) = a_n x^n + a_{n-1} x^{n-1} +...$ is symmetric (if $n$ even) or anti-symmetric (if $n$ odd) around $a=-\frac {a_{n-1}} {n \, a_n}$ (and $b=f(a)$, for anti-symmetry) iff:
$\fbox {$\forall j=n-1, n-3, ... \text{ 2 or 1, }\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$}$

Example
Using $f(x)=(x-4)(x-2)(2x+1)(x+3)(x+5)-209$
which means we have an anti-symmetry around $(a=-\frac 1 2,b=-209)$ $f(x)=2x^5+5x^4-48x^3-77x^2+214x-89$
(The reason why I don't use the example in the question post, is because I would like to have no null coefficient, and $a_n \ne 1$).
$a=-\frac {a_{n-1}} {n\,a_n}=-\frac {5} {5 \times 2}=-\frac 1 2$
$3 n^2 a_n^2 a_{n-3} - 3n(n-2)a_n a_{n-1} a_{n-2}+(n-1)(n-2)a_{n-1}^3 =$
$3 \times 5^2 \times 2^2 \times (-77) - 3 \times 5 \times 3 \times 2 \times 5 \times (-48) + 4 \times 3 \times 5^3=-23100+21600+1500=0$
There is no other condition (the one above is $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ for $j=n-3=2$), so the polynomial is anti-symmetric around $a=-\frac 1 2$ and $b=f(a)=-209$.

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  • $\begingroup$ Thanks a lot for your answer! $\endgroup$ Oct 8, 2022 at 19:11
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    $\begingroup$ @ILikeMathematics Thanks. There was actually a mistake, which is now corrected: the case $j=0$ does not give a condition (because there is $2b$ on the right hand-side). So the example was misleading: the case $3 n^2 a_n^2 a_{n-3} - 3n(n-2)a_n a_{n-1} a_{n-2}+(n-1)(n-2)a_{n-1}^3 =0$ was actually not a condition for anti-symmetry, but the consequence of $b=0$! So I changed the example for one where $b\ne 0$. And a degree $5$ example, so that $3 n^2 a_n^2 a_{n-3} - 3n(n-2)a_n a_{n-1} a_{n-2}+(n-1)(n-2)a_{n-1}^3 =0$ is really a condition for anti-symmetry. $\endgroup$ Oct 8, 2022 at 23:07

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