When $k \in \mathbb N$ and $n \in \mathbb N$

$f(x)$ is a $2k-1$ degree polynomial

$g(x)$ is a $2n$ degree polynomial.

By 'having the same shape at some interval', I mean that when given proper $f(x)$ and $g(x)$ , the graphs of $f(x)$ and $g(x)$ are identical at some interval $[p, q]$.

Can the two functions have the same shape at some interval?


No, this will mean that $f(x)-g(x)=0$ will have infinitely many solutions on the interval $[p,q]$ which is impossible because this equation has at most $\max(2k-1, 2n)$ real roots

  • $\begingroup$ Then what happens when $k, n \to \infty$ ? If then, we can have infinite roots $\endgroup$ – Verthele Aug 1 '19 at 15:36
  • $\begingroup$ $k,n\to \infty$ is a limit; not an actual value. It does not exist in actuality (or even hypothetically). There's no such thing as an infinite degree polynomial. $\endgroup$ – fleablood Aug 1 '19 at 15:40
  • $\begingroup$ .... also we we'd only have countably many roots. $\endgroup$ – fleablood Aug 1 '19 at 15:41

As polynomials, $f$ has $2k$ coefficients, and $g$ has $2n+1$ coefficients. Let $M=\max(2k, 2n+1)$; if $f(x) = g(x)$ at $M+1$ different points, then all their coefficients must be equal, which means $f=g$ everywhere. If they are equal on an interval, then they are equal at infinitely many points, and therefore equal everywhere.

If you take $f$ and $g$ of different degree, or you ensure at least one coefficient is different, then they can never agree on an interval.

Hope this helps!

  • $\begingroup$ "if f(x)=g(x) at M different points, then all their coefficients must be equal" Could you go into more detail? $\endgroup$ – fleablood Aug 1 '19 at 15:46
  • $\begingroup$ Apologies, I meant $M+1$ different points. $\endgroup$ – R_B Aug 1 '19 at 15:52

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