Can $f(x)=ax^{2k-1}+…$ and $g(x)=bx^{2n}+…$ ($k,n \in \mathbb N$) have the same shape at some interval?

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

• Then what happens when $k, n \to \infty$ ? If then, we can have infinite roots – Verthele Aug 1 '19 at 15:36
• $k,n\to \infty$ 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. – fleablood Aug 1 '19 at 15:40
• .... also we we'd only have countably many roots. – 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!

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