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If you were given the graph of a function, without the function's equation, is there a way to determine exact multiplicity (not just parity) of the roots of the function?

M0RF3US
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  • By graph do you mean a plot of the function on a piece of paper? – copper.hat Feb 26 '14 at 22:53
  • Plotting the function in the Cartesian Plane. – M0RF3US Feb 26 '14 at 22:55
  • A polynomial or any function in general? – naslundx Feb 26 '14 at 22:57
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    Is this restricted somehow? Otherwise $x^{2012}$ and $x^{2014}$ might be hard to distinguish... – draks ... Feb 26 '14 at 22:58
  • Mathematicians often use the term graph to mean the set of points ${(x,f(x))}$, in which case you know everything about the function. Also, it is a little ambiguous to talk about the multiplicity, for example, take $f(x) = x$ when $x \ge 0$ and $f(x) = x^2$ when $x <0$. Presumably you are talking about polynomials? – copper.hat Feb 26 '14 at 22:58
  • Exact multiplicity? I'd say no. For example, plot 4 functions, $y = x^2, y=x^3, y = x^4$ and $y = x^5$. Can you say anything about multiplicity of the root $x = 0$? – Kaster Feb 26 '14 at 22:59
  • Yes, Polynomials. And relatively simple ones, think high school precalc.

    I'm leaning towards you can't determine the exact multiplicty of the a zero/root solely by observing the graph...just the parity of the multiplicity of that zero.

     – M0RF3US Feb 26 '14 at 23:02

3 Answers3

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Well, naslundx is correct that we can determine what the equation of a polynomial is if we had a perfectly drawn graph. (We can then use that equation to figure out the multiplicity of the root.)

But if by "determine from the graph" you mean, just by looking at a hand-drawn figure, then no, there is no way. draks makes a good comment that illustrates why; I have better than 20-20 vision, but just by glancing at the graphs of $x^{2012}$ and $x^{2014}$, I could not determine which function was being graphed (unless the scale was huge). In fact, if the scale were small, I probably couldn't even tell the difference between $x^6$ and $x^{10}$ (or $(x+4)^5$ and $(x+4)^7$ etc.). As a result, I could not determine the multiplicity of the root from the graph alone, without the equation.

Andrew Kelley
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  • Yes, I agree. It wouldn't necessarily be a hand drawn graph, but one which you are not guaranteed the exactness of the other points. Thank you! – M0RF3US Feb 26 '14 at 23:14
  • Yeah, no problem, and good point about it not necessarily being a hand-drawn graph. – Andrew Kelley Feb 26 '14 at 23:36
  • nice example... – draks ... Feb 27 '14 at 00:55
  • Agreed. Those first two numbers look quite nice at around this time; when I first read your comment to the question, it reminded me of how some Putnam problems are made. (Eh, by the way, when I first wrote this answer, I was thinking of adding more examples, and so after you made this comment, I edited my answer.) – Andrew Kelley Feb 27 '14 at 14:51
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If we know the function $f$ is a polynomial, and that we can take arbitrarily many exact points along the graph, we can.

From the plot we can pick $n$ points $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ and using a Vandermonde matrix we can solve for all the coefficients, assuming $\deg f = n$.

From there we can 'easily' factorize (since we know the roots from the plot) to find the multiplicity of all roots.

Of course, this all depends on being able to find more or less exact values for the coordinates of the points.

naslundx
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  • My question would be for a high school precalculus class observing zeros, multiplicity and end behavior.

    And all you are given is the graph, the only points being the x/y intercepts.

     – M0RF3US Feb 26 '14 at 23:05
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    I assumed that if you have the graph, you can use points along the graph (not just the x/y intercepts). In that case, if you can solve systems of equations, this method works. If all you have are the x/y intercepts, my answer would be no. – naslundx Feb 26 '14 at 23:10
  • The only points you are given are the x/y intercepts, can't be sure about the rest.

    Thank you!

     – M0RF3US Feb 26 '14 at 23:12
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Yes. Kind of. You can distinguish at least between multiple and simple roots.

The thing is that if a polynomial has a root $a$ of multiplicity $n$, then its derivative has that same root with multiplicity $n-1$. In particular, if (and I think only if) $n>1$ (it's a multiple root), then the derivative is zero at $a$. So, multiple roots are zeros where the graph is flat. A root is simple iff the polynomial "cuts" the $x$-axis.

As an example that should already be familiar to you, a quadratic has a multiple root if and only if it "kisses" $x$-axis.

Also, I believe the first two derivatives of a curve are zero at a point if and only if the curve is flat at that point, but not a local extremum (ie, like $x^3$ at $0$). So you can even tell the difference between roots of multiplicity $2$ and $>2$.

Jack M
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