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Given a real polynomial $p(x)$ of degree $n$ such that $p$ and all its derivatives are nonnegative on some open interval $I$, how many times can $p$ intersect the exponential function (in $I$)? In particular, can they intersect more than twice?

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If (hyper-)osculation counts as multiple intersections, the Taylor polynomials of the exponential function would intersect the exponential more than once... –  J. M. Oct 26 '11 at 4:58
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Let $T_n(x) = \sum_{k=0}^n x^n/n!$ be the Maclaurin polynomial of degree $n$ for $e^x$. Then $Q_n(x) = T_{n-1}(x) + (e - T_{n-1}(1)) x^n$ has $Q_n(1) = e$, $Q_n(x) - e^x = O(x^n)$ as $x \to 0$, and $Q_n(x)$ and all its derivatives are positive at $x=0$ (and therefore for all $x \ge 0$). Let $P_{n,\delta}(x)$ be the Lagrange interpolating polynomial for $e^x$ at the $n+1$ points $j \delta$, $j = 1 \ldots n$ and $1$, where $\delta > 0$. As $\delta \to 0$, $P_{n,\delta}(x)$ converges to $Q_n(x)$ and its derivatives converge to the corresponding derivatives of $Q_n(x)$. In particular, if $\delta > 0$ is sufficiently small, $P_{n,\delta}$ and all its derivatives are positive at $x=0$ (and thus for $x \ge 0$) and it intersects the exponential function in $n+1$ distinct points in $(0, \infty)$.

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Yes, they can intersect more than twice.

The polynomial $p(x) = \frac{1}{2}(x^2+x)+1$ and all its derivatives are nonnegative in the interval $(- \frac{1}{2}, \infty)$, and yet $p(x)$ intersects the function $2^x$ in at least three points: $0$, $1$ and $2$.

If $2^x$ doesn't qualify as the exponential function, then we can modify the above example as follows. Define $q(x) = p(\frac{x}{\ln 2})$, and consider the interval $(- \frac12 \ln 2, \infty)$. Then $q(x)$ intersects $\mathrm{e}^x$ at the points $0$, $\ln 2$ and $2 \ln 2$.

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More generally, if $\deg p = n$, I will not be surprised if a polynomial (satisfying the given side conditions) can be arranged to intersect the exponential function in $n+1$ points. On the other hand, it's an interesting question whether $n+2$ intersections are possible. –  Srivatsan Oct 26 '11 at 5:23
    
Thanks for the fast answer. The $n+1$st derivative of $e^x - p(x)$ will be positive everywhere, so repeated application of Rolle's Theorem tells us the maximum number of intersection points is $n+1$. However, it is not clear to me yet that $n+1$ is always achievable. –  anon Oct 26 '11 at 5:33
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They can intersect at least $n+1$ times. Consider the polynomial of degree $n$ that interpolates values $y_j =e^{-aj}$ at $x_j = -aj$, $j = 0 \ldots n$: $$ p(x) = \sum_{j=0}^n e^{-aj} \prod_{i \ne j} \frac{x + a i}{-a j + a i} $$ The $j = 0$ term is $\prod_{i=1}^n (x + a i)/(a i)$ which has all its coefficients positive, and the factor $e^{-aj}$ for all other $j$ implies that if $a$ is sufficiently large, all coefficients of $p(x)$ are positive, so $p(x)$ and all its derivatives are positive for $x > 0$.

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The polynomial and the derivatives are nonnegative in the interval $I = (0, \infty)$, whereas the intersection points are $x = x_j = -aj$ which are all negative (I am guessing that $a > 0$). So we do not have $n+1$ intersection points inside $I$. Am I missing something here? –  Srivatsan Oct 26 '11 at 6:40
    
The original version of the question, which was the one I answered, did not have that "(in $I$)". –  Robert Israel Oct 26 '11 at 18:34
    
Oh, I did not realize that. Thanks for the clarification. Wish they do not change the question like this without a warning somewhere. –  Srivatsan Oct 26 '11 at 18:35
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