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Boby
Boby
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Let \begin{align} f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c \end{align} where $x_1<x_2<...< x_n$ and $a_i>0$. For some positive constant $c$.

Can we show that $f(t)$ has at most $2n$ zeros?

The intuition here is that if $n=1$ we have a simple bell curve. Then, a horizontal line $c$ can cross it at most $2$ times. Now if $n=2 $, then we have two mixed bell curves and we get at most $4$ crossings.

Let \begin{align} f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c \end{align} where $x_1<x_2<...< x_n$ and $a_i>0$. For some positive constant $c$.

Can we show that has at most $2n$ zeros?

The intuition here is that if $n=1$ we have a simple bell curve. Then, a horizontal line $c$ can cross it at most $2$ times. Now if $n=2 $, then we have two mixed bell curves and we get at most $4$ crossings.

Let \begin{align} f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c \end{align} where $x_1<x_2<...< x_n$ and $a_i>0$. For some positive constant $c$.

Can we show that $f(t)$ has at most $2n$ zeros?

The intuition here is that if $n=1$ we have a simple bell curve. Then, a horizontal line $c$ can cross it at most $2$ times. Now if $n=2 $, then we have two mixed bell curves and we get at most $4$ crossings.

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Boby
Boby
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  • 4
  • 16

Show that $f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c$ has at most $2n$ zeros

Let \begin{align} f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c \end{align} where $x_1<x_2<...< x_n$ and $a_i>0$. For some positive constant $c$.

Can we show that has at most $2n$ zeros?

The intuition here is that if $n=1$ we have a simple bell curve. Then, a horizontal line $c$ can cross it at most $2$ times. Now if $n=2 $, then we have two mixed bell curves and we get at most $4$ crossings.