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Showing that derivative tends to infinity (Poisson solution)

Hi,

In the plot below I have $\displaystyle f(\theta) = sin(\theta)$.

The second graph of the plot shows that, for x = 1, the plot of $\displaystyle P(\theta, x) = f(\theta - x f(\theta))$ has an infinite slope at $\displaystyle \theta = 0$ (called the Poisson non-linear propagation solution).

I'm trying to show this analytically but not getting the right answer. I reproduce my steps below.

We want to show that for $\displaystyle x = 1$, $\displaystyle P(\theta = 0, x)$ has an infinite slope, i.e. $\displaystyle dP/d\theta \rightarrow \inf$.

$\displaystyle P(\theta, x) = f(\theta - x f(\theta))$

we have $\displaystyle f(\theta) = sin(\theta)$, so

$\displaystyle P(\theta, x) = sin(\theta - x sin(\theta))$

$\displaystyle dP/d\theta = cos(\theta - sin(\theta))(1 - x cos(\theta))$

What I'm trying to show is $\displaystyle dP/d\theta \rightarrow \inf$, but that's clearly not possible from my result above.

Keen to hear where I'm going wrong. Thanks :)

Re: Showing that derivative tends to infinity (Poisson solution)

Use the definition of a derivative at a point x=a. where the derivative of some function at a specified point is lim x->a (f(x)-f(a))/(x-a) to solve your problem

Re: Showing that derivative tends to infinity (Poisson solution)

This is a double post of this thread. Please direct your comments to the original thread.

-Dan

Re: Showing that derivative tends to infinity (Poisson solution)

@ShadowKnight8702: I'm doing a similar sort of thing but not getting the right answer. From post #1 I get an answer that won't tend to infinity under any circumstances.

@topsquark: The problem is more concisely put in this thread.