anuttarasammyak's latest activity

A Gaussian wave packet has the property that its width increases with time while its overall shape is preserved. This behavior is...

By the equations of post #4 $$x(0)=1-\frac{h(0)}{f(0)}=0$$ so $$(c_3+c_5)f(0)=c_1c_2$$ Is that correct ? We do not have freedom of...

What is your choice of constant c’s that makes divergence at t=0?

I do not find divergence at t=0 for ##c_3, c_5 \neq 0##. Could you explain it in detail?

You may carry out Taylor expansion $$ f(t) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} t^n = \sum_{n=0}^\infty \frac{g^{(n-1)}(0)}{n!} t^n...

RHS of (1) would be g which is made of f and t.

The cat keeps Schroedinger hear her meow. Quantum Zeno effect saves her life.:wink:

I observe no divergences at t=0 in post #3. Denominator ##V_1=0## at ##t=c_3/c_4## would matter. Say ##f(0)## and function form...

Oh well, it's pretty intricated. I'll write it by steps: $$ \begin{align} & f'(t)= - f(t) \cdot \ \frac{V_1'(t) + c_0 F(t)}{V_1(t)}...

$$ f''=\frac{\partial g}{\partial f} f'+ \frac{\partial g}{\partial t} =\frac{\partial g}{\partial f} g+ \frac{\partial g}{\partial t}...

What actually is ##g(f(t),t)## in your case ?

Yes, this way seems to get the result a little easier. I worked the ## y=4-x^2 ## and ## y=(2x-6)^2 ## problem by this method just...

I got ##x_1=1.8317... \ x_2=3-\frac{x_1}{4}=2.5421...## The equation of ##x_1##: Other two casese : ##(x_1.x_2)=(-1.3444...

Yes, after we check that the curves do not cross at first. We may include not smooth points and end points on the curve in the survey...

Find the two points that share the same normal line. It would be a general way as @pasmith showed.