What I have tried:

Note that my question is that how is it possible that I get $t = \infty$. It does not make sense for me (physically speaking). Where did I get wrong?

Please if there is information you require let me know.

1 vote

What I have tried:

Note that my question is that how is it possible that I get $t = \infty$. It does not make sense for me (physically speaking). Where did I get wrong?

Please if there is information you require let me know.

You have obtained a quartic equation in $z$. This has 4 solutions, not 1. Find some more solutions! Then check which one describes the situation you are trying to find.

In fact there is one other real solution and 2 imaginary solutions. Probably you should solve numerically (this is a physics question, not maths). eg Make a first guess $z$ then reiterate $z'=\frac12 (z^4+1)$. Or apply Newton-Raphson Method.

In fact there is one other real solution and 2 imaginary solutions. Probably you should solve numerically (this is a physics question, not maths). eg Make a first guess $z$ then reiterate $z'=\frac12 (z^4+1)$. Or apply Newton-Raphson Method.

I do not why PasteBoard works with computers connected to the uni server. I did not use it again because the pic was yielded horizontally all the time.

You are right, there is another real solution. Taking logarithms came to my mind:

$$lnz = -\frac{x_0}{4Dt}$$

$$ t = -\frac{-x_0}{4Dlnz}$$

But this gives a negative time so it is not a reliable path to take...

You wrote about reiteration and Newton-Raphson Method. I am not acquainted with those; may you delve into it?

$$lnz = -\frac{x_0}{4Dt}$$

$$ t = -\frac{-x_0}{4Dlnz}$$

But this gives a negative time so it is not a reliable path to take...

You wrote about reiteration and Newton-Raphson Method. I am not acquainted with those; may you delve into it?

http://www.sosmath.com/calculus/diff/der07/der07.html

The quartic equation gives you a value of $z$ which is less than 1, so its logarithm is -ve. The corresponding value of time in your formula is +ve.

The quartic equation gives you a value of $z$ which is less than 1, so its logarithm is -ve. The corresponding value of time in your formula is +ve.

1 vote

Best answer

You have not gone wrong. You simply have not gone far enough.

You have obtained a quartic equation $$z^4-2z+1=0$$ This has 4 solutions, not only the obvious one $z=1$. The solution $z=1$ requires $t=+\infty$ so this is not the solution you require. You need to find some more solutions. Then check which one describes the situation you are trying to find.

In fact there is one other real solution and 2 imaginary solutions. Probably you should solve numerically (this is a physics question, not maths).

One simple method is to make a first guess $z*0$ then reiterate $$z*{n+1}=\frac12 (z_n^4+1)$$ With an initial guess of $z_0=0.5$ I get the following iterations :

0.5

0.53125

0.5398259163

0.5424604827

0.543295467

0.543562654

0.5436484118

0.543675964

0.5436848186

0.5436876646

These converge to 0.5437, though not quickly.

Alternatively apply the Newton-Raphson Method. Using $$f(z)=z^4-2z+1=0$$$$f'(z)=4z^3-2$$ $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}$$ again with $z_0=0.5$ we get

0.5

0.5416666667

0.5436837222

0.5436890127

0.5436890127

which converges much more rapidly.

...

Why didn't you get PasteBoard to work for the 2nd image?

edited Nov 15, 2018 by sammy gerbil