Mr. Robert Long

## 1404 Reputation

14 years, 160 days
Leeds, United Kingdom

## I think this will be problematic due to ...

I think this will be problematic due to the presence of the factor exp(-(sin(phi)^2)) in the integrand.

You should be able to integrate it numerically for given values of x, sigma, theta, k and Q

## pdf...

I might be wrong but I have the feeling that what you have done here is not what you intended. What are you trying to achieve ?

Regarding subs, the error occurs because you are asking Maple to sustitute x<3 into f, which doesn't make sense. It needs to be something=somethingelse. So, subs(p = 1/3, x = 3, f) is OK. Check the help page ?subs

## null list...

The underlying problem seems to be that GG is unassigned when this line is executed.
GG := [op(GG), ......

So, if you assign the null list to GG when you do the same for G that should solve the problem:
G := []: GG := []:

Edit: The reason, I think, Maple 14 gives "too many levels of recursion" is that when GG is unassigned, op(GG) just returns, GG, so GG:=[op(GG),.... becomes an infinite loop. Ths seems to be a more helpful error on Maple 14 than in Maple 15 where is just loses the kernel connection. I'm using Maple 15 and it took me a while to find the problem due to this.

## Int...

It may help to use Int instead of int

## if alpha = x-2 then why do you substitut...

if alpha = x-2 then why do you substitute x = alpha-2 ? Likewise for the other one. If those are typos then I suppose it highlights the problem with doing the substitutions in that way - you have manually manipulated those expressions for the eval and subs. One way to avoid that is by

`restart:ODE := diff(y(x), x) = (y(x)+1)/(x+2)-exp((y(x)+1)/(x+2)):par1:=alpha = x+2:par2:=y(alpha) =  y(x)+1;`
`par1a:=x=solve(par1,x);par2a:=y(x)=solve(par2,y(x));`
`                         x = -2 + alpha                      y(x) = -1 + y(alpha)`
`par2d:=diff(lhs(par2a),x)=diff(rhs(par2a),alpha);`
`                   d            d                              --- y(x) = ------- y(alpha)                   dx         dalpha         `
` `
`subODE:=subs(par2d,lhs(ODE))=eval(rhs(ODE),{par1a,par2a});`
`             d               y(alpha)      /y(alpha)\          ------- y(alpha) = -------- - exp|--------|           dalpha             alpha        \ alpha  /`
`dsolve(subODE);             y(alpha) = -ln(ln(alpha) + _C1) alpha`

Or I maybe I misunderstood what you are doing ?

## One way...

Supposing you import the data as an array, and the array is numeric, this plots the 2nd column (B) vs the third column (C). If the array is called N

plot(N[..,2],N[..,3]);

If the first row contains a header (eg "Temp" etc) then you need to remove that. One way to do that is to convert the array to a matrix  with convert(M,Matrix) and then LinearAlgebra:-DeleteRow then you can use the same command to plot.

To add columns, just make a new plot with the new column and use plots:-display to display them. eg

`p1:=plot(N[..,2],N[..,3]):p2:=plot(N[..,1],N[..,3]):plots:-display(p1,p2);`

## No solution...

Your system has no unique solution.

You have 3 equations in 4 unknowns. If you have 4 equations in 4 unknowns (or 3 equations in 3 unknowns), you may get a result your desire if the system is consisent. For example:

`sys:={ a + b = 1, c = a + b +1, d = a + c, b = d + 1}:solve(sys);`
`                 {a = -1, b = 2, c = 2, d = 1}`
` `
`Edit: Of course, if you use the 2nd system in your post, you get the result:`
`sys:={ a +b = 1, c = a +b +1 }:solve(sys);                   {a = 1 - b, b = b, c = 2}`
` `

## Resize...

Have you tried resizing the plot in Maple ?

This is what I see on my machine, in the attached screencap The arrows look OK to me.

.

As for exporting, I haven't used maple for a while for presentation, but the last time I did, I used rtf format, then edited it in MS Word

Edit: Oh, you mean the export format for the graphic not the worksheet. I don't have much experience with that, but I always found bmp was easy too work with and should be lossless.

## Be careful implicit multiplication...

The problem seems to be caused by this line

PVBC := int((p-p*delta)(SalesBCM)/(1+r)^t, t = 0 .. cmtime)

I believe it should be

PVBC := int((p-p*delta)*(SalesBCM)/(1+r)^t, t = 0 .. cmtime)

In your worksheet, this evaluated to

PVBC := int((50-50*delta((50+50*delta)*(0.3e-1+(.36*(1-exp(-.39*t)))/(12.00000000*exp(-.39*t)+1))*(1-(1-exp(-.39*t))/(12.00000000*exp(-.39*t)+1))))/1.08^t, t = 0 .. -2.564102564*ln((1.+5.*delta)/(53.+5.*delta)))

This resulted in delta((....  being treated as the function delta(.....) rather than delta*((......

This doesn't become apparent until you try to find the stationary points by differentiation, whereupon it becomes clear that the derivatives are not as expected.

Edit:typo

## LinearAlgebra:-Row()...

To extract a Row from a matrix you can use Row For example:

 > with(LinearAlgebra):
 > A:=<<1,2,3>|<4,5,6>|<7,8,9>>;
 (1)
 > Row(A,2);
 (2)
 >

## evalf ?...

Without seeing the rest of your code, I'm not sure, but have you tried using

evalf(fsolve(xi_M[Temp,sig,interfaceenergy)=0.5,T, T=100..400]))

?

It may be preferable if you could post your code, (after removing the output of the worksheet if long expressions are involved)

## Regarding the 2nd question, if F has a n...

Regarding the 2nd question, if F has a normal distribution truncated at [a,b] , it's PDF is

`pdf_tn:=PDF(TruncNormal(mu,sigma,a,b),x);         /                   /          /          2\\//                         |                   | (1/2)    |  (x - mu) || |    (1/2)       piecewise|a          |                   |          |         2 || \                         \                   \          \  2 sigma  //                    /     /          (1/2)\        /          (1/2)\\\   \  |1    |(b - mu) 2     |   1    |(a - mu) 2     |||   |  |- erf|---------------| - - erf|---------------|||, 0|  \2    \    2 sigma    /   2    \    2 sigma    ///   |                                                       /`
`Then, the PDF of a normal distribution with mean mu+t, truncated at [a,b] is`
`subs(x=y-t,pdf_tn);         /                           /          /              2\         |                           | (1/2)    |  (y - t - mu) |piecewise|a          |                           |          |           2   |         \                           \          \    2 sigma    /  \//                /     /          (1/2)\  | |    (1/2)       |1    |(b - mu) 2     |  | |2 Pi      sigma |- erf|---------------|  | \                \2    \    2 sigma    /  /                                                   /          (1/2)\\\   \     1    |(a - mu) 2     |||   |   - - erf|---------------|||, 0|     2    \    2 sigma    ///   |                                /`
`Note that, formally, to compute this PDF, we need to divide by abs(diff(y(x),x)) which is just 1 in this case, so I omitted it.`
` `

## There was a very interesting discussion ...

There was a very interesting discussion on this topic, which included the method shown by epostma above,  recently here:

http://www.mapleprimes.com/questions/123476-Seeking-A-Good-Praxis-In-Solving-Equations

## Lie symmetries...

A good starting point could be here:

http://en.wikipedia.org/wiki/Lie_point_symmetry

## Taylor series...

Since the x range is small, you could use a bivariate taylor expansion of Q. Of course, if you are interested in t>>0  this won't be much use. I got a result, expanding to 8th order around x=0.015 and t=10 and used that in your integral to get a result. I think you would need to look at the taylor polynomial to see how well it resembles Q over the period you are interested in and change the order and t location of the expansion as appropriate.

 >
 >
 >
 (1)

PDE :

 >
 >
 (2)
 >
 (3)
 > plot3d(W11,xi=0..20,tau=0..20);
 >