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These are replies submitted by Hullzie16

@Axel Vogt 

In your language I suppose the zero. In general H is less then G and that is why I want to find when H>G but writiing it as you do then I would want to find when f would be zero. 

My H is

Int(alpha^(3/2)*exp(-1/2*erf(1/2*sqrt(-2*alpha)*t)^2/erf(1/2*sqrt(-2*alpha)*b)^2 - 1/2*alpha*t^2)*I/(Pi*erf(1/2*sqrt(-2*alpha)*b)), t = -infinity .. infinity)and G is

and G is


alpha is in (10^(-9),1) and b is in (1,10^3). My procedure is found in 

I am happy with what it returns and it does it quick, I just don't know if it is the best approach. 

If there is a better way to do it and you want to ammend it I appreciate it. Thank you. 


I did not believe the end game was entirely prevelant for this question, I was originally just interested in the reason for the numeric integration issue but things spiraled from there. 

I also want to avoid the "do my work for me" reputation on this forum, all though I understand that this may be hard for a new user. 

Final sheet attached showing what I want to do. Nonetheless this is simply what I want to find: the lowest b value for a given alpha such that the inequality H(alpha,b)>G(alpha,b) is satisfied where H is some complicated numerical integral and G is just some expression containing both alpha and b that can be evlauted exactly. 

How I would exactly write a procedure I am not quite sure, maybe I will work on that next but this code seems to run smoothly. My only ammend I would like to do is the once the b value is found a given alpha, when bailing out of the inner loop to the next alpha the inner loop starts at previous b value instead of running over my entire loop again. 

Is there a simple procedure I could impliment for this? I assume there is and if I sat down with it I could probably get it quite easily. 


In using both integration methods that were provided by  @acer and @mmcdara it can be seen that the first point in the loop is not the same in the files: - acer - mmcdara

Increasing the bound on the integration with z-substitiuion(mmcdara) will bring the value closer to that obtained through the infinite time integral(acer), however when I set z=1 Maple reports an error. An interesting feature as this does not arise in the integral that sticks with the variable t. 


I took your procedure and implimented it similar to what I mentioned above and that being in a loop, but I was only able to get it to run - as seen in the file - by doing it as presented. 

Is this bad Maple procedure? Is there something less convoluted? This is also a very quick method and I am glad you provided the quick fix. 



Thank you for this respone, I am slightly embarassed I didn't just change the integral myself, I suppose I was wanting to rely on Maple instead of my own brain power. 

Regarding my title, I am interested in running this integral in a doube loop for various values of alpha and b. Since I am doing this the width of this "gaussian" like function will change drastically so I put the bounds as +/- infinity to get the most accurate asnwer. Here is a rough file of what I am interested doing (implimentating your change).


If you have a worksheet you are working with I could respond and help out from there. 

@maryam sadeghi

The figure is extremely small, can you try to edit your post to include a larger sized photo? 

What exactly are you trying to solve here? You are clearly missing an ODE if you have initial condtions, have you tried anything yet? If you post an worksheet with your progress/attempts it may get more responses. 

As well you are defining a function which depends on the functions itself, was that just suppose to be an equal sign? 

Many questions need answers before a viable solution can be attempted. 

@acer Thank you for this thorough explanation.

I am particuarily new to Maple for more sophistcated computations/procedures/code writing so I appreciate this little lesson. Particularily the digits aspect was something I was completely unaware of.

As I will have to do some more complicated computations that are similar to this example I gave I will be make all my further "codes" off of this template. 

Thank you again for all of your help!


Thank you for the quick response, and this does exactly what I want, in a very quick way. To answer your inquiry above, yes I am interested in finding the smallest(negative) upon which this inequality is satisfied. 

I knew there was a more efficent way, and what you said about "stopping once found" is what I was interested in doing I was just not sure how to implement in. 

Now for your modification of my procedure, is the time much quicker simply by the way you modified the integral? 


This is the downside of my idea.. Interpretation needs to be used. If I find a proper solution I will let you know. 


A quick and great response as always. I was unaware of the adaptive option in the plot command, I will keep my eyes out for this in the future.

Much apprecaited.


Thank you for this as I now feel embarassed about the simplicity of the problem that I had not considered.

I will forever remember this, and avoid functions with the same names as coordinates if I use this package. 


For something to be vacuum means there is zero stress-energy. If the metric itself has unkown function(s) you must solve for those functions before you have the full solution, then the solution will have vanishing Ricci and Einstein tensors. If a metric has unkown functions and is a vacuum solution that is simply stating that you must solve Ricci=0. 

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