Recent content by Bill Simpson

If it is too difficult to solve the actual problem then see if anything can be learned from seeing if you can solve a simpler problem. Suppose all the sound is contained in cylindrical tube. One end of the tube is the noise generating speaker and the other end of the tube is the noise cancelling...

If anyone might still be interested in this then I've tried to reformat it as requested by jamesbrazil. In experimenting with that I see that I dropped a couple of ( ) in my post above. I apologize for that and I can't seem to edit that now to correct those mistakes. This will hopefully...

If anyone might still be interested in this then I've tried to reformat it as requested by jamesbrazil a=434972871000000000.0; b=1.488388992*10^-36; c=0.001704000000; d=0.003408000000; e=0.0000004839360000; f=4.227024737*10^-40; g=5.240806311*10^-33; a+ (sqrt(6)*(b*ln(1/12*...

This is certainly not a proof, but if you Plot3D the Re and the Im of either your Bessel or the Abs of your Bessel, say for {delta,-5,5} and {r1,-5,5} then that blowing up to infinity in lots of odd ways strongly hints to me that finding that integral might be really challenging. If you could...

Without any additional information about the variables, the integral is -Hypergeometric2F1[1,-k/(b-k),(b-2*k)/(b-k),-(a*E^((b-k)*x))/c]/(c*E^(k*x)*k) and I don't find any simplifications for that, additional domain info might or might not help. Integrating from 0 to t gives...

Please be very careful with this, there is no safety net to catch you. If all the 100 folders are inside folder x then open a command window and CD into folder x and then erase /p /s *.jpg That /p is going to prompt you to ask whether you want each file erased or not. I would suggest...

To try to imagine how someone would come up with this, look at InputForm[Plot[Sin@x,{x,0,2Pi}]] InputForm presents the data it is given into a less "pretty printed" version that you can look at and use. When you look at that you should see there are some bits adjusting the appearance and...

I think WolframAlpha was originally intended to do "small" problems more easily for students than something like Wolfram Mathematica. Part of the design decisions that went into WolframAlpha appear to have included a line length limit (at least for the free online version, I can't speak about...

Let's suppose, just as an experiment, you wanted the eigenvalues of a matrix Eigenvalues[{{Sin[n w/v1 t],Cos[n w/v1 t],0,0}, {0,0,Sin[n w/v2(2Pi-t)],Cos[n w/v2(2Pi-t)]}, {Sin[n w/v1 Pi],Cos[n w/v1 Pi],-Sin[n w/v2 Pi],-Cos[n w/v2 Pi]}, {n/v1 Cos[n w/v1 Pi],-n/v1 Sin[n w/v1 Pi],n/v2 Cos[n w/v2...

For your final point, yes the values do decrease with every step. It is just they are rounded off to 4 digits and that makes it less obvious that they are consistently decreasing with every step. If instead you round them off to 64 digits then perhaps the pattern will be more obvious...

This takes a fraction of a second to find a solution Clear["Global`````*"]; (*I cannot make it display just and only one grave followed by asterisk*) a={0,0}; b={0,3}; c={4,0}; d={dx,dy}; e={ex,ey}; f={fx,fy}; Reduce[b-d==d-e && a-e==e-f && f-c==2(d-f), {dx,dy,ex,ey,fx,fy}] (*which returns*)...

I don't know if this helps, but Plot[2 JacobiSN[t, k]^2 + JacobiDN[t, k]^2, {t, 0, 10}] does appear to be 1 for all values of t while 4 JacobiSN[t, k]^2 + JacobiDN[t, k]^2 is not. I noticed that by plotting the two functions separately and guessing the needed scale factor.

In Mathematica if I compare the results of NDSolve with the default MachinePrecision versus WorkingPrecision->64 I find the same oscillatory result in the first case and the expected more accurate behavior in the second...

I believe that for some old FORTRAN compilers that is not a syntax error. What you are seeing is called a Hollerith Constant, described here which consist of a positive integer constant followed by H followed by the integer constant number of characters, so 1H( is a one byte constant containing...

If you want to extract the element of a list of complex constants with the smallest real part then SortBy[{1.5 + 1.2 I, 4.8 + 200 I, 0 + 36 I}, Re][[1]] will give you 36 I If you want to order a vector of complex constants by their imaginary components then this SortBy[{1.5 + 1.2 I, 4.8 +...