Wolfram Mathematica

Nasser

00:27

I tried Wolfram alpha, and it could not do it. It asked to try with PRO. So I signed up with PRO. Now it says PRO computation time exceeded. And asking me to try with PRO Premium. If I upgrade to PRO-PREMIUM and it still can't do it, will I then have to upgrade again for more time? May be to PRO-PRMIUM-DELUX? Then upgrade to PRO-PRMIUM-DELUX-ULITIMATE??

halirutan

00:55

@Nasser They should run out of buzz-words pretty fast at Wolfram if they consider this the right product licensing model :)

Wolfram Turbo Alpha Premium Deluxe Ultimate Terra X

5

Searke

If you want to do computations that are kinda serious like that, you should probably be using Wolfram Mathematica Online.

Nasser

@Searke well, I do have Wolfram desktop Mathematica. But Wolfram Alpha Pro can give step-by-step solution which I do not know if Wolfram online will do that. I was told only Wolfram Apha PRO shows step by step if asked. But it seems to run-out of time on some computations. Now I find there is Wolfram Alpha PRO Premium. So I was wondering, what comes after the Premium.

Searke

You're not going to get step by step solutions from a calculation that takes that long.

Nasser

Like with Windows, there is Windows home and windows premium and windows PRO and windows Ultimate and may be more. This is all confusing to users.

Searke

The step by step solutions don't reflect what computations are actually happening to compute the integral. The way a computer solves and integral looks nothing like how you or I would do it on a chalkboard. What happens is the integral is computed and then they search for a way to describe the result using transformations that would be understandable to a calculus student.

So, first thing first. Is there a solution to your integral?

That is, does Mathematica give you a solution? If it doesn't give you a solution in under five seconds then it most likely will not be able to find an understandable step by step solution.

Nasser

01:09

@Searke the step-by-step WOlfram PRO shows are like on paper. I think it is special custom software. Any way, no problem, was just trying things ;)

@Searke Ok, thanks.

Searke

If it doesn't give you step by step solution and you have a solution, you can usually work out how to solve it originally by going thru and differentiating carefully. Otherwise you can often ask the Mathematics stackexchange.

J. M.

01:51

2 days ago, by Searke

@blochwave that's still a one-liner. Also shhhh People's suggestions here get taken seriously whether they're sarcastic or not.

halirutan

@J.M. Uhh... Let's hope no one sees this.

J. M.

Yes, let's.

@Nasser If special functions came out, don't count on Alpha easily generating step-by-step solutions.

Searke

03:49

@J.M. I think the most ridiculous example of this was that someone once tried to color their graphics by changing the font color of the corresponding data. So they thought by making the font red for Point[{0,0}] that should make the actual point in the resulting Graphics Red. Several people asked "why not".

Nasser

06:11

@J.M. You are right. any differential equation which has solution using special functions, do not show the step-by-step. Bessel ODE, Airy, etc... non of these show the step by step. I hope they could add these later on. These are the ones I wanted to see actually.

Shutao Tang

08:07

Guys! Good afternoon:)

Could someone help me test the following code?

Import["https://www.dropbox.com/s/fharox7oep45ymt/pts.xls?dl=0"]

Pickett

It asks me for my Dropbox credentials because I'm not logged in. When I open the address in the browser I do see the file, because I am logged in. If you want to share a file like that you need to put it in the public folder and get its public URL.

Shutao Tang

08:34

@Pickett I cannot open the address in the browser, because I cannot use that Dropbox site in China:(. I uploaded that files via happy fish's help.

user3183724

10:59

I was thinking.. Say you have some list of data, data = Transpose[{x,y}] with the standard interpretation of x and y data. If I wanted to numerically integrate it, I can think of two options. Either I use Total[y] * (x[[-1]]-x[[1]])/Length[x] (which would be summing it up and multiplying it by the 'bin size'), or I could use Integrate[Interpolate[data][x],{x,x[[1]],x[[-1]]}. Is there a substantial difference?

I suppose the first option is essentially the trapezoidal method, except that that one has a small adjustment for the boundary values

Szabolcs

12:21

@halirutan I have been getting many of these in the last few days, and I know that most of these fixes actually happened in the 10.x series, not in 11.0. So there wasn't quite as much delay as it would seem.

halirutan

12:50

@Szabolcs Yes, it was more meant as a joke.

It's 2016 and there are guys that go through the bug-database and write people emails. This is too funny.

JasonB

13:19

@ShutaoTang If you want to import the file directly to MMA, you need to change the url to end with "dl=1" instead of 0, then your code works for me

halirutan

@user3183724 The first method is a Rieman integral

@user3183724 and there is a vast difference between both methods. If you assume that your data is a sample from a smooth underlying function, that it is superior to interpolate and integrate because you will end up with a better result.

@user3183724 But why all the talking. Look at an easy example:

data = Table[{x, Sin[x]}, {x, 0, Pi, Pi/10.}];
ListLinePlot[data]

The true value of the integral is

In[43]:= Integrate[Sin[x], {x, 0, Pi}]

Out[43]= 2

using your approach gives

In[42]:= Function[{x, y}, Total[y]*(x[[-1]] - x[[1]])/Length[x]]@@Transpose[data]

Out[42]= 1.8032

A simple interpolation and using NIntegrate gives

In[44]:= With[{ip = Interpolation[data]},
 NIntegrate[ip[x], {x, 0, Pi}]
 ]

Out[44]= 1.99978

@user3183724 I hope this shows the point.

JasonB

I'm kind of partial to this one: In[21]:= data = Table[{x, Sin[x]}, {x, 0, Pi, Pi/10.}];
Area@DiscretizeGraphics@ListLinePlot[data, Filling -> Axis]

Out[22]= 1.98352

If you give , InterpolationOrder -> 2 to the plot command it comes out to 1.99992

Shutao Tang

13:35

@JasonB THX, I change the url to end with "dl=1", but I got same error:) Maybe In China I cannot use the Dropbox.

JasonB

It must be that firewall again,

In[13]:= Import[
"https://www.dropbox.com/s/fharox7oep45ymt/pts.xls?dl=1",
"XLS"] // Dimensions

Out[13]= {5, 198, 3}

halirutan

@JasonB Hehe :) Discretizing a graphics and calculating an integral by using Area.. I'm pretty sure there is a special place in math-hell for people like you :P

2

(Not to mention what happens if a lost soul tries to integrate to 2Pi)

JasonB

@halirutan Yeah, that definitely fails. But I'm still waiting for the time I can turn a RevolutionPlot3D or a RegionPlot3D into a 3D mesh region and get its volume

halirutan

@JasonB Maybe in the next release.

user3183724

@JasonB Thanks for the elaborate reply. It does prove the point quite nicely. Of course here one generates the data based on a smooth, known function. In reality one often has data that is generated by such a process, but distorted by various sources of noise and inaccuracies. Interpolating is then a bit less reliable, perhaps. I would have to think about it. Also, is there a reason to use NIntegrate for the interpolation and not Integrate?

halirutan

13:53

@user3183724 The point about noise and the true nature of your underlying data is true. It is almost everywhere the same no matter what source you have.

The reason for NIntegrate is that you have an underlying interpolation function, yes.

Use Integrate if there is a chance you can solve an integral analytically. With an interpolation function, the integral is solved with a numerical integration scheme and therefore, you can use NIntegrate.

JasonB

It does seem to work in 11....., but you still have to go through a rigamarole of Plot->MeshRegion->Show->BoundaryMeshRegion->TriangulateMesh->Volume
But it gets a close approximation to the answer

halirutan

@JasonB Well, it would be better if the mesh that is created by Plot could directly be used for tetrahedrisation and volume-calculation.

Michael E2

14:28

@halirutan @user3183724 I would use int = Integrate[ip[x], x] if ip is an InterpolatingFunction. It directly (and speedily) computes the integrals of the interpolating polynomials and returns another InterpolatingFunction. The definite integral would be int[b] - int[a]. However I would use NIntegrate[f[pi[x]], {x, ...}] if the integrand were an expression involving an InterpolatingFunction.

user3183724

14:54

@halirutan Yeah, I just wonder if noise makes interpolation worse than using the trapezoid method in some scenario's, although I couldn't really see why.

JasonB

Why not generate some noisy data and compare the two results to what you should get? Do you want to smooth over the noise before integrating? There are ways to do that

user3183724

Hm yes, the noise is already smoothed over. Its hard to get rid off entirely though. I might end up fitting the curves in question with analytic functions in any case, so I could then just integrate those.

Edmund

17:57

Does anyone recall the function where its parameter is _List->Automatic and it gives you {item1->1 , item2->2, ...itemn->N}?

J. M.

18:21

@Edmund Nearest[]?

Michael E2

@user3183724 Integration would tend to cancel out the noise: integral of (y + error) = (integral of y) + mean error times length of interval. -- BTW, here's a comparison of Integrate vs. NIntegrate:

Hmm, tiny. Well, click on it I guess. In any case Integrate is more than ten times faster and the results are identical.

@J.M. Oh well, I guess not many are interested in Chebyshev interpolation. The power series approach seemed so ineffective to me for what the OP wanted.

J. M.

18:50

@MichaelE2 Hey, I upvoted it. And it's nice; otherwise I wouldn't have brought it up initially. ;)

@user3183724 As a rough rule of thumb: differentiation amplifies noise, while quadrature smooths it out.

anderstood

Hi everybody. Quick question which does not deserve a post:
When assigning a value to a subpart of a matrix, it seems there is two different meanings for =
`m = ConstantArray[0, {10, 10}];
n = m;
m[[;; 5, ;; 5]] = 1;
n[[;; 5, ;; 5]] = ConstantArray[1, {5, 5}];
m == n`
returns `True`

So the = sign is understood either component-wise or for the whole block.

Now if I change the size of the block on the right-hand side (`{4,5}` instead of `{5,5}` for instance), then it becomes elementwise.

Oops, how to format code in the chat?

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