Mathematica 3D Function Graphing

evene · 2016-11-16 02:56:52

Hm... yes, I think that is correct.

I'm basically expanding out two functions and setting coefficients equal to each other...

bobbym · 2016-11-16 06:32:39

I think it will need some reducing down.

evene · 2016-11-16 06:35:27

Is there a way to get Mathematica to solve a system over the integers?

I guess I can reduce down the equations a bit, but I'm not too sure if that'll help that much.

bobbym · 2016-11-16 06:38:10

Yes, M can solve over the Integers, Reals, Complexes, Rationals, Primes and zillions of geometric shapes.

evene · 2016-11-24 13:08:38

What's wrong with this piece of coding

x = NSolve[Sum[h(An+B)/C^n /. A →(3/2)Sqrt[2] /. C→ -2^9 /. h → (2n)!^3 / n!^6, {n, 0, 50}] == 1/Pi, B, 30]

I'm not too sure, Mathematica says that -29/, h4 is protected, along with a couple of other things as protected. And I'm not too sure what it means by Mathematica.

Last edited by evene (2016-11-24 15:39:31)

bobbym · 2016-11-24 14:26:24

Hi;

In Mathematica, C is a reserved constant. There also seems to be a missing bracket on the end. What is 30 supposed to do?

evene · 2016-11-24 15:38:49

I realized that there were some certain mistakes while copying and pasting the function. I have now corrected them.

This isn't my input, I just copied and pasted it from the website and I realized it didn't work. But I'm gonna assume that the 30 means something like 30 decimal places.

So you're saying that I cannot use C when setting it equal to some variable?

Last edited by evene (2016-11-24 15:39:53)

bobbym · 2016-11-24 15:48:05

C is a protected symbol and can not be used for an assignement.

evene · 2016-11-25 01:57:50

But if I replace C with something else, such as X, Mathematica still says that the action cannot be completed with the methods of NSolve, and a whole bunch of other things, such as

is neither a list of replacement rules, nor a valid dispatch table... and others too.

What does this mean??

bobbym · 2016-11-25 07:36:37

Please let see the new command you are running.

evene · 2016-11-27 04:22:19

Never mind, I fixed it by replacing all the capital letters with lower case letters. And placing brackets everywhere.

How would you tell Mathematica to solve this system:

Solve[{-a^2+2 b-3 m x-3 n, b^2-2 a c+2 a^2 m x-4 b m x+3 m x^2+2 a^2 n-4 b n+6 m x n+3 n^2}=={0,0}]

For m and n in terms of a,b,c? I originally thought that you just had to put a comma at the end, then the variable you want, but Mathematica doesn't recognize that.

bobbym · 2016-11-27 06:17:33

Hi;

What about x?

evene · 2016-11-27 06:20:38

Yes, and

.

bobbym · 2016-11-27 06:24:46

Solve[{-a^2 + 2 b - 3 m x - 3 n, 
    b^2 - 2 a c + 2 a^2 m x - 4 b m x + 3 m x^2 + 2 a^2 n - 4 b n + 6 m x n + 3 n^2} == {0, 0}, {m, n}] // FullSimplify

evene · 2016-11-27 06:50:44

How would you do it without

? Just wondering, in case that every comes up...

bobbym · 2016-11-27 07:09:26

I am not seeing how to do that with your question.

That is done with generating functions alot. I usually just set x = 1 or x = 0. It might work or it might not.

Solve[{-a^2 + 2 b - 3 m x - 3 n, 
    b^2 - 2 a c + 2 a^2 m x - 4 b m x + 3 m x^2 + 2 a^2 n - 4 b n + 6 m x n + 3 n^2} == {0, 0}, {m, n}] /. x -> 1

evene · 2016-11-28 12:43:07

Yeah, so I tried to solve for the variable

in the system

in Mathematica and got the interesting output of [math[\left\{\right\}[/math]. -.-

What went wrong? Here's my coding:

Solve[{a + b, a b + c, a c} == {p, q, r}, {a}] // FullSimplify

Last edited by evene (2016-11-28 12:43:19)

bobbym · 2016-11-28 13:46:09

Hi;

One way:

Eliminate[{a + b, a b + c, a c} == {p, q, r}, b];
ans = Eliminate[a p == a^2 - c + q && r == a c, c];
Solve[ans, a]

or

Solve[{a + b, a b + c, a c} == {p, q, r}, {a, b, c}]

evene · 2016-11-28 17:19:25

Cool, you get the cubic formula!

You get that, mostly because

bobbym · 2016-11-28 17:57:11

You mean Mathematica got it.

evene · 2016-11-28 18:29:02

Yes, but that's also implied.. right?

bobbym · 2016-11-29 00:29:15

As you can see that formula is very large and cumbersome.

evene · 2016-11-29 15:48:02

It's the cubic formula, so of course it will be large, cumbersome and not very useful...

bobbym · 2016-11-29 16:34:39

True enough. There is no simple formula that everyone missed.

Math Is Fun Forum

#26 2016-11-16 02:56:52

Re: Mathematica 3D Function Graphing

#27 2016-11-16 06:32:39

Re: Mathematica 3D Function Graphing

#28 2016-11-16 06:35:27

Re: Mathematica 3D Function Graphing

#29 2016-11-16 06:38:10

Re: Mathematica 3D Function Graphing

#30 2016-11-24 13:08:38

Re: Mathematica 3D Function Graphing

#31 2016-11-24 14:26:24

Re: Mathematica 3D Function Graphing

#32 2016-11-24 15:38:49

Re: Mathematica 3D Function Graphing

#33 2016-11-24 15:48:05

Re: Mathematica 3D Function Graphing

#34 2016-11-25 01:57:50

Re: Mathematica 3D Function Graphing

#35 2016-11-25 07:36:37

Re: Mathematica 3D Function Graphing

#36 2016-11-27 04:22:19

Re: Mathematica 3D Function Graphing

#37 2016-11-27 06:17:33

Re: Mathematica 3D Function Graphing

#38 2016-11-27 06:20:38

Re: Mathematica 3D Function Graphing

#39 2016-11-27 06:24:46

Re: Mathematica 3D Function Graphing

#40 2016-11-27 06:50:44

Re: Mathematica 3D Function Graphing

#41 2016-11-27 07:09:26

Re: Mathematica 3D Function Graphing

#42 2016-11-28 12:43:07

Re: Mathematica 3D Function Graphing

#43 2016-11-28 13:46:09

Re: Mathematica 3D Function Graphing

#44 2016-11-28 17:19:25

Re: Mathematica 3D Function Graphing

#45 2016-11-28 17:57:11

Re: Mathematica 3D Function Graphing

#46 2016-11-28 18:29:02

Re: Mathematica 3D Function Graphing

#47 2016-11-29 00:29:15

Re: Mathematica 3D Function Graphing

#48 2016-11-29 15:48:02

Re: Mathematica 3D Function Graphing

#49 2016-11-29 16:34:39

Re: Mathematica 3D Function Graphing

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