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#1 Help Me ! » Finding the coefficient of [math]u^n[/math] » 2017-03-15 10:44:38

evene
Replies: 1

Let's say that I am given the product of two binomials

And I have expanded them via Newton's Binomial Theorem

I'm wondering how to find the coefficient of

in this case. And can this be generalized?

#2 Help Me ! » How to get the LHS into the RHS form? » 2017-01-10 15:17:42

evene
Replies: 0

I have this equation


What substitution would you make to let the equation remain true?

Apparently, substitute

and then substitute
. But how would someone see this?

#3 Re: Computer Math » Mathematica -- Solving a system of equations over a domain » 2017-01-05 13:03:58

Thanks, I got it! big_smile

Another question: Is there a way to solve


For
and
such that the formula for
does not include
and vice versa.

I'm trying to find a substitution for
and
such that when they are substituted into
, it produces an identity such as
.

#4 Re: Computer Math » Mathematica -- Solving a system of equations over a domain » 2016-12-30 04:03:43

No, I think that's it...

Then what would be the code? I tried

Solve[{c^2 f, c f^2, 6 c e f + f^3, e f^2, e^2 f} == {a + c^2 d, a^2 + c d^2, a^3 + 6 a b + d^3 + 6 c d e, a^2 b + d^2 e, a b^2 + d e^2}, Integer]

But the output was just
.

#6 Re: Computer Math » Mathematica -- Solving a system of equations over a domain » 2016-12-29 08:56:05

But Mathematica doesn't have a spleen! tongue

Hm... okay, then this is a bit problematic.

So using an alternative method, I got (I think) a smaller system, but with the same number of variables. Is that considered an improvement?

#7 Re: Computer Math » Mathematica -- Solving a system of equations over a domain » 2016-12-29 07:59:07

Oh... I was kind of hoping that

.

So replace Solve[] with FindInstance[]? Is that what was wrong with my code?

#8 Re: This is Cool » New radical formula? » 2016-12-29 06:04:05

+10.1 big_smile

Sorry, haven't been keeping up with the "this is cool" feed. But nice job!

#9 Computer Math » Mathematica -- Solving a system of equations over a domain » 2016-12-28 16:12:55

evene
Replies: 11

How would you solve a system of equations over a specific domain, such as the reals, integers, rationals, etc.

Specifically, I mean this:


Over the integers (i.e integer solutions). I tried this code:

Solve[{f, f^2 g, f g^2 + f^2 h, g^3 + 6 f g h, g^2 h + f h^2, g h^2, h^3} == {c, a + c^2 d, a^2 + b + c d^2 + c^2 e, a^3 + 6 a b + d^3 + 6 c d e, a^2 b + b^2 + d^2 e + c e^2, a b^2 + d e^2, b^3 + e^3}, Integer]

But that didn't work. Mathematica didn't respond. What went wrong?

#10 Re: Help Me ! » Mathematica 3D Function Graphing » 2016-11-29 15:48:02

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

#11 Re: Help Me ! » Mathematica 3D Function Graphing » 2016-11-28 18:29:02

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

#12 Re: Help Me ! » Mathematica 3D Function Graphing » 2016-11-28 17:19:25

Cool, you get the cubic formula!

You get that, mostly because

#13 Re: Help Me ! » Mathematica 3D Function Graphing » 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

#14 Re: Help Me ! » Mathematica 3D Function Graphing » 2016-11-27 06:50:44

How would you do it without

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

#16 Re: Help Me ! » Mathematica 3D Function Graphing » 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. smile

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.

#17 Re: Help Me ! » Algebra Problems » 2016-11-26 13:29:23

For (3), no immediate ideas come to mind, except for plug in each point separately into

and solve the system.

EDIT: Doing so, we get the system as




Which has solutions

. Thus, the cubic is
and
.

I probably got this wrong though. Mind to check over?

#18 Re: Help Me ! » Can Someone Help Me Solve This » 2016-11-26 08:55:32

Actually, do you think they could mean Roman Numerals? Like I=1 and X=10. The only problem with that is that I don't think there's an S symbol for the Roman Numerals.

#19 Re: Help Me ! » Mathematica 3D Function Graphing » 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??

#20 Re: Help Me ! » Mathematica 3D Function Graphing » 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?

#21 Re: Help Me ! » Mathematica 3D Function Graphing » 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.

#23 Re: Introductions » I hate math » 2016-11-19 12:42:49

Do you have a specific question?

Factoring isn't too difficult. You just need to experiment around with it a bit.

#24 Re: Help Me ! » Summation Theorems » 2016-11-19 05:25:08

No thickhead, the book wrote it as

#25 Help Me ! » Summation Theorems » 2016-11-18 15:13:31

evene
Replies: 5

Given the general form of a series in Geometric Progression

With a=first term, r=common ratio, s=sum of n terms, then


If r be less than 1 and n be infinite.

a,b,c,d, &c are in Harmonic Progression when the reciprocals 1/a,1/b,1/c,1/d,&c are in Arithematic Progression or when a:b::a-b:b-c between any three consecutive terms.

The nth term of the serie is

How would you prove the formulas?

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