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**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?**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?Thanks, I got it!

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 .

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]`

I haven't written/tested the program code yet...

But Mathematica doesn't have a spleen!

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?

Oh... I was kind of hoping that

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

+10.1

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

**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?

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

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

Cool, you get the cubic formula!

You get that, mostly because

Yeah, so I tried to solve for the variable

in the systemin 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`

How would you do it without

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

.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.

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?

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??

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?

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.

That should be right...

Do you have a specific question?

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

No thickhead, the book wrote it as

**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?