# Tag Info

## New answers tagged algebraic-manipulation

1

The following routine tries to eliminate the linear terms by completing the square for arbitrary number of variables: CenterPoly[poly_] := Module[{a, b, c, u, vars}, vars = Variables[poly]; {c, b, a} = {#[[1]], #[[2]]/2, (#[[3]] + Transpose[#[[3]]])/2} &@ Normal@CoefficientArrays[poly, vars]; u = PseudoInverse[a].b; (#\[Transpose].a.#)[[1, ...

0

The operation of completing the square with respect to a specified variable is realized by the function CompleteTheSquare in the Manipulations set of routines from David Park's add-on Presentations (http://home.comcast.net/~djmpark/DrawGraphicsPage.html). In your example: expr = -11 - 2 x + x^2 - 4 y + y^2 - 6 z + z^2; << Presentations` ...

1

fullFactor[f_, x_] := Roots[f == 0, x] /. Equal -> ((#1 - #2) &) /. Or -> Times fullFactor[x^5 - 1, x] (* (-1 + x) ((-1)^(1/5) + x) (-(-1)^(2/5) + x) ((-1)^(3/5) + x) (-(-1)^(4/5) + x) *)

1

This is easier to do with Map, rather than with Reduce. Like this: eq = a == b + c + d - e; Map[Subtract[#, b - e] &, eq] or like this: Map[Plus[#, -b + e] &, eq] In both cases the result is the desired one: a - b + e == c + d

4

Algebraically, the way to construct the inverse of a series is straightforward. The basic iterative step is to add the next degree terms and solve for the coefficients. If we have $u = U(x, y)$, $v = V(x, y)$, then we are looking for $x = X(u, v)$, $y = Y(u, v)$ such that $u = U(X(u,v), Y(u,v))$ and $v = V(X(u,v), Y(u,v))$ identically. If the ...

3

c + d + # & /@ (Subtract @@ (a == b + c + d - e) == 0) // Reverse a == b + c + d - e /. a == c + d + x_ -> c + d == a - x With[{t = c + d}, Reduce[a == b + c + d - e /. t -> x, x] /. x -> t] With[{t = c + d}, Reduce[Eliminate[a == b + c + d - e && t == x, {c, d}], x] /. x -> t] (*c + d == a - b + e*) (*c + d == a - b + e*) ...

1

The combination of Reduce and ReplaceAll may help: Reduce[a == b + c + d - e /. c -> aa - d, aa] /. aa -> c + d (* c + d == a - b + e *) There's a defection in the solution above: it won't work correctly if the original equation contains the variable aa, of course in most cases we can avoid the duplication of name manually without much effort, but ...

1

This isn't really a full answer, but it's too long to fit in the comment box. It sounds like what you want is a Taylor series in 2D. Wikipedia gives this: In order to apply this in your case, you would need to specify the point {a,b} about which you are expanding. Presumably your f[a,b]=0 and the first derivative terms are also zero, so you would look at ...

3

Maybe you could expand the Vin to Fourier series to "normalize" it. For example there are three of them kind: VinSet = {10 Cos[1000 t - π/2], 9 Cos[400 t + π/4], Cos[t + 3.45]}; coeffSet = FourierCoefficient[# /. Times[ω_?NumericQ t] :> t, t, 1] & /@ VinSet $\left\{-5 i,\frac{9 \sqrt[4]{-1}}{2}, -0.476409-0.151771 i\right \}$ {2 Abs[#], ...

1

As for your second question, here's one way that is very similar to what we'd do by hand. Use the exponential form and then identify the phase and magnitude. Clear[A, p, t] Vin = 10 Exp[-Pi/2]*Exp[1000 t I]; {A, p} = Replace[Vin, A_ Exp[p_] -> {A Exp[Re[p]], Im[p]}] A now holds the amplitude and p holds the phase. If you have the trigonometric form you ...

2

If you specify the angle as a real number (rather than an exact integer), it does not do the transformation to Sin. For instance Vin = 10 Cos[1000 t - Pi/2.0] and Vin = 10 Cos[1000 t - 90.0 Degree] both do what you ask.

Top 50 recent answers are included