Solving a Quartic Polynomial using a Trig SubstitutionDate: 12/27/2001 at 15:23:23 From: Kevin Sobieski Subject: Algebra I need help solving this system: y/sqrt(y^2+64) = (x+y)/25 x/sqrt(x^2+64) = (x+y)/20 I tried solving for y in terms of x, but after 1.5 pages, I still don't have my answer. Is there an easier way? Please help if you can. Date: 12/28/2001 at 04:28:29 From: Doctor Pete Subject: Re: Algebra Hi, This is a somewhat challenging and tedious system of equations to solve. The solution set for x is a subset of the roots of a related degree-4 polynomial. Let us write [1] y/Sqrt[y^2+64] = (x+y)/25 [2] x/Sqrt[x^2+64] = (x+y)/20. Clearly, we can see there is the trivial solution (x,y) = (0,0). So we may assume x and y are nonzero. If we were to use a straightforward approach, we would begin by finding y in terms of x in [2], and then substituting this result into [1]. After a great deal of computation you would eventually arrive at a degree-8 polynomial for x, [3] 0 = 196000874496 - 612204544x^2 - 19609344x^4 - 44064x^6 + 81x^8. Obviously this is really a quartic equation, since the powers of x are all even. We could even use a substitution of the form x = Sqrt[(z+408)/3] to obtain a monic degree-4 polynomial in z with rational coefficients, and no term of degree 3. But ultimately this method is not inspiring, and as you observed, it is very tedious. (In fact, I obtained the above using a computer.) So, we want another method. Consider the trigonometric identity Tan[t]^2 + 1 = Sec[t]^2, for any angle t. We are therefore motivated to make the substitutions x = 8*Tan[s], y = 8*Tan[t], and we immediately see that [1,2] become [4] Sin[t] = (8/25)(Tan[s]+Tan[t]) [5] Sin[s] = (2/5)(Tan[s]+Tan[t]). While this looks easy (or easier) to solve, it is actually no better or worse than our original equations. If we multiply [5] by 4/5 and subtract the result from [4], we obtain the relation Sin[t] = (4/5)Sin[s], from which it follows that [6] Sin[s] = (2/5)(Tan[s]+(4/5)Sin[s]/Sqrt[1-(16/25)Sin[s]^2]) = (2/5)Tan[s] + (8/5)Sin[s]/Sqrt[25-16*Sin[s]^2], and since Sin[s] is nonzero, we have [7] 5/2 = 1/Cos[s] + 4/Sqrt[25-16*Sin[s]^2]. If we then let z = 4*Cos[s], we obtain [8] 5/2 = 4/z + 4/Sqrt[9+z^2], or equivalently, [9] (5z-8)Sqrt[9+z^2] = 8z. We readily obtain a quartic in z, [10] 576 - 720z + 225z^2 - 80z^3 + 25z^4 = 0. Since z = 4*Cos[s], and x = 8*Tan[s], it follows that x = 32*Sin[s]/z = 8 Sqrt[16-z^2]/z = 8 Sqrt[(4/z)^2 - 1]. So, we have greatly simplified the algebra by using a trigonometric substitution. But that's not all.... Alternatively, we may proceed from [8] as follows: let z = 3Tan[w]; hence [11] 5/2 = (4/3)Cot[w] + (4/3)Cos[w], or [12] 15/8 = (Cos[w])(1+Sin[w])/Sin[w]. Now let v = Sin[w] to get [13] (15/8)v = Sqrt[1-v^2](1+v) and we now have a new polynomial, [14] -1 - 2v + (225/64)v^2 + 2v^3 + v^4 = 0, where z = 3*Sin[w]/Cos[w] = 3v/Sqrt[1-v^2], so x = (8/3)Sqrt[(4/v)^2 - 25]. That's about the simplest you can get. Of course it is possible to solve all the quartics [3,10,14] in terms of radicals, but this is far too computational to be of much use. Numerically, there are two real solutions for v in [14], approximately v = {-0.3344924259407920018155742, 0.6637031578486982907001578}. Only the second value yields a correct value for x, the other being an extraneous solution. We find approximately, (x,y) = +/- (8.972929477351965431166023, 5.955361629272169366821429). where +/- indicates x and y are both positive or both negative. All said, I'm afraid we are not much better off for our efforts, as ultimately we are still required to solve a quartic polynomial. However, by using some clever trigonometric substitutions, we can at least do two things: (a) Simplify the computation to obtain a quartic, and (b) obtain a simpler quartic polynomial (with smaller coefficients.) - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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