Solve Quartic with Casio FX115MS

11292018, 05:28 PM
(This post was last modified: 08282019 01:17 AM by Albert Chan.)
Post: #1




Solve Quartic with Casio FX115MS
y^4 + A y^2 + B y + C = 0
y^4 + 2D y^2 + D^2 = (2D  A) y^2  B y + (D^2  C) For some value of D, RHS can be made a perfect square: (y^2 + D)^2 = (2D  A) * (y  (B/2)/(2DA))^2 Let E = √(2DA), F = B/(2E), Quartic => (y^2 + E y + D+F)(y^2  E y + DF) = 0 Example: f(x) = x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0 Depress the Quartic (remove cubic term) by substitute x = y + k, where k = b/(4a) We get depressed quartic: y^4 + A y^2 + B y + C k = b/(4a) = 2/4 = 0.5 C = f(k) = 3.5625 B = f(k−1) = 4.0625 A = f(k+1) = 8.0625 B = (A  B) / 2 = 2 A = A  1  B  C = 1.5 Find D such that RHS is perfect square, i.e. discriminant = B^2  4(2D  A)(D^2  C) = 0 We do not want complex values for E, so 2D ≥ A, D^2 ≥ C: D ≥ max(A/2, √C) ≈ max(0.75, 1.904) = 1.904 A trick to force Casio FX115MS to solve for D: 0ABC + B^2  4(2D  A)(D^2  C) D = 1.99130646 (Casio SOLVE, with guessed D = 2) E = √(2DA) = 1.575630959 F = B/(2E) = 0.6346663818 Quadratic Solver, enter a=1, b=+E, c=D+F, roots = −0.78781548 ± 0.857896758 i Quadratic Solver, enter a=1, b=−E, c=DF , roots = +0.78781548 ± 1.416093080 i x roots = y roots + k = 1.28781548 ± 0.857896758 i, 0.28781548 ± 1.416093080 i For calculator without Quad Solver, we can use this: y^2 + E y + (D+F) = 0 ==> y = −E/2 ± √(−F  D/2  A/4) y^2 − E y + (D−F) = 0 ==> y = +E/2 ± √(+F  D/2  A/4) Since C=(D+F)(DF), we might derive 1 from the other, hopefully better accuracy too. 

11292018, 10:31 PM
(This post was last modified: 08232019 11:48 PM by Albert Chan.)
Post: #2




RE: Solve Quartic with Casio FX115MS
(11292018 05:28 PM)Albert Chan Wrote: A trick to force Casio FX115MS to solve for D: 0ABC + B^2  4(2D  A)(D^2  C) Discover a simpler way to solve for D, without the 0ABC + ... hack B^2  4(2D  A)(D^2  C) [SOLVE] Enter the values like before, but scroll back to D= line before pressing [SOLVE] If A, B, C were values already stored, scroll directly to D=, enter guess, scroll up, then [SOLVE] If calculated D too small, below max(A/2, √C), press [SOLVE] again with higher guess. This was vaguely (if followed the keystrokes) stated in the manual ... Oh well Edit: although D can be solved with Cubic Solver, I worried about its bugs. Calculated D must be very accurate, otherwise all 4 roots will be bad. Without the bugs, Cubic Solver is easier. Just enter the coefficients: a = 8, b = 4A, c = 8C, d = B^24AC, we get x1 = 1.99130646 

02042019, 01:39 PM
(This post was last modified: 08282019 01:21 AM by Albert Chan.)
Post: #3




RE: Solve Quartic with Casio FX115MS
(11292018 05:28 PM)Albert Chan Wrote: Example: f(x) = x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0 Another way to calculate shifted polynomial coefficients is repeated synthetic divisions. Shift xaxis to k = 2/(4*1) = 0.5 k> 1 2 3 4 5 k> 1 1.5 2.25 2.875 3.5625 < C k> 1 1.0 1.75 2 < B k> 1 0.5 1.5 < A Repeated differentiations also work: C = f(k) = k^4 + 2 k^3 + 3 k^2 + 4 k + 5 = 3.5625 B = f'(k) = 4k^3 + 6k^2 + 6k + 4 = 2 A = f''(k)/2! = (12k^2 + 12k + 6) / 2! = 1.5 As expected: y^3 coef. = (24k + 12) / 3! = 0, y^4 coef. = (24) / 4! = 1 

08242019, 12:47 PM
(This post was last modified: 08252019 11:14 AM by Albert Chan.)
Post: #4




RE: Solve Quartic with Casio FX115MS
(11292018 05:28 PM)Albert Chan Wrote: Find D such that RHS is perfect square, i.e. discriminant = B^2  4(2D  A)(D^2  C) = 0 For many calculators, [SOLVE] might not be available. We can setup Newtons iterations to get D 2 = Ans  (B²  4(2 Ans  A)(Ans²  C)) ÷ (8(C + Ans(A  3 Ans = → 1.99137931 = → 1.991306465 = → 1.99130646 Note: for Casio FX115MS, it is not necessary to put bracket before 8 However, without the bracket, formula is ambiguous: https://plus.maths.org/content/pemdasparadox 

08302019, 02:02 PM
(This post was last modified: 03132020 06:23 PM by Albert Chan.)
Post: #5




RE: Solve Quartic with Casio FX115MS
An old algorithm for solving Quartics: Brown's method
(source: W.S. Brown, "Solution of Biquadratic Equations", Aircraft Engineering, 16,14 (1944) Quartics: x^4 + A x^3 + B x^2 + C x + D = 0 Step 1: get Y, the largest real root of resolvent cubics y^3  B y^2 + (AC4D) y + D(4BA²)C² = 0 it will work for any root of cubics, but later quadratics might involve complex numbers. Step 2: solve 2 quadratics c^2  A c + (BY) = 0 d^2  Y d + D = 0 Step 3: make sure roots, c1 d2 + c2 d1 = C. If not, swap d1, d2. This may required calculation of c1 d1 + c2 d2 as well, due to numerical imprecision. I have a slight improvement in this step, see example below Step 4: solve 2 more quadratics, the 4 roots is the roots of Quartics: x^2 + c1 x + d1 = 0 x^2 + c2 x + d2 = 0 Example: solve x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0 A = 2 B = 3 C = 4 D = 5 E = AC4D = 12 F = D(4BA²)C² = 24 y^3  B y^2 + E y + F = 0 → Y = 4.482612919 c^2  A c + (BY) = 0 → [c1,c2] = [0.575630959, 2.575630959] d^2  Y d + D = 0 → [d1,d2] = [2.088157364, 2.394455555] if B and Y nearly equal, we may use another form for BY Y^2 (YB) + E Y + F = 0 → BY = (E Y + F) / Y² Calculate AY/2 If AY/2 > C, make sure roots of c,d sorted the same way If AY/2 < C, make sure roots of c,d sorted opposite way For this case, AY/2 = Y > C, thus c,d roots sorted the same way. x^2 + c1 x + d1 = 0 → x = 0.2878154795 ± 1.416093081 i x^2 + c2 x + d2 = 0 → x = 1.287815480 ± 0.8578967571 i 

09022019, 02:35 PM
(This post was last modified: 09022019 02:37 PM by Albert Chan.)
Post: #6




RE: Solve Quartic with Casio FX115MS
(08302019 02:02 PM)Albert Chan Wrote: An old algorithm for solving Quartics: Brown's method ... To show AY/2 trick work, let roots, [c1,c2] = [p+q, pq], [d1,d2] = [r+s,rs] C = c1 d2 + c2 d1 = (p+q)(rs) + (pq)(r+s) = (pr  ps + qr  qs) + (pr + ps  qr  qs) = 2pr  2qs = 2(A/2)(Y/2)  2qs = AY/2  2qs If AY/2 = C, then 2qs = 0, sort order does not matter. If AY/2 > C, then 2qs > 0, thus c,d roots sorted the same way. If AY/2 < C, then 2qs < 0, thus c,d roots sorted opposite way. 

02012020, 12:27 PM
(This post was last modified: 03132020 06:24 PM by Albert Chan.)
Post: #7




RE: Solve Quartic with Casio FX115MS
Found a Quartic Solver from https://thomasokken.com/free42/
It uses the same resolvent cubics as Brown's method, without sorting the roots. (08302019 02:02 PM)Albert Chan Wrote: Example: solve x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0 Upto calculations of Y is identical E = √(A²/4  B + Y) = 1.575630959 F = √(Y²/4  D) = 0.153149095 → [c1,c2] = A/2 ± E = [2.575630959, 0.575630959] → [d1,d2] = Y/2 ∓ F = [2.088157364, 2.394455555] Unfortunately, the roots are in wrong order ! Solving (x^2 + c1 x + d1)(x^2 + c2 x + d2) = 0, gives all wrong roots. Can anyone try the actual code, or am I mistaken somewhere ? Source in https://thomasokken.com/free42/42progs/, named G4G 

03132020, 06:30 PM
Post: #8




RE: Solve Quartic with Casio FX115MS
(02012020 12:27 PM)Albert Chan Wrote: Unfortunately, the roots are in wrong order ! Tried above program just now, for Quartic x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0 2 Enter 3 Enter 4 Enter 5 XEQ "G4G" Output "X=I,Z=I" ⇒ roots = Y ± X*I, T ± Z*I ≈ 1.2878 ± 0.6555*I, 0.2878 ± 1.5204*I Correct root should be 1.2878 ± 0.8579*I, 0.2878 ± 1.416*I G4G code indeed gives the wrong roots 

03142020, 01:28 AM
(This post was last modified: 03142020 01:34 AM by SlideRule.)
Post: #9




RE: Solve Quartic with Casio FX115MS
(08242019 12:47 PM)Albert Chan Wrote: Note: … without the bracket, formula is ambiguous: https://plus.maths.org/content/pemdasparadox Most excellent article! I deployed 'GEMA' in my technical math classes, placing an additional emphasis on 'equality' { = ≠ ≡ ≈ < ≤ > ≥ := etc. } notation. The article parallels (for me anyway) the many aspects relating fallacy and ambiguity in language as documented in Fallacies Arising from Ambiguity (APPLIED LOGIC SERIES), DOUGLAS WALTON, Department of Philosophy, University of Winnipeg, Canada, © 1996, ISBN 9789401586320 (eBook). SlideRule 

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