## Monday, 10 November 2008

### Calculator program that will be useful in Mathematics lesson

Factorization (100 bytes) Factorizing Ax^2+Bx+C and find it’s solution. Mem clear: ?→A: ?→B: ?→C: B2 - 4AC→C: . 5( A-1(√C - B→X◢ . 5( A-1(- √C - B→Y◢ Lbl 0: 1M+: XM: Fix 0: Rnd: Norm 1: XM - Ans => Goto 0: A÷M→A: MM-◢ - XAns◢ Y→X: D=0→D => Goto 0: A e.g. factorize 2x^2+8x+6. Select program 2 EXE 8 EXE 6 EXE (show -1, 1st root) EXE (show -3 3rd root) EXE (show 1, coeff. of x in the first factor) EXE (show 1, constant term), EXE (1) EXE (3) EXE 2 (constant factor) So that 2x^2+8x+6 = 2(x+1)(x+3) Simplification of square roots (47 bytes) MM-: ?→A: ?→C: AC→A: Sci 5: Lbl 0: 1M+: √AM-1→ B: Rnd: B - Ans => Goto 0: Norm 1: B┘C◢ M e.g. simplify sqrt(8/27) select program, input, 8, 27, show 2/9 and 6 so that sqrt(8/27)=(2/9)sqrt(6) e.g.2, sqrt(18), input 18, 1 show 3 and 2. sqrt(18)=3sqrt(2). e.g.3 sqrt(0.75), input 0.75, 1 show 0.5 and 3, sqrt(0.75)=0.5sqrt(3) Another program for simplification of square root (64 bytes) ?→A: ?→D: AD→A: 1: Fix 0: Lbl 0: Rnd: √A Ans-1 - . 5: Rnd: Ans→B: A ÷ B2→C: Rnd: C - Ans => C + . 5 => Goto 0: Norm 1: B┘D◢ C It is much faster when the sqrt no. is very big. Simplification of nth roots (77 bytes) ?→D: ?→A: ?→X: AX^(D - 1→A: 1: Fix 0: Lbl 0: Rnd: D x√ ( A ÷Ans ) - . 5: Rnd: Ans→B: A ÷ B^D→C: Rnd: C - Ans => C + . 5 => Goto 0: Norm 1: B┘X◢ C Input nth root, and the fraction, output (no. outside) and (no. inside the root) Decimal to square root Mem clear: Ans→D: Ans2→A: Ans→B: 1→C: Fix 0: Lbl 0: A: Rnd: Ans→X: A - X => Ans-1→A: YX + C→X: Y→C: X→Y: BY: Sci 9: Rnd: Ans→X: Fix 0: Rnd: X - Ans => Goto 0: XY →X: 1: Lbl 1: Rnd: √X Ans-1 - . 5: Rnd: Ans→M: X ÷ M2→C: Rnd: C - Ans => C + . 5 => Goto 1: Norm 1: M┘(D√Y2 ÷ √D2◢ C First input Ans then press program e.g. tan30. Press tan30 EXE P1, show 1/3, 3. tan 30 = 1/3 sqrt(3) Recurring decimal to fraction (29 bytes) ?→A: ?→B: ?→C: ?→M: B┘(10xA - 1M+: M┘10xC Input (no. of place of non- recurring part), (value of non- recurring part), (no. of place of recurring part) and (value of recurring part). e.g. 0.23333…, input 1, 2, 1, 3 get 7/30. No. base (2-16) to decimal system (BASE) Mem clear: Hex: ?→D: ?→A: 10→B: 1→Y: D => A→B => 10→A => Dec: ?→C: Lbl 0: Dec: C÷B→X: Y(C - BX M+: AY→Y: X→C => Goto 0: D => Hex: M e.g. 15 under base 7 to decimal, input 0 (means that (2-16) to decimal), 7, 15, get 12. e.g. 29 under decimal to base 15, input 1 (means that decimal to (2-16)), 15, 29, get 1E.