Squeezing the most out of the casio fx-570ms for electrical

Proceedings of the International Conferenceon Computational and Mathematical Methodsin Science and Engineering, CMMSE 200813–16 June 2008.

Squeezing the most out of the casio

fx-570ms for electrical/electronics engineers∗

Pablo Guerrero-Garcıa?1 and Angel Santos-Palomo1

1 Department of Applied Mathematics, University of Malaga,Complejo Tecnologico, Campus de Teatinos s/n, 29071 Malaga (Spain)

emails: [email protected], [email protected]

Abstract

A collection of non-trivial keystroke sequences is offered to highlight that,in order to design illustrative examples for a whole course on numerical meth-ods like those envisaged for the Education Reform at an undergraduate electri-cal/electronics engineering level, all we need is a humble standard scientific calcu-lator like the casio fx-570ms when a computer lab or a graphic calculator are notaffordable. The only requirement is a carefully chosen schedule of both the datastructures and the computational ordering.

Key words: standard scientific calculator, numerical methods, teaching methods,classroom techniques

MSC 2000: 65-01, 97D40, 47N70

Outline

There are situations in which a computer lab or a graphic calculator are not affordable,mainly when some kind of massive evaluation is involved. The avoidance of electronicchatting via wireless devices and the lack of labs large enough to accommodate allpupils at once have led to the academic authorities to prohibit the very same resourceswhose use they encourage for teaching purposes. This lead to inconsistencies like thatof teachers using Matlab or a classpad 330 in the classroom, but pupils having tomanage themselves with a standard scientific calculator for evaluation purposes.

The main aim of this contribution is to provide a collection of non-trivial keystrokesequences to design illustrative examples for a whole course on numerical methods with

∗Technical Report MA-08/01, 30th March 2008, http://www.satd.uma.es/matap/investig.htm. Con-tribution to be presented by the starred author (to whom correspondence should be addressed) at the8th CMMSE International Conference, Special Session on Mathematics, the Education Reform and theuse of New Teaching Resources, La Manga del Mar Menor (Murcia, Spain), June 2008.

@ CMMSE Volume I Page 307 of 720 ISBN: 978-84-612-1982-7

Squeezing the most out of the casio fx-570ms

a humble standard scientific calculator like the casio fx-570ms [1]. Our teaching expe-rience with electrical/electronics engineers revealed that pupils can only deal with directexpressions as opposite to iterative procedures, and they are not aware of most of thefeatures of their inexpensive standard scientific calculators. A carefully chosen scheduleof both the data structures and the computational ordering allow them to discover allthe computational power they had been missing, and now they acquire this skillfulnessin a specific ad-hoc course entitled The standard scientific calculator at our university,cf. page 2 of http://hs.sci.uma.es:8070/ht/ProgramacionDocente Centro 306.pdf.

Unfortunately, to collect a handful of specific examples to illustrate every aspectof a numerical course is a tedious task, but the outcomes are highly satisfactory. Thevariety of strategies employed implies that the adaptation is by no means trivial: it hasto be done in a concise manner, because each example must be explained and solvedin no more than twenty minutes since we do not want to spend more than ten hours intotal throughout the ad-hoc course. The examples, which have been taken from [2, 3, 4]and the references therein, can be classified as follows:

1. Expressions and nonlinear algebra

• Floating-point arithmetic model

• Determinant of the product of two matrices

• Computing roots of algebraic equations

• Function tables and fixed point iterations

• Newton’s method for nonlinear equations

• Nonlinear equations and multiple roots

• Birge-Vietta’s method for algebraic equations

2. Linear algebra

• Premultiplying by elementary matrices

• Solving a system of linear equations (SLE)

• Jacobi’s iterative method for solving an SLE

• Gauss-Seidel’s iterative method for solving an SLE

• Deflation in algebraic equations of degree greater than 3

• Power method

3. Interpolation and approximation

• Osculatory interpolation (general scheme)

• Osculatory interpolation (intermediate computations)

• Linear fitting (quadratic)

• Nonlinear fitting (exponential)

• Nonlinear fitting (hyperbolic)


P. Guerrero-Garcıa and A. Santos-Palomo

• DFT as a sum of complex exponentials

• IDFT as a sum of complex exponentials

4. Differential problems

• Choice of step length in numerical differentiation

• Capacitor charge under current matching

• Romberg’s extrapolation (general scheme)

• Romberg’s extrapolation (intermediate computations)

• Euler’s method

• Predictor-corrector method

This battery could be helpful for other numerical teachers during their lesson prepa-ration to highlight some specific numerical issues that increase the broadcasting levelof the way in which a numerical analyst works even when a computer is not at hand,in the same way that a magician has her own bouquet of enchanting tricks. In thefollowing pages you can find our twenty-six cards. Which ones are yours?

References

[1] Casio, fx-570ms user’s guide and additional functions, 2003.

[2] P.Guerrero-Garcıa, Slides for a Course on Numerical Methods (Spanish),Dpto. Matematica Aplicada, Universidad de Malaga, December 2003.

[3] P.Guerrero-Garcıa and A. Santos-Palomo, Motivational numeri-cal examples for electrical/electronics engineers, Technical Report MA-06/01, Dpto.Matematica Aplicada, Universidad de Malaga, March 2006,http://www.satd.uma.es/matap/investig.htm. Poster presented at the Interna-tional Congress of Mathematicians, Madrid 2006, pp. 184–185.

[4] A. Santos-Palomo, Slides for a Course on Numerical Methods (Spanish),Dpto. Matematica Aplicada, Universidad de Malaga, September 2003.



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Floating-point arithmetic model

To compute x1 ⊕ (x2 ⊕ (x3 ⊕ x4)) in rounded 4-digit arithmetic,

x1 = 1.2342, x2 = 0.34291, x3 = 0.012896, x4 = 0.0009875,

we proceed as follows:

MODE52 (Sci) 4

.0009875 =

+ .01290 = SHIFTRnd

+ .3429 = SHIFTRnd

+ 1.234 = SHIFTRnd

1.591×1000


Determinant of the product of two matrices

To compute det(MT1 M1) first input M1 as A into memory,

M1 =

2 0 3−1 1 2

1 0 2

MODE32 (Mat)

SHIFTMAT 1 (Dim) 1 (A) 3 = 3 =

2 = = 3 = −1 = 1 = 2 = 1 = = 2 = AC

and then proceed as follows:

SHIFTMAT � 2 (Trn) SHIFTMAT 3 (Mat) 1 (A)

× SHIFTMAT 3 (Mat) 1 (A) =

6 � -1 � 6 � -1 � 1 � 2 � 6 � 2 � 17

SHIFTMAT � 1 (Det) SHIFTMAT 3 (Mat) 4 (Ans) =

1




Computing roots of algebraic equations

To compute the roots of λ2 − 19λ + 45 = 0 we did:

MODE31 (Eqn) � 2

1 = −19 = 45 =

16.22681202 � 2.773187976

whereas to compute those of −λ3 + 24λ2 − 84λ + 1 = 0 we did:

MODE31 (Eqn) � 3

−1 = 24 = −84 = 1 =

19.74923447 � 0.011945511 � 4.238820018


Function tables and fixed point iterations

To construct a table with the values of g(x) = x exp(x):

ALPHAX × SHIFTex ALPHAX

CALC −3 = CALC −1 = CALC 0 =

-0.14936 -0.36788 0.0000

CALC 1 = CALC 3 = CALC −.5 =

2.7183 60.257 -0.30327

The same technique can be used to implement fixed point iterations forg(x) =

√10 − x3/2 starting from x0 = 1.5 until x∗ ≈ 1.3652:

√( 10− ALPHAX SHIFTx3 ) ÷ 2 MODE

52 (Sci) 5

CALC 1.5 = CALC Ans = CALC Ans = CALC Ans = SHIFTRnd

1.2870 1.4025 1.3455

CALC Ans = CALC Ans = CALC Ans =

1.3752 1.3601 1.3678

Using 5-digit arithmetic, the only difference is 1.3751 rather than 1.3752.




Newton’s method for nonlinear equations

Nonlinear equation solver (Newton) allow us to obtain (in case of conver-gence) the zeros of f (x) one at a time starting from good starting points:

f (λ) = −λ3 + 24λ2 − 84λ + 1 = 0

− ALPHAX SHIFTx3 + 24 × ALPHAX x2 − 84 × ALPHAX + 1

SHIFTSOLVE 18 = SHIFTSOLVE

19.74923447

� SHIFTSOLVE 0 = SHIFTSOLVE

0.011945511


4.238820018

� SHIFTSOLVE SHIFT10x 8 = SHIFTSOLVE

Can’t solve


Nonlinear equations and multiple roots

To tune the resistance R of a potentiometer for energy dissipation implies

f (R) = exp(A ·R) · cos(B · √C −D ·R2) − E = 0,

where A = −0.005, B = 0.05, C = 2000, D = 0.01 y E = 0.01. Thus,

MODE42 (Rad) SHIFTex ( ALPHAA × ALPHAX ) ×

cos ( ALPHAB × √( ALPHAC − ALPHAD × ALPHAX x2 ) ) − ALPHAE

SHIFTSOLVE −.005 = � .05 = 2000 = .01 = .01 = � � � � SHIFTSOLVE

328.1514291

with 0 as starting point. But the solver cannot compute the roots of

f (x) = (x− 1)7 = 0

( ALPHAX − 1 ) ∧ 7

SHIFTSOLVE 0 = SHIFTSOLVE � SHIFTSOLVE 2 = SHIFTSOLVE � SHIFTSOLVE

Can’t solve (0.9982) Can’t solve (1.002)




Birge-Vietta’s method for algebraic equations

We get into trouble when the starting point is a zero of P ′(x), where

P (x) = x3 − 3x2 − 7

ALPHAX SHIFTx3 − 3 × ALPHAX x2 − 7

SHIFTSOLVE 2 = SHIFTSOLVE � SHIFTSOLVE 1 = SHIFTSOLVE

Can’t solve Can’t solve

� SHIFTSOLVE 0 = SHIFTSOLVE � SHIFTSOLVE 4 = SHIFTSOLVE

Can’t solve 3.554149219

To compute the roots of P (x) = 0 we did:

MODE31 (Eqn) � 3

1 = −3 = = −7 =

3.554 � SHIFTRe↔Im −0.2771 + 1.376i � SHIFTRe↔Im −0.2771− 1.376i


Premultiplying by elementary matrices

We want to premultiply by B = F32(−1) a given matrix A: 1 0 0

0 1 00 −1 1

1 2 3

4 5 67 8 9

.

First input A and B into memory,

MODE32 (Mat)


1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = AC

SHIFTMAT 1 (Dim) 2 (B) 3 = 3 =

1 = = = = 1 = = = −1 = 1 = AC


SHIFTMAT 3 (Mat) 2 (B) × SHIFTMAT 3 (Mat) 1 (A) =

1 � 2 � 3 � 4 � 5 � 6 � 3 � 3 � 3




Solving a system of linear equations (SLE)

Given ε = 10−4, we want to solve the linear systems[1 1 + ε

1 − ε 1

] [x1x2

]=

[1 + ε + ε2

1

],

[1 − ε 1 − ε2

1 ε + ε2

] [y1y2

]=

[11

],

whose solutions are readily checked to be

x =

[1ε

], y =

[(1 − 2ε)(1 − ε)−2

ε(1 + ε)−1(1 − ε)−2

].

We proceed as follows, using Cramer’s rule:

MODE31 (Eqn) 2 (Unknowns)

1 = 1.0001 = 1.00010001 = .9999 = 1 = 1 =

1 � 0 AC

.9999 = .99999999 = 1 = 1 = .00010001 = 1 =

0.99999999 � 1.00010002×10−4

The condition number of the former is 4 · 108 and δ(x) = 10−4, whereasthe condition number of the latter is 2.6 and δ(y) = 2 · 10−12.


Jacobi’s iterative method for solving an SLE

We want to carry out two iterations of Jacobi’s method for

1 · x1 + 2 · x2 + 3 · x3 = 94 · x1 + 5 · x2 + 6 · x3 = 127 · x1 + 8 · x2 + 10 · x3 = 13

A = ( −2 ×B − 3 × C + 9) ÷ 1B = (−4 × A − 6 × C + 12) ÷ 5C = (−7 × A− 8 ×B + 13) ÷ 10

and we obtain x(0)J =

0

00

; x

(1)J =

9

2.41.3

; x

(2)J =

0.3−6.36−6.92

.

ALPHAA ALPHA= ( −2 × ALPHAB −3 × ALPHAC + 9 ) ÷ 1 =

ALPHAB ALPHA= ( −4 × ALPHAA −6 × ALPHAC + 12 ) ÷ 5 =

ALPHAC ALPHA= ( −7 × ALPHAA −8 × ALPHAB + 13 ) ÷ 10 =

� � CALC 0 = 0 = 9 � � CALC 2.4 = = 0.3

� � CALC 0 = 0 = 2.4 � � CALC 9 = = −6.36

� � CALC 0 = 0 = 1.3 � � CALC = 2.4 = −6.92




Gauss-Seidel’s iterative method for solving an SLE

We want to carry out two iterations of Gauss-Seidel’s method for

1 · x1 + 2 · x2 + 3 · x3 = 94 · x1 + 5 · x2 + 6 · x3 = 127 · x1 + 8 · x2 + 10 · x3 = 13

A = ( −2 ×B − 3 × C + 9) ÷ 1B = (−4 × A − 6 × C + 12) ÷ 5C = (−7 × A− 8 ×B + 13) ÷ 10

and we obtain x(0)GS =

0

00

; x

(1)GS =

9

−4.8−1.16

; x

(2)GS =

22.08−13.872−3.0584

.




� � CALC 0 = 0 = 9 � � CALC = = 22.08

� � CALC = 0 = −4.8 � � CALC = = −13.872

� � CALC = = −1.16 � � CALC = = −3.0584


Deflation in algebraic equations of degree greater than 3

Compute poles of system with tranfer function H(s) = N(s)/D(s), where

D(s) = s(s + 3)(s2 + 2s + 2) + (s + 2) = s4 + 5s3 + 8s2 + 7s + 2,

ALPHAX ∧ 4 + 5 × ALPHAX SHIFTx3 + 8 × ALPHAX x2 + 7 × ALPHAX + 2

SHIFTSOLVE −5 = SHIFTSOLVE � SHIFTSOLVE 5 = SHIFTSOLVE

−3.065994892 −0.472901719

Deflating D(s) in 4-digit arithmetic by −0.4729 we obtain remainder 0and E(s) = s3 + 4.527s2 + 5.859s + 4.229, whose zeros are

MODE31 (Eqn) � 3 1 = 4.527 = 5.859 = 4.229 =

−3.066� SHIFTRe↔Im−0.7306 + 0.9196i� SHIFTRe↔Im−0.7306− 0.9196i

However, if we deflate E(s) in same arithmetic by −3.066 then we obtainF (s) = s2 + 1.461s + 1.380 but remainder −2 · 10−3, whose zeros are

MODE31 (Eqn) � 2 1 = 1.461 = 1.380 =

SHIFTRe↔Im −0.7305 + 0.9200i � SHIFTRe↔Im −0.7305 − 0.9200i




Power method

Starting from x(0) = [1; 1; 1] = B, to apply the power method to

A =

−4 14 0−5 13 0−1 0 2

,

first input A and B into memory, and let MatAns have B:

MODE32 (Mat) SHIFTMAT 1 (Dim) 1 (A) 3 = 3 =

−4 = 14 = = −5 = 13 = = −1 = = 2 = AC SHIFTMAT 1 (Dim) 2 (B) 3 = =

1 = 1 = 1 = AC SHIFTMAT 3 (Mat) 2 (B) ÷ 1 =


SHIFTMAT 3 (Mat) 1 (A) × SHIFTMAT 3 (Mat) 4 (Ans) ÷ 1 = 10 � 8 � 1 AC �= 72 � 54 � −8 AC � = 468 � 342 � −88 AC � = 2916 � 2106 � −644

To normalize in �∞, divide into 10 (rather than into 1) at the end of thefirst line above (then into 7.2, next into 6.5 and so on).


Osculatory interpolation (general scheme)

We want to compute the osculatory interpolation polynomial in 2-digitarithmetic for:

i xi f (xi) f ′(xi) f ′′(xi)0 1 1 0.5 21 2 2 −1 •

The suitable scheme to compute Neville’s divided diferences is:

i zi f [zi] f [zi−1, zi] f [zi−2, . . . , zi] f [zi−3, . . . , zi] f [zi−4, . . . , zi]0 1 1.0 → A1 1 1.0 → B 0.50 → B2 1 1.0 → C 0.50 → C 1.0 → C3 2 2.0 → D 1.0 → D 0.50 → D −0.50 → D4 2 2.0 → E −1.0 → E −2.0 → E −2.5 → E −2.0 → E

and hence the osculatory interpolation polynomial can be written as

P4(x) = A + B(x− 1) + C(x− 1)2 + D(x− 1)3 + E(x− 1)3(x− 2)

where intermediate results (from bottom to top) were obtained as follows:




Osculatory interpolation (intermediate computations)

MODE52 (Sci) 2

−1.0 = SHIFTRnd Ans SHIFTSTO E =

2.0 = SHIFTRnd Ans SHIFTSTO D = 1.0 = SHIFTRnd Ans SHIFTSTO C =

( ALPHAD − ALPHAC ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO D =

0.50 = SHIFTRnd Ans SHIFTSTO C =

0.50 = SHIFTRnd Ans SHIFTSTO B =

1.0 = SHIFTRnd Ans SHIFTSTO A =

( ALPHAE − ALPHAD ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO E = �2


2.0 ÷ 2 = SHIFTRnd Ans SHIFTSTO C = �8


( ALPHAD − ALPHAC ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO D = �5

( ALPHAE − ALPHAD ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO E =

ALPHAA = ALPHAB = ALPHAC = ALPHAD = ALPHAE =


Linear fitting (quadratic)

To find the best parabola ϕ(x) = a0 + a1x + a2x2 to fit

i 0 1 2 3 4xi 0 2 3 5 6

f (xi) 1 3 3 5 4

we proceed as follows:

MODE22 (Reg) � 3 (Quad) SHIFTCLR 1 (Scl) =

0 , 1 DT 2 , 3 DT 3 , 3 DT 5 , 5 DT 6 , 4 DT

SHIFTS-SUM 1 (∑

x2) = SHIFTS-SUM � 3 (∑

xy) =

74 64

SHIFTS-SUM � � 1 (∑

x3) = SHIFTS-SUM � � 2 (∑

x2y) =

376 308

SHIFTS-VAR � � 1 (A) = SHIFTS-VAR � � 2 (B) = SHIFTS-VAR � � 3 (C) =

0.95238 1.1429 − 0.095238




Nonlinear fitting (exponential)

To perform an exponential fitting for the data of the following table:

i 0 1 2 3 4xi 1.00 1.25 1.50 1.75 2.00

f (xi) 5.10 5.79 6.53 7.45 8.46

we proceed as described below:

MODE22 (Reg) 3 (Exp) SHIFTCLR 1 (Scl) =

1 , 5.1 DT 1.25 , 5.79 DT 1.5 , 6.53 DT 1.75 , 7.45 DT 2 , 8.46 DT

SHIFTS-SUM 1 (∑


xy) =

11.875 14.424

SHIFTS-SUM � 1 (∑

y2) = SHIFTS-SUM � 2 (∑

y) =

17.852 9.4053

SHIFTS-VAR � � 1 (A) = SHIFTS-VAR � � 2 (B) =

3.0725 0.50572


Nonlinear fitting (hyperbolic)

To perform an hyperbolic fitting as y = a/(c + bx) where a = g(b, c) =exp(c + b) for the data (with 6 correct significant digits)

xi 0.00000 1.00000 2.00000 3.00000yi −0.250000 −0.142857 −0.100000 −0.0769231

we first carry out a linear fitting for 1/y = C +Bx,

MODE22 (Reg) � 1 (Lin) SHIFTCLR 1 (Scl) =

0 , −1÷ .25 DT 1 , −1÷ .142857 DT 2 , −1÷ .1 DT 3 , −1÷ .0769231 DT


−4.00000358 − 2.99999813

Since c = Ca and b = Ba, to obtain a it suffices to solve the NLE

a = exp(c + b) = exp(Ca +Ba) = exp((C +B)a) = g(a);

ALPHAX − SHIFTex ( − 7 × ALPHAX ) SHIFTSOLVE SHIFTSOLVE 0.2177636

Hence in this case a ≈ 0.217764 is the only fixed point in [0, 1] for g(a) =exp(−7a), therefore c = −4a ≈ −0.871054 and b = −3a ≈ −0.653291.




DFT as a sum of complex exponentials

The discrete Fourier transform for y = [2;−1; 1; 4] is:

ck =

3∑r=0

yre−ikxr = 2·e−ikx0+(−1)·e−ikx1+1·e−ikx2+4·e−ikx3, k ∈ 0 : 3,

where the xr are 4 equally spaced nodes in [0, 2π), hence

c0 = 2 · e−i00 + (−1) · e−i0π/2 + 1 · e−i0π + 4 · e−i03π/2 = 6

c1 = 2 · e−i10 + (−1) · e−i1π/2 + 1 · e−i1π + 4 · e−i13π/2 = 1 + 5i

c2 = 2 · e−i20 + (−1) · e−i2π/2 + 1 · e−i2π + 4 · e−i23π/2 = 0

c3 = 2 · e−i30 + (−1) · e−i3π/2 + 1 · e−i3π + 4 · e−i33π/2 = 1 − 5i

MODE42 (Rad) MODE 2 (Cmplx)

2 − 1 SFT∠ ( − 0 π ÷ 2 ) + 1 SFT∠ ( − 0 π ) + 4 SFT∠ ( − 0 × 3 π ÷ 2 ) = SFTRe↔Im





IDFT as a sum of complex exponentials

The inverse discrete Fourier transform for c = [6; 1 + 5i; 0; 1 − 5i] is:

yr =1

4

3∑k=0

ck·eikxr =1

4·(6·ei0xr+(1+5i)·ei1xr+0·ei2xr+(1−5i)·ei3xr), r ∈ 0 : 3,

where the xr are 4 equally spaced nodes in [0, 2π), hence

y0 = (6 · ei0x0 + (1 + 5i) · ei1x0 + 0 · ei2x0 + (1 − 5i) · ei3x0)/4 = 2

y1 = (6 · ei0x1 + (1 + 5i) · ei1x1 + 0 · ei2x1 + (1 − 5i) · ei3x1)/4 = −1



MODE42 (Rad) MODE 2 (Cmplx) 1 + 5 SFTi SFTSTO A =

( 6 + AHAA × 1 SFT∠ ( 0 π ÷ 2 ) + SFTConjg AHAA × 1 SFT∠ ( 3 × 0 π ÷ 2 ) ) ÷ 4 =







Choice of step length in numerical differentiation

Given f (x) = ln(x), we want to approximate f ′(2.00) by using

f ′(a) ≈ −3 · f (a) + 4 · f (a + h) − f (a + 2h)

2 · hwhen the following values with 5 correct decimals are known:

i 0 1 2 3 4xi 2.00 2.01 2.02 2.06 2.12

f (xi) 0.69315 0.69813 0.70310 0.72271 0.75142

( −3 × ALPHAA + 4 × ALPHAB − ALPHAC ) ÷ ( 2 × ALPHAD )

CALC .69315 = .69813 = .70310 = .01 =

0.49850

CALC = .72271 = .75142 = .06 =

0.49975

SHIFTd/dx ln ALPHAX , 2 , .01 ) = SHIFTd/dx ln ALPHAX , 2 , .06 ) =

0.500000010 0.499999979


Capacitor charge under current matching

We want to estimate Q(t∗) =∫ t∗

0 ϕ(t)dt with Simpson’s formula, where

ϕ(t) = I1(t) − I2(t), I1(t) = sin(t), I2(t) = exp(t− 1)t,

for t∗ > 0 root of f (t) = g(t)− t = sin(t)exp(1− t)− t computed with 7correct significant digits by using tk+1 = g(tk) starting from t0 = 0.5:

MODE42 (Rad) sin ALPHAX × SHIFTex ( 1 − ALPHAX )

CALC .5 = CALC Ans = CALC Ans = · · · CALC Ans =

0.7904391 0.8763436 0.8695468 · · · 0.8703867

sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX

CALC .8703867 ÷ 2 = × .8703867 × 4 ÷ 6

0.10108∫dx sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX , 0 , .8703867 , 1 )

Math error

� ∫dx sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX , 0 , .8703867 , 4 )

0.10145




Romberg’s extrapolation (general scheme)

Given f (x) = exp(−x2), we want to approximate∫ 1.5

1 f (x)dx with Rom-berg’s extrapolation in 7-digit arithmetic:

MODE52 (Sci) 7 SHIFTex − ALPHAX x2

CALC 1 = SHIFTRnd Ans SHIFTSTO A = �2 f (1) ≈ 3.678794 × 10−01 → A

CALC 1.125 = SHIFTRnd Ans SHIFTSTO B = �2 f (1.125) ≈ 2.820630×10−01 → B

CALC 1.25 = SHIFTRnd Ans SHIFTSTO C = �2 f (1.25) ≈ 2.096114× 10−01 → C

CALC 1.375 = SHIFTRnd Ans SHIFTSTO D = �2 f (1.375) ≈ 1.509774×10−01 → D

CALC 1.5 = SHIFTRnd Ans SHIFTSTO E = f (1.5) ≈ 1.053992×10−01 → E

The suitable scheme for this deferred approach to the limit is:

T1(h) = 0.1183197 → BT2(h) = 0.1093104 → C

T1(h/2) = 0.1115627 → C T3(h) = 0.1093644T2(h/2) = 0.1093610 → D

T1(h/4) = 0.1099114 → D

where intermediate results (from bottom to top) were obtained as follows:


Romberg’s extrapolation (intermediate computations)

( ALPHAA + ALPHAE + 2 × ( ALPHAC + ALPHAB + ALPHAD ) ) ÷ 16

= SHIFTRnd Ans SHIFTSTO D = �2 T1(h/4) = 1.099114×10−01 → D

( ALPHAA + ALPHAE + 2 × ALPHAC ) ÷ 8

= SHIFTRnd Ans SHIFTSTO C = �2 T1(h/2) = 1.115627×10−01 → C

( ALPHAA + ALPHAE ) ÷ 4

= SHIFTRnd Ans SHIFTSTO B = �2 T1(h) = 1.183197 × 10−01 → B

( 4 × ALPHAD − ALPHAC ) ÷ 3


( 4 × ALPHAC − ALPHAB ) ÷ 3

= SHIFTRnd Ans SHIFTSTO C = �5 T2(h) = 1.093104 × 10−01 → C

( 16 × ALPHAD − ALPHAC ) ÷ 15 = T3(h) = 1.093644×10−01

SHIFTCLR 2 (Mode) = AC∫dx SHIFTex − ALPHAX x2 , 1 , 1.5 , 5 )

0.1093643 Simpson option 5 (i.e., 33 points)




Euler’s method

To solve y′(x) = sin(x · y(x)) with y(0) = 3 in the interval 0 ≤ x ≤ 5 byusing Euler’s method with h = 1 in 4-digit arithmetic:

MODE42 (Rad) MODE

52 (Sci) 4

ALPHAY + 1 × sin ( ALPHAX × ALPHAY )

CALC 3 = 0 = SHIFTRnd

3.000×1000

CALC Ans = 1 = SHIFTRnd

3.141×1000


3.140×1000


3.145×1000


3.159×1000


Predictor-corrector method

To solve y′(x) = (x− y(x))/2 with y(0) = 2 in 0 ≤ x ≤ 1 by using (withh = 0.5 and 4-digit arithmetic) the predictor-corrector method

k1 = h·fi → A, k2 = h·f (xi+h/2, wi+k1/2) → B, wi+1 = wi+k2 → C,

wPi+1 = wi +

h

2(3fi − fi−1) → D, wi+1 =

4

3wi − 1

3wi−1 +

2h

3fP

i+1 → E

MODE52 (Sci) 4 ( ALPHAX − ALPHAY ) ÷ 2 CALC 0 = 2 = SHIFTRnd × .5 =

SHIFTRnd Ans SHIFTSTO A = -5.000×10−01

�3CALC 0 + .5 ÷ 2 = 2 + ALPHAA ÷ 2 = SHIFTRnd × .5 =

SHIFTRnd Ans SHIFTSTO B = -3.750×10−01

2 + ALPHAB = SHIFTRnd Ans SHIFTSTO C = 1.625×1000

�6CALC .5 = ALPHAC = SHIFTRnd -5.625×10−01

ALPHAC + .5 × ( 3 × Ans + 1 ) ÷ 2 = SHIFTRnd Ans SHIFTSTO D = 1.453×1000

�3CALC 1 = ALPHAD = SHIFTRnd -2.265×10−01

( 4 × ALPHAC − 2 + 2 × .5 × Ans ) ÷ 3 = SHIFTRnd Ans SHIFTSTO E = 1.425×1000


Proceedings of the International Conferenceon Computational and Mathematical Methodsin Science and Engineering, CMMSE 200813–16 June 2008.

Sacandole partido a la casio fx-570ms

en sistemas electronicos de telecomunicacion*

Pablo Guerrero-Garcıa?1 y Angel Santos-Palomo1

1 Departmento de Matematica Aplicada, Universidad de Malaga,Complejo Tecnologico, Campus de Teatinos s/n, 29071 Malaga (Espana)

emails: [email protected], [email protected]

Resumen

Se ofrece una coleccion de secuencias no triviales de tecleado para enfatizarque, a la hora de disenar ejemplos ilustrativos para un curso completo de metodosnumericos como los que se preveen para la Reforma Educativa en los cursos previosa la obtencion del grado en sistemas electronicos de telecomunicacion, todo loque se necesita es una humilde calculadora cientıfica estandar como la casio fx-570ms cuando un laboratorio de ordenadores o una calculadora grafica no sonasequibles. El unico requisito es una planificacion cuidadosamente elegida tanto delas estructuras de datos como de la ordenacion computacional.

Palabras clave: calculadora cientıfica estandar, metodos numericos, metodos deensenanza, tecnicas para el aula

MSC 2000: 65-01, 97D40, 47N70

Semblanza

Hay situaciones en las que un laboratorio de ordenadores o una calculadora graficano son asequibles, principalmente cuando hay implicada algun tipo de evaluacion ma-siva. El tratar de evitar el copieteo electronico a traves de dispositivos inalambricos yla asusencia de laboratorios lo suficientemente grandes como para acomodar a todoslos alumnos a la vez ha llevado a las autoridades academicas a prohibir precisamentelos mismos recursos cuyo uso promueven cuando se trata de ensenar. Esto lleva a in-consistencias como la de que los profesores usan Matlab o una classpad 330 en el

*Informe Tecnico MA-08/01, 30 Marzo 2008, http://www.satd.uma.es/matap/investig.htm. Contri-bucion a presentar por el autor indicado con una estrella (a quien debe dirigirse la correspondencia) enel 8th CMMSE International Conference, Special Session on Mathematics, the Education Reform andthe use of New Teaching Resources, La Manga del Mar Menor (Murcia, Espana), Junio 2008.



aula, pero los alumnos tienen que apanarselas con una calculadora cientıfica estandarcuando tienen que evaluarse.

El principal objetivo de esta contribucion es proporcionar una coleccion de secuen-cias no triviales de tecleado para disenar ejemplos ilustrativos para un curso completode metodos numericos con una humilde calculadora cientıfica estandar como la casiofx-570ms [1]. Nuestra experiencia docente con ingenieros en sistemas electronicos detelecomunicacion nos revelo que los alumnos solo son capaces de manejar expresionesdirectas en contraposicion con procedimientos iterativos, y que ellos no estan al tantode la mayorıa de las caracterısticas de sus baratas calculadoras cientıficas estandar.Una planificacion cuidadosamente elegida tanto de las estructuras de datos como dela ordenacion computacional les permite descubrir toda la potencia computacional quese habıan estado perdiendo, y ahora ellos adquieren esta destreza en un curso ad-hocespecıfico titulado La calculadora cientıfica estandar impartido en nuestra universidad,cf. pagina 2 de http://hs.sci.uma.es:8070/ht/ProgramacionDocente Centro 306.pdf.

Desgraciadamente, recopilar una coleccion de ejemplos especıficos para ilustrar to-dos los aspectos de un curso numerico es una tarea tediosa, pero los resultados sonampliamente satisfactorios. La variedad de estrategias empleadas implica que la adap-tacion no es ni mucho menos trivial: debe hacerse de manera concisa, ya que cadaejemplo debe explicarse y resolverse en menos de veinte minutos ya que no se quierededicar mas de diez horas en total del curso ad-hoc completo. Los ejemplos, que hansido extraıdos de [2, 3, 4] y referencias citadas en ellos, pueden clasificarse como sigue:

1. Evaluacion y algebra no lineal

• Modelo aritmetico de punto flotante

• Determinante del producto de dos matrices

• Calculo de raıces de ecuaciones algebraicas

• Tablas de funciones e iteraciones de punto fijo

• Metodo de Newton para ecuaciones no lineales

• Ecuaciones no lineales y raıces multiples

• Metodo de Birge-Vietta para ecuaciones algebraicas

2. Algebra lineal

• Premultiplicado por matrices elementales

• Resolucion de sistemas de ecuaciones lineales

• Metodo iterativo de Jacobi para resolver un SEL

• Metodo iterativo de Gauss-Seidel para resolver un SEL

• Deflaccion en ecuaciones algebraicas de grado mayor que 3

• Metodo de las potencias

3. Interpolacion y aproximacion


P. Guerrero-Garcıa y A. Santos-Palomo

• Interpolacion osculatoria (esquema general)

• Interpolacion osculatoria (calculos intermedios)

• Ajuste lineal (cuadratico)

• Ajuste no lineal (exponencial)

• Ajuste no lineal (hiperbolico)

• TFD como suma de exponenciales complejas

• TIFD como suma de exponenciales complejas

4. Problemas diferenciales

• Eleccion del tamano de paso en derivacion numerica

• Carga de condensador bajo coincidencia de intensidades

• Extrapolacion de Romberg (esquema general)

• Extrapolacion de Romberg (calculos intermedios)

• Metodo de Euler

• Metodo predictor-corrector

Este muestrario podrıa ayudar a otros profesores numericos en la preparacion desus clases para enfatizar ciertos aspectos numericos especıficos que incrementen el nivelde difusion de la forma de trabajar del analista numerico cuando no tiene a mano unordenador, de la misma forma que un mago tiene su propio ramillete de trucos. En laspaginas siguientes puedes encontrar nuestras veintiseis cartas. ¿Cuales son las tuyas?

Referencias

[1] Casio, fx-570ms guıa del usuario y funciones adicionales, 2003.

[2] P.Guerrero-Garcıa, Transparencias de Metodos Numericos, Dpto.MatematicaAplicada, Universidad de Malaga, Diciembre 2003.

[3] P.Guerrero-Garcıa y A. Santos-Palomo, Ejemplos numericos de mo-tivacion sobre sistemas electronicos de telecomunicacion, Informe TecnicoMA-06/01, Dpto. Matematica Aplicada, Universidad de Malaga, Marzo 2006,http://www.satd.uma.es/matap/investig.htm. Version en ingles presentada comoposter en el International Congress of Mathematicians, Madrid 2006, pp. 184–185.

[4] A. Santos-Palomo, Transparencias de Metodos Numericos, Dpto.MatematicaAplicada, Universidad de Malaga, Septiembre 2003.




Modelo aritmetico de punto flotante

Para calcular x1 ⊕ (x2 ⊕ (x3 ⊕ x4)) en aritmetica de 4 dıgitos y redondeo,

x1 = 1.2342, x2 = 0.34291, x3 = 0.012896, x4 = 0.0009875,

hay que hacer:

MODE52 (Sci) 4

.0009875 =

+ .01290 = SHIFTRnd

+ .3429 = SHIFTRnd

+ 1.234 = SHIFTRnd

1.591×1000


Determinante del producto de dos matrices

Para calcular det(MT1 M1) primero introducir M1 como A en memoria,

M1 =

2 0 3−1 1 21 0 2

MODE32 (Mat)


2 = = 3 = −1 = 1 = 2 = 1 = = 2 = AC

y despues proceder como sigue:

SHIFTMAT � 2 (Trn) SHIFTMAT 3 (Mat) 1 (A)

× SHIFTMAT 3 (Mat) 1 (A) =

6 � -1 � 6 � -1 � 1 � 2 � 6 � 2 � 17

SHIFTMAT � 1 (Det) SHIFTMAT 3 (Mat) 4 (Ans) =

1




Calculo de raıces de ecuaciones algebraicas

Para calcular las raıces de λ2 − 19λ + 45 = 0 hacemos:

MODE31 (Eqn) � 2

1 = −19 = 45 =

16.22681202 � 2.773187976

mientras que para calcular las de −λ3 + 24λ2 − 84λ + 1 = 0 hacemos:

MODE31 (Eqn) � 3

−1 = 24 = −84 = 1 =

19.74923447 � 0.011945511 � 4.238820018


Tablas de funciones e iteraciones de punto fijo

Para hacer una tabla de valores para g(x) = x exp(x):

ALPHAX × SHIFTex ALPHAX

CALC −3 = CALC −1 = CALC 0 =

-0.14936 -0.36788 0.0000

CALC 1 = CALC 3 = CALC −.5 =

2.7183 60.257 -0.30327

La misma tecnica puede usarse para implementar iteraciones de punto fijocon g(x) =

√10− x3/2 empezando desde x0 = 1.5 hasta x∗ ≈ 1.3652:

√( 10− ALPHAX SHIFTx3 ) ÷ 2 MODE

52 (Sci) 5

CALC 1.5 = CALC Ans = CALC Ans = CALC Ans = SHIFTRnd

1.2870 1.4025 1.3455

CALC Ans = CALC Ans = CALC Ans =

1.3752 1.3601 1.3678

En aritmetica de 5 dıgitos la unica diferencia es 1.3751 en vez de 1.3752.




Metodo de Newton para ecuaciones no lineales

El resolutor de ecuaciones no lineales (Newton) permite obtener (si con-verge) los ceros de f (x) uno a uno empezando de un buen punto inicial:

f (λ) = −λ3 + 24λ2 − 84λ + 1 = 0

− ALPHAX SHIFTx3 + 24 × ALPHAX x2 − 84 × ALPHAX + 1

SHIFTSOLVE 18 = SHIFTSOLVE

19.74923447


0.011945511


4.238820018

� SHIFTSOLVE SHIFT10x 8 = SHIFTSOLVE

Can’t solve


Ecuaciones no lineales y raıces multiples

Ajustar la resistencia R del potenciometro disipador de energıa requiere

f (R) = exp(A ·R) · cos(B · √C −D ·R2)− E = 0,

siendo A = −0.005, B = 0.05, C = 2000, D = 0.01 y E = 0.01. Ası,

MODE42 (Rad) SHIFTex ( ALPHAA × ALPHAX ) ×

cos ( ALPHAB × √( ALPHAC − ALPHAD × ALPHAX x2 ) ) − ALPHAE

SHIFTSOLVE −.005 = � .05 = 2000 = .01 = .01 = � � � � SHIFTSOLVE

328.1514291

con 0 como punto inicial. Pero el resolutor no puede calcular las raıces de

f (x) = (x− 1)7 = 0

( ALPHAX − 1 ) ∧ 7

SHIFTSOLVE 0 = SHIFTSOLVE � SHIFTSOLVE 2 = SHIFTSOLVE � SHIFTSOLVE

Can’t solve (0.9982) Can’t solve (1.002)




Metodo de Birge-Vietta para ecuaciones algebraicas

Surgen problemas cuando el punto inicial es un cero de P ′(x), siendo

P (x) = x3 − 3x2 − 7

ALPHAX SHIFTx3 − 3 × ALPHAX x2 − 7

SHIFTSOLVE 2 = SHIFTSOLVE � SHIFTSOLVE 1 = SHIFTSOLVE

Can’t solve Can’t solve

� SHIFTSOLVE 0 = SHIFTSOLVE � SHIFTSOLVE 4 = SHIFTSOLVE

Can’t solve 3.554149219

Para calcular todas las raıces de P (x) = 0 hacemos:

MODE31 (Eqn) � 3

1 = −3 = = −7 =

3.554 � SHIFTRe↔Im −0.2771 + 1.376i � SHIFTRe↔Im −0.2771− 1.376i


Premultiplicado por matrices elementales

Se quiere premultiplicar por B = F32(−1) una matriz A: 1 0 0

0 1 00 −1 1

1 2 3

4 5 67 8 9

.

Primero introducir A y B en memoria,

MODE32 (Mat)


1 = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 9 = AC

SHIFTMAT 1 (Dim) 2 (B) 3 = 3 =

1 = = = = 1 = = = −1 = 1 = AC


SHIFTMAT 3 (Mat) 2 (B) × SHIFTMAT 3 (Mat) 1 (A) =

1 � 2 � 3 � 4 � 5 � 6 � 3 � 3 � 3




Resolucion de sistemas de ecuaciones lineales

Siendo ε = 10−4, se quieren resolver los sistemas lineales[1 1 + ε

1− ε 1

] [x1x2

]=

[1 + ε + ε2

1

],

[1− ε 1− ε2

1 ε + ε2

] [y1y2

]=

[11

],

cuyas soluciones es facil comprobar que son

x =

[1ε

], y =

[(1− 2ε)(1− ε)−2

ε(1 + ε)−1(1− ε)−2

].

Procedemos de la forma siguiente, utilizando la regla de Cramer:

MODE31 (Eqn) 2 (Unknowns)

1 = 1.0001 = 1.00010001 = .9999 = 1 = 1 =

1 � 0 AC

.9999 = .99999999 = 1 = 1 = .00010001 = 1 =

0.99999999 � 1.00010002×10−4

El numero de condicion del primero es 4 · 108 y δ(x) = 10−4, mientras queel numero de condicion del segundo es 2.6 y δ(y) = 2 · 10−12.


Metodo iterativo de Jacobi para resolver un SEL

Se quieren realizar un par de iteraciones del metodo de Jacobi para

1 · x1 + 2 · x2 + 3 · x3 = 94 · x1 + 5 · x2 + 6 · x3 = 127 · x1 + 8 · x2 + 10 · x3 = 13

A = ( −2×B − 3× C + 9)÷ 1B = (−4× A − 6× C + 12)÷ 5C = (−7× A− 8×B + 13)÷ 10

obteniendose x(0)J =

0

00

; x

(1)J =

9

2.41.3

; x

(2)J =

0.3−6.36−6.92

.




� � CALC 0 = 0 = 9 � � CALC 2.4 = = 0.3

� � CALC 0 = 0 = 2.4 � � CALC 9 = = −6.36

� � CALC 0 = 0 = 1.3 � � CALC = 2.4 = −6.92




Metodo iterativo de Gauss-Seidel para resolver un SEL

Se quieren realizar un par de iteraciones del metodo de Gauss-Seidel para

1 · x1 + 2 · x2 + 3 · x3 = 94 · x1 + 5 · x2 + 6 · x3 = 127 · x1 + 8 · x2 + 10 · x3 = 13

A = ( −2×B − 3× C + 9)÷ 1B = (−4× A − 6× C + 12)÷ 5C = (−7× A− 8×B + 13)÷ 10

obteniendose x(0)GS =

0

00

; x

(1)GS =

9

−4.8−1.16

; x

(2)GS =

22.08−13.872−3.0584

.




� � CALC 0 = 0 = 9 � � CALC = = 22.08

� � CALC = 0 = −4.8 � � CALC = = −13.872

� � CALC = = −1.16 � � CALC = = −3.0584


Deflaccion en ecuaciones algebraicas de grado mayor que 3

Los polos de un sistema con func. transferenciaH(s) = N(s)/D(s), donde

D(s) = s(s + 3)(s2 + 2s + 2) + (s + 2) = s4 + 5s3 + 8s2 + 7s + 2,

ALPHAX ∧ 4 + 5 × ALPHAX SHIFTx3 + 8 × ALPHAX x2 + 7 × ALPHAX + 2

SHIFTSOLVE −5 = SHIFTSOLVE � SHIFTSOLVE 5 = SHIFTSOLVE

−3.065994892 −0.472901719

Deflaccionando D(s) en aritmetica de 4 dıgitos con −0.4729 se obtieneresto 0 y E(s) = s3 + 4.527s2 + 5.859s + 4.229, cuyas raıces son

MODE31 (Eqn) � 3 1 = 4.527 = 5.859 = 4.229 =

−3.066� SHIFTRe↔Im−0.7306 + 0.9196i� SHIFTRe↔Im−0.7306− 0.9196i

Sin embargo, al deflaccionar E(s) en la misma aritmetica con −3.066 seobtiene F (s) = s2 + 1.461s+ 1.380 pero resto −2 · 10−3, cuyas raıces son

MODE31 (Eqn) � 2 1 = 1.461 = 1.380 =

SHIFTRe↔Im −0.7305 + 0.9200i � SHIFTRe↔Im −0.7305− 0.9200i




Metodo de las potencias

Empezando desde x(0) = [1; 1; 1] = B, para aplicar metodo potencias a

A =

−4 14 0−5 13 0−1 0 2

,

primero introducir A y B en memoria, y hacer que MatAns tenga B:

MODE32 (Mat) SHIFTMAT 1 (Dim) 1 (A) 3 = 3 =

−4 = 14 = = −5 = 13 = = −1 = = 2 = AC SHIFTMAT 1 (Dim) 2 (B) 3 = =

1 = 1 = 1 = AC SHIFTMAT 3 (Mat) 2 (B) ÷ 1 =


SHIFTMAT 3 (Mat) 1 (A) × SHIFTMAT 3 (Mat) 4 (Ans) ÷ 1 = 10 � 8 � 1 AC �= 72 � 54 � −8 AC � = 468 � 342 � −88 AC � = 2916 � 2106 � −644

Para normalizacion en �∞, dividir por 10 (en vez de por 1) al final de laprimera lınea anterior (luego por 7.2, despues por 6.5 y ası sucesivamente).


Interpolacion osculatoria (esquema general)

Se desea obtener el polinomio de interpolacion osculatoria en aritmeticade 2 dıgitos para:

i xi f (xi) f ′(xi) f ′′(xi)0 1 1 0.5 21 2 2 −1 •

El esquema a seguir para calcular diferencias divididas estilo Neville es:

i zi f [zi] f [zi−1, zi] f [zi−2, . . . , zi] f [zi−3, . . . , zi] f [zi−4, . . . , zi]0 1 1.0 → A1 1 1.0 → B 0.50 → B2 1 1.0 → C 0.50 → C 1.0 → C3 2 2.0 → D 1.0 → D 0.50 → D −0.50 → D4 2 2.0 → E −1.0 → E −2.0 → E −2.5 → E −2.0 → E

y entonces el polinomio de interpolacion osculatoria queda:

P4(x) = A + B(x− 1) + C(x− 1)2 + D(x− 1)3 + E(x− 1)3(x− 2)

donde los calculos intermedios (de abajo hacia arriba) se hicieron ası:




Interpolacion osculatoria (calculos intermedios)

MODE52 (Sci) 2

−1.0 = SHIFTRnd Ans SHIFTSTO E =

2.0 = SHIFTRnd Ans SHIFTSTO D = 1.0 = SHIFTRnd Ans SHIFTSTO C =


0.50 = SHIFTRnd Ans SHIFTSTO C =

0.50 = SHIFTRnd Ans SHIFTSTO B =

1.0 = SHIFTRnd Ans SHIFTSTO A =



2.0 ÷ 2 = SHIFTRnd Ans SHIFTSTO C = �8


( ALPHAD − ALPHAC ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO D = �5

( ALPHAE − ALPHAD ) ÷ (2 − 1) = SHIFTRnd Ans SHIFTSTO E =

ALPHAA = ALPHAB = ALPHAC = ALPHAD = ALPHAE =


Ajuste lineal (cuadratico)

Para encontrar la parabola ϕ(x) = a0 + a1x+ a2x2 que mejor aproxima a

i 0 1 2 3 4xi 0 2 3 5 6

f (xi) 1 3 3 5 4

procedemos de la forma siguiente:

MODE22 (Reg) � 3 (Quad) SHIFTCLR 1 (Scl) =

0 , 1 DT 2 , 3 DT 3 , 3 DT 5 , 5 DT 6 , 4 DT

SHIFTS-SUM 1 (∑


xy) =

74 64

SHIFTS-SUM � � 1 (∑

x3) = SHIFTS-SUM � � 2 (∑

x2y) =

376 308

SHIFTS-VAR � � 1 (A) = SHIFTS-VAR � � 2 (B) = SHIFTS-VAR � � 3 (C) =

0.95238 1.1429 − 0.095238




Ajuste no lineal (exponencial)

Para realizar un ajuste exponencial a los datos de la siguiente tabla:

i 0 1 2 3 4xi 1.00 1.25 1.50 1.75 2.00

f (xi) 5.10 5.79 6.53 7.45 8.46

procedemos como se describe a continuacion:

MODE22 (Reg) 3 (Exp) SHIFTCLR 1 (Scl) =

1 , 5.1 DT 1.25 , 5.79 DT 1.5 , 6.53 DT 1.75 , 7.45 DT 2 , 8.46 DT

SHIFTS-SUM 1 (∑


xy) =

11.875 14.424

SHIFTS-SUM � 1 (∑

y2) = SHIFTS-SUM � 2 (∑

y) =

17.852 9.4053


3.0725 0.50572


Ajuste no lineal (hiperbolico)

Para realizar un ajuste hiperbolico de la forma y = a/(c + bx) con a =g(b, c) = exp(c + b) a los datos (con 6 dıgitos significativos correctos) de

xi 0.00000 1.00000 2.00000 3.00000yi −0.250000 −0.142857 −0.100000 −0.0769231

hacemos primero un ajuste lineal para 1/y = C +Bx,

MODE22 (Reg) � 1 (Lin) SHIFTCLR 1 (Scl) =

0 , −1÷ .25 DT 1 , −1÷ .142857 DT 2 , −1÷ .1 DT 3 , −1÷ .0769231 DT


−4.00000358 − 2.99999813

Dado que c = Ca y b = Ba, para obtener a basta resolver la ENL

a = exp(c + b) = exp(Ca +Ba) = exp((C +B)a) = g(a);

ALPHAX − SHIFTex ( − 7 × ALPHAX ) SHIFTSOLVE SHIFTSOLVE 0.2177636

Luego en este caso a ≈ 0.217764 es el unico punto fijo en [0, 1] de g(a) =exp(−7a), con lo cual c = −4a ≈ −0.871054 y b = −3a ≈ −0.653291.




TFD como suma de exponenciales complejas

La transformada de Fourier discreta de y = [2;−1; 1; 4] es:

ck =

3∑r=0

yre−ikxr = 2·e−ikx0+(−1)·e−ikx1+1·e−ikx2+4·e−ikx3, k ∈ 0 : 3,

donde los xr son 4 nodos equiespaciados en [0, 2π), luego

c0 = 2 · e−i00 + (−1) · e−i0π/2 + 1 · e−i0π + 4 · e−i03π/2 = 6

c1 = 2 · e−i10 + (−1) · e−i1π/2 + 1 · e−i1π + 4 · e−i13π/2 = 1 + 5i

c2 = 2 · e−i20 + (−1) · e−i2π/2 + 1 · e−i2π + 4 · e−i23π/2 = 0

c3 = 2 · e−i30 + (−1) · e−i3π/2 + 1 · e−i3π + 4 · e−i33π/2 = 1− 5i

MODE42 (Rad) MODE 2 (Cmplx)






TIFD como suma de exponenciales complejas

La transformada inversa de Fourier discreta de c = [6; 1 + 5i; 0; 1− 5i] es:

yr =1

4

3∑k=0

ck·eikxr =1

4·(6·ei0xr+(1+5i)·ei1xr+0·ei2xr+(1−5i)·ei3xr), r ∈ 0 : 3,

donde los xr son 4 nodos equiespaciados en [0, 2π), luego

y0 = (6 · ei0x0 + (1 + 5i) · ei1x0 + 0 · ei2x0 + (1− 5i) · ei3x0)/4 = 2

y1 = (6 · ei0x1 + (1 + 5i) · ei1x1 + 0 · ei2x1 + (1− 5i) · ei3x1)/4 = −1



MODE42 (Rad) MODE 2 (Cmplx) 1 + 5 SFTi SFTSTO A =








Eleccion del tamano de paso en derivacion numerica

Para f (x) = ln(x), se quiere aproximar f ′(2.00) mediante

f ′(a) ≈ −3 · f (a) + 4 · f (a + h)− f (a + 2h)

2 · hconocidos los siguientes valores con 5 decimales correctos:

i 0 1 2 3 4xi 2.00 2.01 2.02 2.06 2.12

f (xi) 0.69315 0.69813 0.70310 0.72271 0.75142

( −3 × ALPHAA + 4 × ALPHAB − ALPHAC ) ÷ ( 2 × ALPHAD )

CALC .69315 = .69813 = .70310 = .01 =

0.49850

CALC = .72271 = .75142 = .06 =

0.49975

SHIFTd/dx ln ALPHAX , 2 , .01 ) = SHIFTd/dx ln ALPHAX , 2 , .06 ) =

0.500000010 0.499999979


Carga de condensador bajo coincidencia de intensidades

Se desea estimar Q(t∗) =∫ t∗

0 ϕ(t)dt con la formula de Simpson, siendo

ϕ(t) = I1(t)− I2(t), I1(t) = sen(t), I2(t) = exp(t− 1)t,

para t∗ > 0 raız de f (t) = g(t)− t = sen(t)exp(1− t)− t calculada con 7dıgitos significativos correctos mediante tk+1 = g(tk) a partir de t0 = 0.5:

MODE42 (Rad) sin ALPHAX × SHIFTex ( 1 − ALPHAX )

CALC .5 = CALC Ans = CALC Ans = · · · CALC Ans =

0.7904391 0.8763436 0.8695468 · · · 0.8703867

sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX

CALC .8703867÷ 2 = × .8703867 × 4 ÷ 6

0.10108∫dx sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX , 0 , .8703867 , 1 )

Math error

� ∫dx sin ALPHAX − SHIFTex ( ALPHAX − 1 ) × ALPHAX , 0 , .8703867 , 4 )

0.10145




Extrapolacion de Romberg (esquema general)

Siendo f (x) = exp(−x2), se desea aproximar∫ 1.5

1 f (x)dx por extrapola-cion de Romberg en aritmetica de 7 dıgitos:

MODE52 (Sci) 7 SHIFTex − ALPHAX x2

CALC 1 = SHIFTRnd Ans SHIFTSTO A = �2 f (1) ≈ 3.678794× 10−01 → A

CALC 1.125 = SHIFTRnd Ans SHIFTSTO B = �2 f (1.125) ≈ 2.820630×10−01 → B

CALC 1.25 = SHIFTRnd Ans SHIFTSTO C = �2 f (1.25) ≈ 2.096114× 10−01 → C

CALC 1.375 = SHIFTRnd Ans SHIFTSTO D = �2 f (1.375) ≈ 1.509774×10−01 → D

CALC 1.5 = SHIFTRnd Ans SHIFTSTO E = f (1.5) ≈ 1.053992×10−01 → E

El esquema a seguir en este proceso de extrapolacion por paso al lımite es:

T1(h) = 0.1183197 → BT2(h) = 0.1093104 → C

T1(h/2) = 0.1115627 → C T3(h) = 0.1093644T2(h/2) = 0.1093610 → D

T1(h/4) = 0.1099114 → D

donde los calculos intermedios (de abajo hacia arriba) se hicieron ası:


Extrapolacion de Romberg (calculos intermedios)

( ALPHAA + ALPHAE + 2 × ( ALPHAC + ALPHAB + ALPHAD ) ) ÷ 16


( ALPHAA + ALPHAE + 2 × ALPHAC ) ÷ 8

= SHIFTRnd Ans SHIFTSTO C = �2 T1(h/2) = 1.115627×10−01 → C

( ALPHAA + ALPHAE ) ÷ 4

= SHIFTRnd Ans SHIFTSTO B = �2 T1(h) = 1.183197× 10−01 → B

( 4 × ALPHAD − ALPHAC ) ÷ 3


( 4 × ALPHAC − ALPHAB ) ÷ 3

= SHIFTRnd Ans SHIFTSTO C = �5 T2(h) = 1.093104× 10−01 → C

( 16 × ALPHAD − ALPHAC ) ÷ 15 = T3(h) = 1.093644×10−01

SHIFTCLR 2 (Mode) = AC∫dx SHIFTex − ALPHAX x2 , 1 , 1.5 , 5 )

0.1093643 Simpson opcion 5 (i.e., 33 puntos)




Metodo de Euler

Para resolver y′(x) = sen(x · y(x)) con y(0) = 3 en el intervalo 0 ≤ x ≤ 5usando el metodo de Euler con h = 1 en aritmetica de 4 dıgitos:

MODE42 (Rad) MODE5

2 (Sci) 4

ALPHAY + 1 × sin ( ALPHAX × ALPHAY )

CALC 3 = 0 = SHIFTRnd

3.000×1000


3.141×1000


3.140×1000


3.145×1000


3.159×1000


Metodo predictor-corrector

Para resolver y′(x) = (x − y(x))/2 con y(0) = 2 en 0 ≤ x ≤ 1 usando(con h = 0.5 y aritmetica de 4 dıgitos) el metodo predictor-corrector

k1 = h·fi → A, k2 = h·f (xi+h/2, wi+k1/2) → B, wi+1 = wi+k2 → C,

wPi+1 = wi +

h

2(3fi − fi−1) → D, wi+1 =

4

3wi − 1

3wi−1 +

2h

3fP

i+1 → E

MODE52 (Sci) 4 ( ALPHAX − ALPHAY ) ÷ 2 CALC 0 = 2 = SHIFTRnd × .5 =

SHIFTRnd Ans SHIFTSTO A = -5.000×10−01

�3CALC 0 + .5 ÷ 2 = 2 + ALPHAA ÷ 2 = SHIFTRnd × .5 =

SHIFTRnd Ans SHIFTSTO B = -3.750×10−01

2 + ALPHAB = SHIFTRnd Ans SHIFTSTO C = 1.625×1000

�6CALC .5 = ALPHAC = SHIFTRnd -5.625×10−01

ALPHAC + .5 × ( 3 × Ans + 1 ) ÷ 2 = SHIFTRnd Ans SHIFTSTO D = 1.453×1000

�3CALC 1 = ALPHAD = SHIFTRnd -2.265×10−01

( 4 × ALPHAC − 2 + 2 × .5 × Ans ) ÷ 3 = SHIFTRnd Ans SHIFTSTO E = 1.425×1000


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