The functions e^x, e^x,
e^(x^2), erf(x) and Taylor Series
Accurate digits are
highlighted in green. Calculations are used with a TI 84 Plus CE.
e^x = 1 + x + x^2/2! +
x^3/3! + x^4/4 + … = Σ(x^n/n!, from n = 0 to ∞)
x =

e^x

10 terms

25 terms

50 terms

1

2.718281828

2.718281801

2.718281828

2.718281828

3

20.08553692

20.07966518

20.08553692

20.08553692

5

148.4131591

146.380601

148.4131591

148.4131591

9.9

19930.37044

11869.50538

19930.07221

19930.37044

e^(x) = 1 – x + x^2/2! –
x^3/3! + x^4/4  … = Σ( (x)^n/n!, from n = 0 to ∞)
x =

e^(x)

10 terms

25 terms

50 terms

1

0.3678794412

0.3678794643

0.3678794412

0.3678794412

3

0.0497870684

0.0533258929

0.0497870684

0.0497870684

5

0.006737947

0.8640390763

0.0067379439

0.006737947

9.9

5.017468206E5

1207.799663

0.1392914019

5.017463241E5

I think you know where I’m
going.
e^(x^2) = 1 – x^2 + x^4/2!
– x^6/3! + x^8/4! = Σ( (x)^(2*n)/n!, from n = 0 to ∞)
x =

e^(x^2)

10 terms

25 terms

50 terms

1

0.3678794412

0.3678794643

0.3678794412

0.3678794412

3

1.234098041E4

442.2750223

0.0118646275

1.234194001E4

5

1.38879439E11

18613495.8

2834107793

85689.40174

9.9

2.72143414E43

2.04347238E13

3.10254183E24

7.951057508E34

(Something really goes
bonkers as x increases and n increases)
Error Function
erf(x) = 2/√π * ∫(e^(t^2)
dt, 0, x)
= 2/√π * (x – x^3/3 +
x^5/(5*2!) – x^7/(7*3!) + x^9/(9*4!)  ...)
= 2/√π * Σ(
(x^(2n+1)/((2n+1)*n!) from n = 0 to ∞ )
x =

erf(x)

10 terms

25 terms

50 terms

1

0.8427007929

0.8427007941

0.8427007929

0.8427007929

3

0.9999779095

68.58627744

0.9992050426

0.9999779095

5

1

4853382.901

3070260210.4

4724.331354

9.9

1

1.076461715E13

6.7395908E23

*overflows during
calculation*
(Result: 8.73442E33 from
WolframAlpha)
(erf(x) is practically 1
for x > 3)

Note: 9.9^(2*50+1) ≈ 3.623E100
Thoughts:
* Taylor series are great when x is near its
center point. In the all the cases
above, the center point is x = 0.
* The more simple the expression, the better
range of accuracy with less terms.
* Before you recommend a Taylor Series to
approximate f(x), check the accuracy and the range. A cautionary tale.
Eddie
This blog is property of
Edward Shore, 2016.
Ola Edward,
ReplyDeleteMuito legal o blog, Parabéns!
Notei que informa varias dicas, poderia me auxiliar na programação da Hp prime?
Quero aplicar a Formula de transformação gasosa: P1*V1/T1=P2*V2/V2
Qual seria a melhor logica para retornar com o reultado?
obrigado!!!