Why is `evalf` not working?

Question

Why is evalf not changing $\pi$ and $e$ to floating point in the following:

h := proc (x) options operator, arrow; exp(x)*cos(x) end proc

Tay := taylor(h(x), x = (1/6)*pi, 2)

P2 := convert(Tay, polynom)

evalf(P2(.5))

The output is(exp((1/6)*Pi))(.5)*(cos((1/6)*Pi))(.5)+(-1.*(exp((1/6)*Pi))(.5)*(sin((1/6)*Pi))(.5)+(exp((1/6)*Pi))(.5)*(cos((1/6)*Pi))(.5))*(x(.5)-.1666666667*Pi(.5)). For my calculation, I need $\pi$ and $e$ to be approximations.

Did you try using placeholder variables like mypi=3.14159... and mye=2.71828...? — abiessu, Sep 30 '13 at 20:55
I did not, but evalf(pi) and evalf(exp(1)) usually evaluate to floating point. — Sujaan Kunalan, Sep 30 '13 at 20:58

score 6 · Accepted Answer · answered Sep 30 '13 at 21:00

6

To Maple, Pi is $\pi$, but pi is just a variable.

Also, seeing the result, I think you shouldn't do "evalf(P2(0.5))", but "evalf(subs(x=0.5,P2))".

answered Sep 30 '13 at 21:00

Magdiragdag

15,049

Leox · Answer 2 · 2013-10-20T19:48:57.353

2

You should use the capital letter for 'Pi-

Tay := taylor(h(x), x = (1/6) *Pi, 2);

It works.

edited Oct 20 '13 at 19:48

answered Sep 30 '13 at 21:02

Leox

8,120

Why is `evalf` not working?

2 Answers2