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Derivatives on HP 42S
08-26-2018, 04:47 PM
Post: #21
RE: Derivatives on HP 42S
Hi Thomas,

The solver for dx=0 of some function fx, and then the running of fx to find f(x) of the extremum found by dx is quite successful. When I run dx for, say, pi/4 (let's be simple and use fx=sin x), I get ~0 instead of ~.707

Why is dx successful in solve, but not as a means to find dx of a specific x value?
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08-26-2018, 08:00 PM
Post: #22
RE: Derivatives on HP 42S
(08-26-2018 04:47 PM)lrdheat Wrote:  Why is dx successful in solve, but not as a means to find dx of a specific x value?

(08-21-2018 12:34 AM)Thomas Klemm Wrote:  For a small discrete h, this can be approximated by

\(\frac{\partial f}{\partial x}\approx\frac{\Im[f(x + ih)]}{h}\)

Since we search for stationary points we set:

\(\frac{\partial f}{\partial x}=0\)

Which we approximate with just:

\(\Im[f(x + ih)]=0\)

To return an approximation for the derivative we have to divide \(\Im[f(x + ih)]\) by \(h\):
Code:
LBL "dF"
MVAR "x"
MVAR "h"
RCL "x"
RCL "h"
COMPLEX
XEQ "Fx"
COMPLEX
RCL÷ "h"
END

With your example:
Code:
LBL "Fx"
SIN
END

We can calculate the value of \(\frac{d}{dx}\sin(x)\) at \(x=1\):

1E-6
STO "h"
1
STO "x"
XEQ "dF"

x: 5.40302305868e-1


Compare this with \(\cos(1)\):

1
COS

x: 5.40302305868e-1


Best regards
Thomas
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08-26-2018, 08:39 PM
Post: #23
RE: Derivatives on HP 42S
(08-26-2018 04:47 PM)lrdheat Wrote:  When I run dx for, say, pi/4 (let's be simple and use fx=sin x), I get ~0 instead of ~.707

my guess is you forget to divide by h:

df(x) ~ Im((f(x + i*h)) / h

Code:
>>> from cmath import *
>>> h = 1e-6
>>> sin(complex(pi/4, h).imag
7.0710678118666542e-07
>>> _ / h
0.70710678118666548
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08-29-2018, 01:52 PM (This post was last modified: 08-27-2019 10:07 PM by Albert Chan.)
Post: #24
RE: Derivatives on HP 42S
A blog from the inventor of Complex Step Differentiation Algorithm.
It had a fully worked out example, showing very stable and accurate estimates:

https://blogs.mathworks.com/cleve/2013/1...entiation/

Real part of f(x + h*I) is also useful, for getting second derivative (involve subtraction):

http://ancs.eng.buffalo.edu/pdf/ancs_pap..._gnc06.pdf

Since we already have the value of y = f(x + h*I) of approximate extremum x, we can improve it:

Code:
from cmath import *
f = lambda x: sin(x) ** exp(x)  # rework post 3
x, h = 14.137167, 1e-6          # try to improve extremum x
y = f(complex(x, h))

# Newton's method, using f''(x) = 2/(h*h) * (f(x).real - y.real)
x -= h/2 * y.imag / (f(x).real - y.real)

2nd order x = 14.137167
3rd order x = 14.1371669412
    Actual x = 14.137166942003920 ...
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