Derivatives on HP 42S - Printable Version

Derivatives on HP 42S - Printable Version

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Derivatives on HP 42S - lrdheat - 08-20-2018 03:03 AM

Awhile back, a gem of a way to produce d/dx using the complex number capabilities of the HP 15LE was posted. It worked wonderfully, and could be used in solve to find extremums. I am stumbling over how to implement this on my HP 42S.

I attached the image of a sample program for the HP 15LE to take d/dx of (sin x)/x and the use of it in solve. How should I accomplish this on the HP 42S?

RE: Derivatives on HP 42S - Thomas Klemm - 08-20-2018 04:38 AM

I've inlined the function:

Code:

LBL "dF"

MVAR "x"

MVAR "h"

RCL "x"

RCL "h"

COMPLEX

SIN

LASTX

÷

COMPLEX

END

Set MODES to RAD.
Start SOLVER.
Select Solve Program: dF
Set small value h: 1E-5 h
Set lower guess: π x
Set upper guess: π 2 × x
Solve for x: x

x=4.49341

Kind regards
Thomas

PS: Link to original post

RE: Derivatives on HP 42S - Thomas Klemm - 08-20-2018 07:43 AM

Here's another example:

\(f(x)=\sin(x)^{e^x}\)

Search for \(\max(f(x))\) for \(x\in[14.13, 14.14]\).

Code:

LBL "dF"

MVAR "x"

MVAR "h"

RCL "x"

RCL "h"

COMPLEX

SIN

LASTX

E↑X

Y↑X

COMPLEX

END

Set MODES to RAD.
Start SOLVER.
Select Solve Program: dF
Set small value h: 1E-6 h
Set lower guess: 14.13 x
Set upper guess: 14.14 x
Solve for x: x

x=14.137167

Cheers
Thomas

RE: Derivatives on HP 42S - lrdheat - 08-20-2018 10:57 PM

Hi Thomas,

When I try to XEQ the function in it's own little program, say "FX" within the derivative program sandwiched between the 2 complex commands, the correct answer is returned with a minute complex component. (I wish to have the function that I am finding the extremum point as it's own program that the derivative program can call...this allows me to then run the function program to evaluate f(x) of the extremum)

I can then do a switch of x and y stack levels, but was wondering why I cannot simply get the real part of the extremum to be delivered to the x stack level.

Richard

RE: Derivatives on HP 42S - Thomas Klemm - 08-20-2018 11:43 PM

Not sure if I understand you correctly but you can split the program into:

Function

Code:

LBL "Fx"

SIN

LASTX

÷

END

Derivative

Code:

LBL "dF"

MVAR "x"

MVAR "h"

RCL "x"

RCL "h"

COMPLEX

XEQ "Fx"

COMPLEX

END

Set MODES to RAD.
Start SOLVER.
Select Solve Program: dF
Set small value h: 1E-6 h
Set lower guess: π x
Set upper guess: π 2 × x
Solve for x: x

x=4.493409

Exit the solver with:
EXIT
EXIT

Now you can run:
XEQ "Fx"

x: -0.217234

Is this what you wanted to achieve?

Kind regards
Thomas

RE: Derivatives on HP 42S - Albert Chan - 08-20-2018 11:54 PM

(08-20-2018 07:43 AM)Thomas Klemm Wrote: Here's another example:

\(f(x)=\sin(x)^{e^x}\)

Search for \(\max(f(x))\) for \(x\in[14.13, 14.14]\).

x=14.137167

Why would search for slope of 0 get the maximum ?

Should it get the minimum, at x = 14.13 ?
Slope at 14.13 = 5.2e-12

Also, Why use complex numbers to search real extremums ?

RE: Derivatives on HP 42S - Thomas Klemm - 08-21-2018 12:34 AM

(08-20-2018 11:54 PM)Albert Chan Wrote: Also, Why use complex numbers to search real extremums ?

From The Complex-Step Derivative Approximation:

Quote:For a small discrete h, this can be approximated by

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

We call this the complex-step derivative approximation. This estimate is not
subject to subtractive cancellation errors, since it does not involve a difference
operation. This constitutes a tremendous advantage over the finite-difference
approximations.

Since we search for stationary points we set:

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

Which we approximate with just:

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

Cheers
Thomas

This is why it's a pain point if a calculator claims to support complex numbers but does not implement all functions. Yes, I'm looking at you HP-35S!

RE: Derivatives on HP 42S - Thomas Klemm - 08-21-2018 01:35 AM

(08-20-2018 11:54 PM)Albert Chan Wrote: Why would search for slope of 0 get the maximum ?

Cf. Maxima and minima

We search for a critical point in the interval [14.13,14.14], that is x where f'(x) = 0.

Quote:Should it get the minimum, at x = 14.13 ?

The function is somewhat pathological in that it's close to 0 most of the time and only 1 at the maximum which is where \(\sin(x)=1\).
That means \(x=\frac{\pi}{2}+k\cdot2\pi\) for \(k\in\mathbb{Z}\).

It's not defined (or then takes complex values) where \(\sin(x)<0\).
Thus the minimal value is 0 when \(x=k\cdot\pi\) for \(k\in\mathbb{Z}\).

So the minimal values in that section are \((4\pi, 0)\) and \((5\pi, 0)\) and the maximal value is \((\frac{9}{2}\pi, 1)\).

The flatness of the function makes it hard to numerically find the stationary points.

Cheers
Thomas

Since I used Free42 instead of the HP-42S it could very well be that the results don't agree.

RE: Derivatives on HP 42S - lrdheat - 08-21-2018 02:24 AM

Yes, it becomes very hard to find the critical points as one gets further away from 0: f(x) approaches it's maximum of 1 over smaller and smaller intervals, requiring an impossibly tight interval to enter into the search far from zero!

Your program was just what I wanted. I was very close, but your program was what I was after.

Thanks again...never was exposed to the relationship between derivatives and complex numbers before you posted several years ago!

Richard

RE: Derivatives on HP 42S - Thomas Klemm - 08-21-2018 06:14 AM

(08-21-2018 01:35 AM)Thomas Klemm Wrote: The flatness of the function makes it hard to numerically find the stationary points.

Here's another example: \(f(x)=\sin(x+\cos(x))\)

This is the power series expansion for \(f\) about the point \(x=\frac{\pi}{2}\):

\(1 - \frac{1}{72} (x - \frac{\pi}{2})^6 + \frac{1}{720} (x - \frac{\pi}{2})^8 + O((x - \frac{\pi}{2})^{10})\)

You should start with the initial values 1 and 2.

Kind regards
Thomas

RE: Derivatives on HP 42S - RMollov - 08-23-2018 12:58 PM

a bit out of topic, but here it goes: Big thank you, Thomas, for your posts here, I really enjoy reading them, so enlightening they are!
Cheers,

RE: Derivatives on HP 42S - lrdheat - 08-24-2018 02:51 AM

Hi Thomas,

When trying a more complicated equation: x^(4/3) + 4(x^(1/3)), I get a correct f(x) for all values of x (x=-6, f(x)=3.63, x=-4, f(x)=0 x=0, f(x)=0 x=2, f(x)=7.56)

Instead of a min at x=-1, f(x)= -3, my dx program comes up with a min at x=-3.

I know that this is probably not a concise program for FX, but it does work for this example. Why is DX not working?

LBL FX
STO X
4
Y^X
3
1/X
Y^X
STO B
RCL X
SIGN
STO S
RCL X
*
3
1/X
Y^X
4
*
ABS
RCL S
*
RCL B
+
END

LBL DX
MVAR A
MVAR H
RCL A
RCL H
COMPLEX
XEQ FX
COMPLEX
END

Thanks for your time!

RE: Derivatives on HP 42S - Thomas Klemm - 08-24-2018 05:52 AM

(08-24-2018 02:51 AM)lrdheat Wrote: When trying a more complicated equation: x^(4/3) + 4(x^(1/3)), I get a correct f(x) for all values of x (x=-6, f(x)=3.63, x=-4, f(x)=0 x=0, f(x)=0 x=2, f(x)=7.56)

The function can be written as

\(y=\sqrt[3]{x}(x+4)\)

Even though you might think otherwise the domain of this function is

\(\{x\in\mathbb{R}:x\geqslant0\}\)

The problem with the cubic root of negative numbers is which value to choose as the principal value. The advantage of choosing the negative, real value is that you don't have to deal with complex numbers. That's probably the reason for this schoolbook definition.

The problem however is that this choice is not continuous: as soon as the exponent is slightly off from \(\frac{1}{3}\) you get a complex value anyway and then the principal value jumps to the branch of the value with the smallest argument.

The HP-42S doesn't provide a cubic root function and thus we have to use \(y^x\) with an approximation to \(\frac{1}{3}\) which is the reason you get a complex result for a negative value.

This is the program I used to calculate the values of the function:

Function

Code:

LBL "Fx"

RCL ST X

3

1/X

Y↑X

X<>Y

4

+

*

END

It returns complex values for negative real values.
Nothing can hinder us to still try to find a critical point for which I got:

x=-4.00000057735

But for \(x=-4\) the function will return

0 i0

Which is the reason we get that result.

I haven't stressed that but there are two conditions that the function has to meet:

it has to be real valued
it must be analytical

The 2nd condition means you can't use operations like ABS or SIGN in your program to define the function. These functions aren't analytical.

For the given function the minimum is at (0, 0).

HTH
Thomas

RE: Derivatives on HP 42S - lrdheat - 08-25-2018 05:19 PM

This is why, if just 1 function could be added to the HP 42S, I would choose xROOTy !

RE: Derivatives on HP 42S - Thomas Klemm - 08-25-2018 06:05 PM

You can still deliberately choose the other branch that makes the result real for negative input:

Code:

LBL "Fx"

RCL ST X

+/-

3

1/X

Y↑X

+/-

X<>Y

4

+

*

END

With initial guesses -4 and 0 you get what you expected:

x=-1

But now the function returns complex values for positive input.

Kind regards
Thomas

RE: Derivatives on HP 42S - Albert Chan - 08-25-2018 07:03 PM

(08-25-2018 05:19 PM)lrdheat Wrote: This is why, if just 1 function could be added to the HP 42S, I would choose xROOTy !

seems this is trivial to implement.

If x < 0 and y is odd integer, return -(-x) ^ (1/y), else x ^ (1/y)

If you knew y is odd integer, it is even simpler, just check sign of x.
This avoids the extremum search returning complex results.

Just add the check straight into Fx

Edit: complex number don't understand x < 0, may be real part of x instead ?

RE: Derivatives on HP 42S - Thomas Klemm - 08-25-2018 08:00 PM

(08-25-2018 07:03 PM)Albert Chan Wrote: seems this is trivial to implement.

This sounds a bit like: this is left as an exercise to the dear reader.

Quote:If x < 0 and y is odd integer, return -(-x) ^ (1/y), else x ^ (1/y)

Unfortunately \(x=\Re[z]\) and \(y=\Im[z]\) aren't analytic functions. So anything based on these values isn't analytic as well.

Quote:Just add the check straight into Fx

This violates the precondition of this algorithm: the function must be analytical

You might get away in this case but it won't work if a stationary point happens to be at the place where you stitch the functions together.

Cheers
Thomas

RE: Derivatives on HP 42S - Albert Chan - 08-25-2018 09:20 PM

(08-25-2018 08:00 PM)Thomas Klemm Wrote:
(08-25-2018 07:03 PM)Albert Chan Wrote: If x < 0 and y is odd integer, return -(-x) ^ (1/y), else x ^ (1/y)
Unfortunately \(x=\Re[z]\) and \(y=\Im[z]\) aren't analytic functions.
So anything based on these values isn't analytic as well.

Can you explain the word analytic ?
Is this the cost of using complex numbers ? (even using the real part not allowed ?)

Is f(x) = x^(1/3) an analytic function ?
If x is complex, is it true that f(x) same as -f(-x) ?

Googling "function not analytical":

A complex function is said to be analytic on a region if it is complex differentiable at every point in.
The terms holomorphic function, differentiable function, and complex differentiable function are
sometimes used interchangeably with "analytic function" (Krantz 1999, p. 16).

After above googling, click the graph, you get wikipedia explanation:

non-analytic smooth function is a smooth function which is nowhere real analytic Sad

RE: Derivatives on HP 42S - Thomas Klemm - 08-26-2018 04:54 AM

(08-25-2018 09:20 PM)Albert Chan Wrote: Can you explain the word analytic ?

Consider the complex valued function \(w=z^2\).
You can calculate both the real and imaginary part of \(w=u+iv\):

\(\begin{aligned}
u &= x^2-y^2 \\
v &= 2xy
\end{aligned}\)

But these functions \(u(x, y)\) and \(v(x, y)\) are not independent.
Instead the Cauchy–Riemann equations hold true:

\(\begin{aligned}
\frac {\partial u}{\partial x}&=\frac{\partial v}{\partial y} \\
\frac {\partial u}{\partial y}&=-\frac{\partial v}{\partial x}
\end{aligned}\)

And indeed:

\(\begin{aligned}
u_x &=2x=v_y \\
u_y &=-2y=-v_x
\end{aligned}\)

Thus this function is analytic.

However this function isn't:

\(\begin{aligned}
u &= x^2+x-y^2 \\
v &= 2xy
\end{aligned}\)

Because \(u_x=2x+1\) but still \(v_y=2x\).

For the other function that I mentioned, e.g. \(\Re[z]\) we have:

\(\begin{aligned}
u &= x \\
v &= 0
\end{aligned}\)

This isn't analytic since \(u_x=1\) but \(v_y=0\).

In short: If you define the function in terms of \(z\) the function is most probably analytic. However if you try to stitch together a complex function based on \(x\) and \(y\) chances are high that it's not analytic.

Quote:Is f(x) = x^(1/3) an analytic function ?

Yes. Its derivative is:

\(\frac{d}{dz}\left(\sqrt[3]{z}\right)=\frac{1}{3z^{\frac{2}{3}}}\)

Quote:If x is complex, is it true that f(x) same as -f(-x) ?

No. Consider \(f(z)=z^2\). Here we have \(f(-z)=f(z)\).

Cheers
Thomas

RE: Derivatives on HP 42S - Thomas Okken - 08-26-2018 01:54 PM

(08-26-2018 04:54 AM)Thomas Klemm Wrote:
(08-25-2018 09:20 PM)Albert Chan Wrote: If x is complex, is it true that f(x) same as -f(-x) ?

No. Consider \(f(z)=z^2\). Here we have \(f(-z)=f(z)\).

Functions for which f(x) = f(-x) are called even functions, and functions for which f(x) = -f(-x) are called odd functions, on account of the fact that their Taylor series contain only even or odd powers of x, respectively. The resulting symmetry applies for complex numbers as well as it does for real ones.