I thought my initial post posted (it didn't). Is this the best way to find extremums? This finds "x". I then need to have the original function available to plug the "x" in order to find f(x) at the extremum.
There are better ways to estimate a functions derivative at a point. The WP 34S uses a four, six or ten point method e.g.
At the very least use (f(x+h) - f(x-h)) / (2h).
- Pauli
Edited: 21 May 2013, 7:51 a.m. after one or more responses were posted
if you just look for f'==0 you can save the division.
Example
f(x) = sin(x)/x
Find the extremum between and 2.
The program for the function
LBL 0
SIN
LSTx
/
RTN
The program to calculate the derivative
LBL 1
RCL 0
I
GSB 0
Re<>Im
RTN
Solution
We use a small value h in register 0:
0.00001
STO 0
ENTER
ENTER
+
SOLVE 14.49341
GSB 0
-0.21723
Kind regards
Thomas
Addendum
Using calculus we end up with: x = tan(x). We can solve that with a simple fixed-point iteration. Just fill the stack with , enter 4 as starting guess and repeat the following steps:
TAN-1This list is the result:
+
4.00000
4.46741
4.49218
4.49335
4.49341
Edited: 20 May 2013, 10:40 a.m.
Your link to Wikipedia is broken.
There is a great little article (very readable) on the complex-step method of calculating numerical derivatives, entitled "Complex-Step Derivatives: How Did I Miss This?" by Michael Sherman (Stanford University): http://biomedicalcomputationreview.org/2/3/8.pdf
Edited: 20 May 2013, 1:45 p.m.
That is neat! Your example worked great. Next post with article nice as well. I'm not getting the syntax for other examples, however. How would I enter as my LBL 0 f(x)= x^2 - 3*x -10
An extremum should be at x=1.5. f(x)=-12.25
Thanks!
f(x)= x^2 - 3*x -10
LBL 0
x2
LSTx
3
*
-
10
-
RTN
Cheers
Thomas
If you're missing the fact that the stack is filled with
x you could use the following:
LBL 1This makes using Horner's scheme to evaluate polynomials somewhat easier:
RCL 0
I
ENTER
ENTER
ENTER
GSB 0
Re<>Im
RTN
LBL 0
3
-
*
10
-
RTN
Edited: 20 May 2013, 3:10 p.m.
It's done exactly the same way as in the previous example.
- Define your function f(x) as usual. For instance like this:
LBL 0
x^2
LstX
3
x
-
1
0
-
RTN
- Then simply call f(x) for your (real) x plus a small imaginary part h.
Finally get the derivative by dividing the imaginary part of the result by h.First, store a suitable value for h in R0, e.g. 10-10:
1 EEX 10 CHS STO 0
Enter the routine for the first derivative:
LBL 1
RCL 0 ; now x holds the complex number x + h*i
I ; and the 15C switches to complex mode
GSB 0 ; simply call f(x) for x+hi
Re<>Im ; Get the imaginary part of the result
RCL/ 0 ; and divide it by h to obtain the derivative
CF 8 ; optional clean-up: quit complex mode
RTN
-
Now you can solve directly for f'(x)=0, using LBL 1 as the function to solve. Assume the minimum is somewhere between x = 0 and 5.
0 ENTER 5
f SOLVE 1 => 1,5000 local minimum x
GSB 0 => -12,2500 corresponding y
Edited: 20 May 2013, 3:33 p.m.
Hi Thomas,
When I enter your steps, I get 0 for solve 1 (and solve 0 incorrectly gives a root of 4 for intial guesses of 0,12 instead of 5). If I go:
LBL 0
2
y^x
x exchange y
3
*
-
1
0
-
RTN
I get the correct root of 5 for solve 0 (solve f(x)=0)
Solve 1 gives me 0 when using
LBL 1
RCL 0
I
GSB 0
Re exchange Im
RTN
where .00001 is stored in "0"
Can you reproduce my experience? How do I get the correct results that you came up with?
still get 0 for solve 1 when I add
RCL 0
/
after
Re exchange Im
First, please do not use 2 y^x when you want to get x^2. There is a dedicated x^2 key on virtually any scientific HP, which works much faster, preserves x in LastX and is even potentially more accurate.
Your approach assumes that x fills the whole stack so that it can be recalled with x<>y in your LBL 0 routine. However, in LBL 1 it's this routine that provides the x-value, so you will have to enter a few ENTER commands that fill the stack before GSB 0 is called. Also the LBL 1 routine lacks the final division by h. Which usually (but not always) does not matter much if you want to solve for f'(x)=0, but I think you should add that division there.
With these adjustments everything works fine. All this can be done shorter and faster if the function is coded the way Thomas and I proposed. Also see my detailled post below.
0 ENTER 12 f SOLVE 0 => 5,0000 f(5) = 0
0 ENTER 5 f SOLVE 1 => 1,5000 f'(1,5) = 0
Dieter
If you expect that the stack is filled with x, use this program instead:
LBL 1
RCL 0
I
ENTER
ENTER
ENTER
GSB 0
Re<>Im
RTN
HTH
Thomas
PS: You can use the [pre]-tag for listings: just click on the button [Preformatted] and REPLACE THIS TEXT WITH YOUR LISTING.
Blind as a bat re x^2 key.
Last question...(and thanks for the last x discussion)...how do I implement that in this example...the HP 15C manual has a solve example:
f(x)=5000*(1-e^(-x/20))-200*x
using 0,12 a local max should occur at about x=4.4629,f(x)=107.4258
I've modified the expression a little:
5000*(1-e^(-x/20))-200*x = -200*(25*(e-x/20 - 1) + x)
LBL 0Be careful when using [CHS]: it changes only the real part of a complex number! Make sure to use it just with real constants as in -20. It took me some time to figure that out.
20
CHS
/
ex
1
-
25
*
+
200
CHS
*
RTN
4 ENTER 5
SOLVE 1
=> 4.426871026GSB 0
=> 107.4257948
Cheers
Thomas
Edited: 20 May 2013, 6:29 p.m.
Quote:You do it the same way for each and every arbitrary function you come across. Just define your function at LBL 0 and leave the rest as it is (be sure to add those three ENTERs in the LBL 1 routine).
how do I implement that in this example
So simply rewrite your new function at LBL 0 - that's all. If you have problems implementing simple functions in 15C code, please take a closer look at the basic chapters on programming in the 15C manual.
A literal implementation of your last case may look like this:
LBL 0Please note the already mentioned special behaviour of CHS with complex numbers, which is crucial here. Therefore CHS was only used to enter a negative real constant -20, but not for changing the sign of a computed result.
ENTER ; may be omitted
ENTER ; if the stack already
ENTER ; is filled with x
2
0
CHS
/
e^x
1
x<>y
-
5
EEX
3
x
x<>y
2
0
0
x
-
RTN
0 ENTER 12Dieter
f SOLVE 1 => 4,4629
GSB 0 => 107,4258
Oops, fixed now.
- Pauli
Quote:
0 ENTER 12
And now I suddenly understand this:
Quote:
using 0,12 a local max should occur
Quote:This doesn't happen with the HP-42S though.
special behaviour of CHS with complex numbers, which is crucial here
And if somebody is still in doubt why it would have been a good idea to provide complex variants for all functions of the HP-35S: that's why. You can still use this neat trick with polynomials but it fails with trigonometric functions or a simple square root.
Cheers
Thomas
Edited: 21 May 2013, 8:41 a.m.
A most interesting discussion. Excellent! Amazing what happens when one (me) isn't afraid to uncover some lack of knowledge!
I'm dangerous...I get an error 8 when I try solve 1 on x^(3/5) * (4-x)
GSB 0 works for values that should represent max, i.e. 1.5 which should yield a max of 3.1886.
Any way around this?
Thanks again!
LBL 0
.6
Y^X
X<>Y
4
X<>Y
-
*
RTN
1 ENTER 2
SOLVE 1
=> 1.500000000
GSB 0
=> 3.188561254
Works for me.
Cheers
Thomas
For what it's worth you can shave a step by replacing X<>Y 4 X<>Y with 4 RollUp.
Quote:Then you must be doing something wrong.
I get an error 8 when I try solve 1 on x^(3/5) * (4-x)
LBL 0Where's the problem?
4
x<>y
-
LstX
,
6
y^x
*
RTN0 ENTER 10
f SOLVE 1 => 1,5000
GSB 0 => 3,1886
Dieter
It's always worth to save a byte here or there. Thanks for pointing that out.
Best regards
Thomas
Quote:
I'm dangerous...
Just wondered what your next example will be. Maybe something like Valentin's extremely flat maximum that is difficult to calculate numerically? I faintly remember a cos was involved but I might be wrong.
Ok, found it in message #52 of
hp 35s not very impressive:
Quote:
Numerically find a root of
Sin(x + Cos(x)) = 1
for x in [0, Pi].
LBL 0
COS
+
SIN
RTN
1 ENTER 2
SOLVE 1
=> 1.569585997
Quote:
Of course the exact root isx = Pi/2 = 1.57079632679489661923132169163975144209858469968755291...
Cheers
Thomas
PS: Marcus von Cube posted in message #67:
Quote:
I did some playing around with the function. It's not only flat at the root, it's damn flat: the first 5 derivatives of the function are all zero at the root.
Edited: 21 May 2013, 8:37 p.m.
I was inefficient like a novice...but don't know why it works as subroutine 0, but generates error 8 on solve 1:
LBL 1
enter
enter
enter
CHS
4
+
x exchange y
3
enter
5
/
y^x
*
RTN
You forgot the special behaviour of the CHS function for complex values, as already discussed in this thread. So do not use CHS to change the sign of an already computed result in X, since it will change only the sign of the real part of X, but not that of the imaginary part!
That's why your CHS after the three ENTERs is causing the problem here. Your implementation
CHSwill work fine for reals (as in GSB 0), but not for complex numbers (as used at LBL 1). You have to use
4
+
4here, and everything will work fine. Both for real and for complex values.
x<>y
-
Dieter
Edited: 21 May 2013, 4:44 p.m.
Again, thanks, and thanks for the patience!
Homework
1. Single-step through GSB 1 both with a value a little smaller and a little bigger than 1.5, (e.g. 1.49 and 1.51). Use the (i)-key to monitor the imaginary part of the intermediate results. Is there a change in the sign of the derivatives? How is that related to the Error 8 you encountered?
2. Repeat the steps above with Kiyoshi's solution:
LBL 0
.6
Y^X
4
R^
-
*
RTN
3. Edit your post. Insert [pre] before and [/pre] after your listing. Remove the additional line-feed between the steps.
Kind regards
Thomas
PS: Bonus points if you succeed in writing yx instead of y^x.