Second derivative with complex numbers

[peacecalc]

Hello all,

I'm shure, in former threads has discussed that topic. After programming the first derivative with complex numbers in my DM42, I tried the same with the second derivative.

For rememberance, the taylor-series for the first derivative (truncating after the first derivative):

\[ f(x+i\cdot h) \approx f(x) + f'(\tau)|_{\tau = x}\cdot i h \quad \hbox{ take the imaginary part with}\quad \Im(f(x)) = 0 \quad \hbox{ because it is only real, one gets:} \quad
f'(x) \approx \frac{1}{h}\Im(f(x+ih)) \]

Now let us see the taylor series until the second derivative (and then truncating further terms):

\[ f(x+i\cdot h) \approx f(x) + f'(\tau)|_{\tau = x}\cdot i h + \frac{1}{2} f''(\tau)|_{\tau = x}\cdot (i h)^2 \quad \hbox{with} \quad i^2 = - 1 \Rightarrow f(x+i\cdot h) \approx f(x) + f'(\tau)|_{\tau = x}\cdot i h - \frac{1}{2} f''(\tau)|_{\tau = x}\cdot h^2 \]
now we take the real part:
\[ \Re(f(x+i\cdot h)) \approx f(x) - \frac{1}{2} f''(\tau)|_{\tau = x}\cdot h^2 \quad\hbox{the term:} \quad f'(\tau)|_{\tau = x}\cdot i h \quad \hbox{vanish because it is only imagenary and one gets:} \quad f''(x) \approx - \frac{2}{h^2} \cdot \left( \Re(f(x+i\cdot h)) - f(x) \right) \]

Maybe it is an advantage, because you only have to calculate one difference instead of three if you calculate the first order derivative without doing that in the complex plane. Please pay attention to, only the real part is taken from f(x+ih).

Sincerely peacecalc

[Albert Chan]

Here is another way to obtain the derivative formulas

\(\displaystyle f'(x) ≈ {f(x+ih) - f(x) \over ih}\)

\(\displaystyle f''(x) ≈ {f'(x) - f'(x-ih) \over ih}
≈ \frac{{{f(x+ih) - f(x) \over ih}} - {{f(x) - f(x-ih) \over ih}}}{ih}
= {f(x+ih) - 2 f(x) + f(x-ih) \over -h^2} \)

Weierstrass approximation theorem: we can assume f(x) is polynomial.
Conjugate of polynomial is polynomial of conjugate: f(x-hi) = conj(f(x+hi))

If f(x) is smooth and real, so does its derivatives → RHS must be real.

\(\displaystyle f'(x) ≈ {\Im(f(x+ih)) \over h} \)

\(\displaystyle f''(x) ≈ {\Re(f(x+ih) - f(x)) \over -h^2/2} \)

---

Using central difference derivative formula, we also get the same f'(x) formula.
Note: below does not assume real f'(x), RHS imaginary parts really cancelled out.

\(\displaystyle f'(x)
≈ {f(x+ih) - f(x-ih) \over 2ih}
= {\Im(f(x+ih)) - \Im(f(x-ih)) \over 2h}
= {\Im(f(x+ih)) \over h} \)

→ both derivative formulas should give accuracy similar to central difference formulas.

XCas> f(x) := x*(x^3+5*x^2-21*x)                    // example from here

XCas> f'(4)                                                         → 328
XCas> (f(x+h) - f(x-h)) / (2h) | x=4, h=1e-3        → 328.000021
XCas> im(f(x+h*i)) / h           | x=4, h=1e-3        → 327.999979

XCas> f''(4)                                                         → 270
XCas> (f(x+h)-2*f(x)+f(x-h))/h^2 | x=4, h=1e-3  → 270.000002132
XCas> re(f(x+h*i)-f(x)) / (-h^2/2) | x=4, h=1e-3  → 269.999998125

[peacecalc]

Hello Albert,

thank you for your lucide comments about that topic. In deed I always thought that the way into the complex plane is an advantage in precision, but it seems not. What a pity!

[Werner]

It may avoid roundoff if you want high accuracy.
Try the above examples on a real 42S, with h=1e-11.
The complex first derivative gets 328 exactly, the difference function gets 300.

Cheers, Werner

[Albert Chan]

(02-11-2021 12:36 PM)Werner Wrote:  It may avoid roundoff if you want high accuracy.

Yes, but only with first derivative. (assumed im(f(x+hi)) does not hit with catastrophic cancellation)
For second derivative, you still hit with subtraction cancellation errors.

XCas> f(x) := x*(x^3+5*x^2-21*x)

XCas> f1(x,h) := re((f(x+h)-f(x-h))/(2h))                // central difference 1st derivative
XCas> f'(x) .- [f1(x,h), f1(x,h*i)] | x=4, h=1e-3       → [-2.10000508787e-05, 2.10000000607e-05]
XCas> f'(x) .- [f1(x,h), f1(x,h*i)] | x=4, h=1e-6       → [-1.0320036381e-08, 2.09752215596e-11]

XCas> f2(x,h) := re((f(x+h)-2*f(x)+f(x-h))/h^2)     // central difference 2nd derivative
XCas> f''(x) .- [f2(x,h), f2(x,h*i)] | x=4, h=1e-3      → [-2.13206476474e-06, 1.87539626495e-06]
XCas> f''(x) .- [f2(x,h), f2(x,h*i)] | x=4, h=1e-6      → [0.0221821205923, 0.0506038300227]

---

If |h| is not too small, we avoided catastrophic cancellation, error = O(h^2)
We may take advantage of it, with h = ε*√i, which make h^2 purely imaginary.

XCas> f'(x) - f1(x,h) | x=4, h=1e-3*√i        → -2.44426701101e-12
XCas> f''(x) - f2(x,h) | x=4, h=1e-3*√i       → -4.78621586808e-11

[Werner]

You never cease to amaze me, Albert!
One more for my files, complex derivatives with h^2 purely imaginary..
Werner

[Werner]

A Free42 program for both first and second derivative.
I took h=sqrt(i), that seems to work as well, and then the division by h^2 for the second derivative is no longer necessary ;-)
Update: whenever I think I'm being smart, it turns out I make things worse. So the program below uses h = sqrt(i)/1e3..
Update 2 : shorter

Instructions: put the name of the function to derive in the ALPHA reg, X-value in X, press either F' or F" for first and second derivative, respectively.

Code:

00 { 86-Byte Prgm }
01▸LBL 02
02 RCL+ "X"
03 ASTO ST L
04 GTO IND ST L
05▸LBL "F'"
06 AOFF
07 GTO 00
08▸LBL "F""
09 AON
10▸LBL 00
11 LSTO "X"
12 -1ᴇ-12
13 SQRT
14 SQRT
15 LSTO "h"
16 XEQ 02
17 LSTO "."
18 RCL "h"
19 +/-
20 XEQ 02
21 FC? 48
22 GTO 00
23 STO+ "."
24 CLX
25 XEQ 02
26 STO+ ST X
27▸LBL 00
28 +/-
29 RCL+ "."
30 RCL "h"
31 FC? 48
32 STO+ ST X
33 FS? 48
34 X^2
35 ÷
36 COMPLEX
37 R↓
38 AOFF
39 END

With the polynomial above:

Code:

00 { 19-Byte Prgm }
01▸LBL "FX"
02 ENTER
03 ENTER
04 ENTER
05 5
06 +
07 ×
08 21
09 -
10 ×
11 ×
12 END

"FX"
4 XEQ "F'" -> 328 (-2.91e-29)
4 XEQ "F"" -> 270 (+2e-28)

Cheers, Werner

[Werner]

I updated the routine above, because I was wrong..
Werner

[Albert Chan]

(02-11-2021 02:13 PM)Albert Chan Wrote:  XCas> f2(x,h) := re((f(x+h)-2*f(x)+f(x-h))/h^2)     // central difference 2nd derivative
...
If |h| is not too small, we avoided catastrophic cancellation, error = O(h^2)
We may take advantage of it, with h = ε*√i, which make h^2 purely imaginary.

With h^2 purely imaginary, we can optimized away evaluation of f(x)
For h = ε*√i, all formulas below should have error = O(ε^4)

f(x) ≈ re((f(x+h) + f(x-h))/2)
f'(x) ≈ re((f(x+h) - f(x-h))/(2h))
f''(x) ≈ re((f(x+h) + f(x-h))/h^2)

[peacecalc]

hmm... a very serious contribution!
   

[Werner]

That calls for a new rewrite, simpler again.
Another attempt at improvement, I take h = ε*√(2*i) = (ε,ε) - then both h and h^2 are exact (in a decimal calculator. This is the hpmuseum, after all)

Code:

00 { 71-Byte Prgm }
01▸LBL 02
02 +
03 ASTO ST L
04 GTO IND ST L
05▸LBL "F'"
06 AOFF
07 GTO 00
08▸LBL "F""
09 AON
10▸LBL 00
11 LSTO "X"
12 1ᴇ-3
13 ENTER
14 COMPLEX
15 LSTO "h"
16 XEQ 02
17 X<> "X"
18 RCL "h"
19 +/-
20 XEQ 02
21 FC? 48
22 +/-
23 RCL+ "."
24 RCL "h"
25 FC? 48
26 STO+ ST X
27 FS? 48
28 X^2
29 ÷
30 COMPLEX
31 R↓
32 AOFF
33 END

Still, for F', I favour the f'(x) = Im(f(x+ih))/h formula, as it needs only one function evaluation.
Then the routines become:

Code:

00 { 77-Byte Prgm }
01▸LBL 02
02 +
03 ASTO ST L
04 GTO IND ST L
05▸LBL "F""
06 LSTO "X"
07 1ᴇ-3
08 ENTER
09 COMPLEX
10 LSTO "h"
11 XEQ 02
12 X<> "X"
13 RCL "h"
14 +/-
15 XEQ 02
16 RCL+ "X"
17 RCL "h"
18 X^2
19 ÷
20 COMPLEX
21 R↓
22 RTN
23▸LBL "F'"
24 1ᴇ-99
25 COMPLEX
26 ASTO ST L
27 XEQ IND ST L
28 COMPLEX
29 1ᴇ99
30 ×
31 END

Cheers, Werner