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Finding 2022nd derivative (Posted on 2023-07-07) Difficulty: 4 of 5
Let us consider this function:

f(x) = x13*cos2(x1000)*ex^3.

Determine f(2022)(0), that is, evaluating the 2022nd derivative of f(x) at x=0.

Note: Computer program simulations are certainly acceptable , but an semi-analytic method would be much better.

No Solution Yet Submitted by K Sengupta    
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patterns are elusive | Comment 1 of 15
syms x
f=x^13 * (cos(x^1000))^2 * exp(x^3)
f1=diff(f)
f2=diff(f1)
f3=diff(f2)
x=0;
eval(f)
eval(f1)
eval(f2)
eval(f3)

finds

f1 =
13*x^12*cos(x^1000)^2*exp(x^3) + 3*x^15*cos(x^1000)^2*exp(x^3) - 2000*x^1012*cos(x^1000)*sin(x^1000)*exp(x^3)
f2 =
156*x^11*cos(x^1000)^2*exp(x^3) + 84*x^14*cos(x^1000)^2*exp(x^3) + 9*x^17*cos(x^1000)^2*exp(x^3) - 2000000*x^2011*cos(x^1000)^2*exp(x^3) + 2000000*x^2011*sin(x^1000)^2*exp(x^3) - 2050000*x^1011*cos(x^1000)*sin(x^1000)*exp(x^3) - 12000*x^1014*cos(x^1000)*sin(x^1000)*exp(x^3)
f3 =
1716*x^10*cos(x^1000)^2*exp(x^3) + 1644*x^13*cos(x^1000)^2*exp(x^3) + 405*x^16*cos(x^1000)^2*exp(x^3) + 27*x^19*cos(x^1000)^2*exp(x^3) - 6072000000*x^2010*cos(x^1000)^2*exp(x^3) - 18000000*x^2013*cos(x^1000)^2*exp(x^3) + 6072000000*x^2010*sin(x^1000)^2*exp(x^3) + 18000000*x^2013*sin(x^1000)^2*exp(x^3) - 2072862000*x^1010*cos(x^1000)*sin(x^1000)*exp(x^3) - 18486000*x^1013*cos(x^1000)*sin(x^1000)*exp(x^3) - 54000*x^1016*cos(x^1000)*sin(x^1000)*exp(x^3) + 8000000000*x^3010*cos(x^1000)*sin(x^1000)*exp(x^3)
ans =
     0
ans =
     0
ans =
     0
ans =
     0
     
     
so the successive evaluations of derivatives at x=0 all seem to be zero, and each term in each successive derivative has a factor of x to a power.

However:

syms x
f(1)=x^13 * (cos(x^1000))^2 * exp(x^3)

for i=2:16
  f(i)=diff(f(i-1))
end
x=sym(0);
for i=1:16
  disp(i-1)
  eval(f(i))
endsyms x
f(1)=x^13 * (cos(x^1000))^2 * exp(x^3)

for i=2:16
  f(i)=diff(f(i-1))
end
x=sym(0);
for i=1:16
  disp(i-1)
  eval(f(i))
end

finds that the 13th derivative evaluates to 6227020800 at x=0. The 14th and 15th derivatives revert to zeros.

Further results show that every third level of differentiation evaluates to nonzero thereafter. Per level of differentiation, the value of that derivative at x = 0 is:

 1 0                             
 2 0                             
 3 0                             
 4 0                             
 5 0                             
 6 0                             
 7 0                             
 8 0                             
 9 0                             
10 0                             
11 0                             
12 0                             
13 6227020800                    
14 0                             
15 0                             
16 20922789888000                
17 0                             
18 0                             
19 60822550204416000             
20 0                             
21 0                             
22 9223372036854775807           
23 0                             
24 0                             
25 9223372036854775807           
26 0                  
27 0                             
28 9223372036854775807           
29 0                             
30 0                             
31 9223372036854775807           
32 0                             
33 0                             
34 9223372036854775807           
35 0                             
36 0                             
37 9223372036854775807           
38 0                             
39 0   

The non-zero values have settled down to one value. The non-zeros show up when the differentiation level is 1 mod 3. The puzzle asks for level 2022, which is 0 mod 3, so the answer seems to be zero. But that's certainly not definitive, as it's based on patterns which change along the way.



  Posted by Charlie on 2023-07-07 12:16:36
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