A simple question2

hypsin_0 · 2025-07-12 18:30:31

Try to solve f(3)

Last edited by hypsin_0 (2025-07-12 18:32:21)

ktesla39 · 2025-07-12 22:52:05

Hey dude, I calculated 100s of values and the final approximate answer was 0 as shown in the figure. You can see the complete execution at https://ktesla.infy.uk/math.html But only open if you are on a PC as this has various nested loops that might hang your computer a bit (3 GB RAM)

hypsin_0 · 2025-07-12 23:14:52

The correct answer:

This function has it name,but I don't know it's english name,maybe has a word"fusible"
It's chinese name:伪燃烧数函数
1/f(10) is greater than graham's number.
关于它的定理在皮亚诺算术公理体系中不可证(I don't know how to translate)

ktesla39 · 2025-07-12 23:29:46

Do you mean

Both are nearly 0

This version used JavaScript for calculation. I was originally going to use Python and it's SymPy module to find exact answer but python denied the recursive loop as it reached it's recursion limit.

But how did you know this? Can you tell me your grade?

hypsin_0 · 2025-07-12 23:46:52

visit this website: https://zhuanlan.zhihu.com/p/26493979529

ktesla39 · 2025-07-12 23:52:07

Ok that's strange and I had never seen such thing before

SyncZeta! · 2025-07-17 20:12:52

ktesla39 wrote:

Do you mean
Both are nearly 0
This version used JavaScript for calculation. I was originally going to use Python and it's SymPy module to find exact answer but python denied the recursive loop as it reached it's recursion limit.
But how did you know this? Can you tell me your grade?

2^{-1541023937} exactly.
Just try to calculate on your own. It's not that difficult and you can try to figure out it in, I guess, a few hour?
You can start from calculating f(2) and f(2-(1/2^a)-(1-2^b)-...), I think.
If you can finally get the value of sth. like f(2-(1/2^10)-(1/2^23)-(1/2^34)), you can start calculating f(3).

Bob · 2025-07-23 07:55:18

hypsin_0 wrote:

The correct answer:
2^-1541023937

I've been working on this since you posted it.

If x is negative, it's easy to work out f(x).

When positive each f(x) depends on f(x-1) so I reasoned that it should be possible to create a chain of f(y) where each y is smaller than the one before so that eventually y is negative and we can work that one out and hence all those in the chain.

On paper I kept making errors so I put the whole lot into a spreadsheet with a LOOKUP table with 2 columns, x and f(x)

That spreadsheet is growing and growing as the chains get longer and longer. MS Excel is not very kind to me. If I use the lookup for an x that isn't yet in the table, Excel gives me the nearest value anyway rather than saying it cannot do it. As a result I'm having to check everything by hand anyway. At the moment my table has 26 entries and I think there's some errors.

I'm considering writing a program with a subprocedure that creates the chain. That way I should be able to get correct answers and may have an answer to your puzzle sometime in the next two weeks.

Great puzzle! Simple? No it isn't!

Bob

Bob · 2025-07-29 03:10:59

I had my doubts about BBC Basic being able to handle nested function calls, but it worked better than I had hoped:

5 count = 0
10 INPUT x
20 y = FNf(x)
30 PRINT x," ",y
40
50
60 DEF FNf(x)
65 count = count + 1:PRINT "x="x,"count="count
70 IF x<0 THEN = -x ELSE = 0.5*FNf(x-FNf(x-1))

I used it to confirm my table in Excel:

The cells highlighted in yellow I'm fairly sure are correct.

The accuracy may drop off where the number of iterations is large.

Tried x = 3 without much hope. Gave up waiting at count = 100000.

Bob

Math Is Fun Forum

#1 2025-07-12 18:30:31

A simple question2

#2 2025-07-12 22:52:05

Re: A simple question2

#3 2025-07-12 23:14:52

Re: A simple question2

#4 2025-07-12 23:29:46

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#5 2025-07-12 23:46:52

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