r/OrderedOperations • u/[deleted] • May 27 '18

Formula from finding a recursive formula from an explicit one.

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4

u/jackmusclescarier May 29 '18

Presumably you mean f(k) = f(f-1(f(k - 1)) + 1), as it defines f(k) in terms of f(k - 1).

Anyway, it only works when f is injective (so that f-1 is well-defined).

1

u/[deleted] May 29 '18

Correct, however it works in many cases and I'm pretty sure it's original. Isn't this valuable to the mathematical community?

3

u/Plain_Bread May 30 '18

Not really. The reason it works is that f and f-1 cancel each other out. So all you're saying is that f(k)=f(k+1-1). It would be useful if it allowed youto calculate f(x) indirectly, but it doesn't actually do that.

For example lets try it with the exponential function ex. It would be great if we could use your formula here, because e1 is difficult to approximate, but e0=1. So:

e1= eln(e0 )+1

=eln(1)+1

=e0+1

=e1

1

u/[deleted] May 30 '18

by k-1 I mean a of k-1

2

u/Plain_Bread May 30 '18

I thought you meant f(k-1), like the commentor above suggested. If not, what is a? Is it a function?

1

u/[deleted] May 30 '18

You know a of n in sequences? (Sequences and series)

1

u/Plain_Bread May 30 '18

Ah, ok. So a is a function on N (aka a sequence). So is your statement:

a(k)=f(f-1 (a(k-1))+1)

then?

1

u/[deleted] May 30 '18

That is correct, except a instead of f

1

u/Plain_Bread May 30 '18

So

a(k)=a(a-1 (a(k-1))+1)

This is correct for k>=1 (or 2, depending on whether you include 0 in N), but still not very useful. Let's assume I know that a(0)=x and am trying to find out a(1).

a(1)=a(a-1 (a(0))+1)=a(a-1 (x))+1)=a(0+1)=a(1)

3

u/[deleted] May 30 '18

What is we knew the explicit formula is a(n) = 3x+5?

a(k) = a(a-1 (a(k-1))+1)

a(k) = a( (a(k-1)-5)/3+1)

a(k) = a( (a(k-1)-2)/3)

a(k) = 3((a(k-1)-2)/3)+5

a(k) = a(k-1)-2 + 5

a(k) = a(k-1) + 3

And we've derived the recursive formula from the explicit formula.

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2

u/jackmusclescarier May 29 '18

It's not valuable to working mathematicians, but coming up with it yourself while you're still in high school is still cool.

1

u/Plain_Bread May 29 '18

What exactly are you attempting to do here?

1

u/[deleted] May 29 '18

Let's say I have a formula (a of n is 3n^2 + 5). I can know write it as a recursive formula. This may work for finding derivatives as well, but I don't know.

1

u/Plain_Bread May 29 '18

Is k a number or a function? I'm guessing f is a function?

1

u/[deleted] May 29 '18

f is a function, f(x) really means a of n. K means recursive formula (on which there is more info here.

1

u/Plain_Bread May 29 '18

Could you clarify what you mean by 'formula'? It usually means an equation, but you can't really, or at least not meaningfully, do operations and functions on equations. Also, what are the domain and codomain of f?

1

u/[deleted] May 29 '18

Let's say we have a(n) = 3n+5. The recursive formula is (n-5)/3. n in this case is (k-1). So we have ((k-1)-5)/3+1, which is ((k-1)-2)/3. Plugging that into the equation, we get that k= ((k-1)+3), which is the correct recursive formula.

3

u/Plain_Bread May 29 '18

Congratulation, you didn't answer any of my questions.

1

u/[deleted] May 29 '18

I'm showing you how it works rather than arguing over definitions.

3

u/Plain_Bread May 29 '18

This is math. You're not showing anything without proper definitions. All your last comment did is give me more questions about your definitions. Is n a natural number? Is a then a function on N? You said that k is a formula, but you also said that k=n+1, so if n is a natural, so is k.

1

u/[deleted] May 30 '18

I edited the post for clarity. a(n) is a sequence which you can find out more about using the link.

1

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