OT math help.

[George Herold]

Hi all, I was helping my daughter with algebra last night.
One set of problems had you find the generating function for a list of integers.
For instance,
list was 1,4,16,64...
Which I wrote as F(n) = 4*F(n-1), F(1) = 1
The HW started counting at n=1 rather than zero... but that hardly matters.
Now my way to solve this is just to guess at the answer. (not very helpful to my daughter.) So I had her play around with different guesses till we 'found' the answer.
F(n) = 4^(n-1).
But it seems there should be a more formal way to arrive at the answer.
So what is it?
A solution for the above would be fine... or web link, or just tell me what this type of problem is called and I can go find the answer on my own.

Thanks George (and Elsie) H.

 

[George Herold]

Back in February thread regarding difference equations, Tim Wescott

wrote...



"The guy who taught my second term of diff eqs clearly wanted us to

remember this for all time, because he made this a mantra. Nearly

every

class meeting he would put some new form of differential equation up

on

the board, and he'd say "Now, how do we solve this differential

equation?" then (because we didn't all shout it out in unison) he'd

answer himself: "We guess, and prove that we're right!"



That was 30 years ago. It's stuck with me, so I guess he met his goal

in

my case."



The, _much_ later in the thread, devoted mostly to guessing solutions,

Euthymios Kappos wrote,



"Suppose f(1) = A, then we have



f(1) = A

f(2) = k * f(1)

f(3) = k * f(2)

f(4) = k * f(3)

...

f(N) = k * f(N-1)

=====================

then the product of all the terms on the left equals

the product of all the terms on the right,

and after eliminating f(1), f(2) ... which occur on both the

left and the right of the equaltio, we have



f(N) = A * k^(N-1), for N=1, 2, 3, ...



So, this is your formula.



Euthymios Kappos"



Does that help, or add to the confusion?

Thanks Jim, that helps. After posting the question I went and 'guessed' a solution but with unknown coefficients, and then plugged the guess back into the
formula and 'found' the coef. So that at least works. The math 'savy' comes in making a good guess!

Is it correct to think of these types of problems as a 'digital' form of a differential equation?

George H.

 

[George Herold]

The problem doesn't have a unique solution unless you possess the entire
list, or a rule for generating the list. For instance,

F(n) = 4**(n-1)+(n-1)(n-2)(n-3)(n-4)*n!

That matches the terms given, but grows a lot faster than 4**N.

Since there's no unique solution given any finite list, it's a heuristic
problem rather than a strictly mathematical one.

Phil, do you every get the feeling that you're too smart for your own good?
(He asked with a smile on his face.)

Did you miss the three little dots? 16,64... so 256, 1024... (etc.)

So given F(n) = 4*F(n-1) and F(1) = 1, is making a good guess the best way to get to a solution? (with unknown coefficients.)

George H.

 

[George Herold]

I'm sure the pedagogues (our crazy language, eh?) do have term for it
(no pun intended) but I always just called these "series problems."

Fun, I never heard of pedagogue. Hmm if I tell her math teacher I hear she's a good pedagogue will I be slapped?

 

[Jasen Betts]

Hi all, I was helping my daughter with algebra last night.
One set of problems had you find the generating function for a list of integers.
For instance,
list was 1,4,16,64...
Which I wrote as F(n) = 4*F(n-1), F(1) = 1
The HW started counting at n=1 rather than zero... but that hardly matters.
Now my way to solve this is just to guess at the answer. (not very helpful to my daughter.) So I had her play around with different guesses till we 'found' the answer.
F(n) = 4^(n-1).
But it seems there should be a more formal way to arrive at the answer.
So what is it?

The "easy" way to automate it is to find a polynomial
solution by means on simultaneousl equations.

That won't find 4^(n-1) but something else that fits those points.



the other way is to type them into google, with quotes around them
you'll get a result from OEIS

I remeber being given sequences like this 4,3,3,5,4,4,3,5,5,4,3,6,6,8
or 7,8,4,5,3,4,4 in highschool.. I suspect these where put on the end keep the
fast students quiet.

 

[George Herold]

1, 4, 16, 64, 42, ...

x_n = 4 * x_{n-1) - 86 x_{n-4}

So there.

Feeling chastised.
Lowers head, kicks dirt with shoe, "Sorry guys".

(I just hope they don't see the little smirk on my face :^)

George H.

 

[George Herold]

I remembered this...





<http://www.analog-innovations.com/SED/Difference_Equations-Earnshaw.pdf>

Interesting. (in a math sort of way) I've never heard of falling factorials.
(or maybe I was asleep that day in class) reminds me of statistics.. I didn't like that course.

George H.