Question

Find the explicit function for this recursive equation:
\[
y = 2y+5
\]

Is this possible? What do you guys think?

Answer 1

\[
y = 2(2y+5)+5 = 4y + 15
\]

Answer 2

One solution is \(y=-5\). A constant function

Answer 3

I think that's the only solution :/

Answer 4

How about just \[
y=2y
\]

Answer 5

\(y\) would have to be 0 for that to be true.

Answer 6

No functions have this property then

Answer 7

other than 0

Answer 8

Well, constant functions are still functions, just not very interesting ones.

Answer 9

I'm thinking maybe some periodic function might work, but then you'd have to restrict the domain.

Answer 10

How about something like: \[
y(x-1) = 2 y(x) +5
\]

Answer 11

\[
y(x-2) = 2(2y(x)+5)+5
\]

Answer 12

\[
y(x-n) = 2^ny(x) + \sum_{k=1}^{n}2^{k-1}(5)
\]

Answer 13

I kinda wonder how something like this might be solved.

Answer 14

By the way, did you manage to get quoting to work?

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @wio
By the way, did you manage to get quoting to work?
\(\color{blue}{\text{End of Quote}}\)
Indeed I did.

Answer 16

Anyway, I don't know if there's a rigorous way to prove it, but I think that, generally, a function shifted horizontally by one unit can never be the same as the original function scaled by a factor of 2 and shifted vertically by five units.

Answer 17

unless it's constant, like the first two you posted.

Answer 18

**an appropriate constant**

Answer 19

Well, if you seeded it like said... \(y(0) = 0\), would the rest of it sort of fall into play?

Answer 20

\[
y(x-1) = 2y(x) +5
\]So \[
y(x) = \frac{y(x-1)-5}{2}
\]And \[
y(1) = \frac{y(0)-5}{2}=\frac {-5}2
\]

Answer 21

You'd be able to list out a bunch of inputs and values, so it would be a discrete function of sorts, so shouldn't there be some way to describe this function, even if not algebraically?

Answer 22

Ah, right, you could express every output in terms of \(y(0)\)... I get the feeling this would give you a linear function.

Answer 23

Or maybe not: In this case, \(y(x)=-\dfrac{5}{2}x\), but \(y(x-1)\not=2y(x)+5\) when you plug it in.

Answer 24

\[
y(2) = \frac{\frac{-5}2-5}{2} = \frac{\frac{-15}2}{2} = \frac{-15}4
\]

Answer 25

You could find a pattern if you kept doing this.

Answer 26

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
y(x-n) = 2^ny(x) + \sum_{k=1}^{n}2^{k-1}(5)
\]
\(\color{blue}{\text{End of Quote}}\)


Inverting this:\[
y(x+n) = \frac{y(x)- 5\sum_{k=1}^{n}2^{k-1}}{2^n}
\]Now letting \(x=0\) and \(n=x\):\[
y(0+x) = y(x) = \frac{y(0)- 5\sum_{k=1}^{x}2^{k-1}}{2^x} = \frac{- 5\sum_{k=1}^{x}2^{k-1}}{2^x}
\]

Answer 27

Since it is a geometric series, the summation can be simplified into an algebraic expression.

Answer 28

But you're only dealing with integeres \(x\), right? A real-valued function would be difficult to describe in this way.

Answer 29

Well, the recursive equation provides a definition to verify integer domain. Any real valued function which hits all the integer values correctly would be a valid extension onto the real domain.

Answer 30

In the same way that \(\Gamma(n+1)\) extends \(n!\)

Answer 31

Hmm, can't say I know enough about this stuff to be able to contribute here, but it's interesting nonetheless.