anonymous
  • anonymous
Find the explicit function for this recursive equation: \[ y = 2y+5 \] Is this possible? What do you guys think?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[ y = 2(2y+5)+5 = 4y + 15 \]
anonymous
  • anonymous
One solution is \(y=-5\). A constant function
anonymous
  • anonymous
I think that's the only solution :/

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anonymous
  • anonymous
How about just \[ y=2y \]
anonymous
  • anonymous
\(y\) would have to be 0 for that to be true.
anonymous
  • anonymous
No functions have this property then
anonymous
  • anonymous
other than 0
anonymous
  • anonymous
Well, constant functions are still functions, just not very interesting ones.
anonymous
  • anonymous
I'm thinking maybe some periodic function might work, but then you'd have to restrict the domain.
anonymous
  • anonymous
How about something like: \[ y(x-1) = 2 y(x) +5 \]
anonymous
  • anonymous
\[ y(x-2) = 2(2y(x)+5)+5 \]
anonymous
  • anonymous
\[ y(x-n) = 2^ny(x) + \sum_{k=1}^{n}2^{k-1}(5) \]
anonymous
  • anonymous
I kinda wonder how something like this might be solved.
anonymous
  • anonymous
By the way, did you manage to get quoting to work?
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @wio By the way, did you manage to get quoting to work? \(\color{blue}{\text{End of Quote}}\) Indeed I did.
anonymous
  • anonymous
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.
anonymous
  • anonymous
unless it's constant, like the first two you posted.
anonymous
  • anonymous
**an appropriate constant**
anonymous
  • anonymous
Well, if you seeded it like said... \(y(0) = 0\), would the rest of it sort of fall into play?
anonymous
  • anonymous
\[ y(x-1) = 2y(x) +5 \]So \[ y(x) = \frac{y(x-1)-5}{2} \]And \[ y(1) = \frac{y(0)-5}{2}=\frac {-5}2 \]
anonymous
  • anonymous
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?
anonymous
  • anonymous
Ah, right, you could express every output in terms of \(y(0)\)... I get the feeling this would give you a linear function.
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[ y(2) = \frac{\frac{-5}2-5}{2} = \frac{\frac{-15}2}{2} = \frac{-15}4 \]
anonymous
  • anonymous
You could find a pattern if you kept doing this.
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @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 \(n=x\):\[ y(0+x) = y(x) = \frac{y(0)- 5\sum_{k=1}^{x}2^{k-1}}{2^x} = \frac{- 5\sum_{k=1}^{x}2^{k-1}}{2^x} \]
anonymous
  • anonymous
Since it is a geometric series, the summation can be simplified into an algebraic expression.
anonymous
  • anonymous
But you're only dealing with integeres \(x\), right? A real-valued function would be difficult to describe in this way.
anonymous
  • anonymous
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.
anonymous
  • anonymous
In the same way that \(\Gamma(n+1)\) extends \(n!\)
anonymous
  • anonymous
Hmm, can't say I know enough about this stuff to be able to contribute here, but it's interesting nonetheless.

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