Recursive sequence: h1 = x, hn = 1 = 2hn-1

[lunabee]

h1(x) = x, h n(x) = 1 = 2hn-1(x),n ≥ 2

 

[skeeter]

Re: Recursive sequences

h1(x) = x, h n(x) = 1 = 2hn-1(x),n ≥ 2

I assume you mean (and, please correct me if I'm mistaken) ...

h₁(x) = x, h_n(x) = 1 + 2h_n-1(x) for n > 2

I just have one more question ... what are you supposed to do with it?

 

[soroban]

Hello, lunabee!

I assume we're asked to find the \(\displaystyle n^{th}\) term of the sequence.

h1(x) = x, h n(x) = 1 = 2hn-1(x),n ≥ 2

The notation is very clumsy.
\(\displaystyle \;\;h\) is a function of n . . . not of x.

We have: \(\displaystyle \,h(1)\,=\,x,\;h(n)\:=\:1\,-\,2\cdot h(n-1),\,\, n\,\geq\,2\)

Crank out the first few terms:

\(\displaystyle h(1)\;=\;x\)
\(\displaystyle h(2)\;=\;1\,-\,2x\)
\(\displaystyle h(3)\;=\;\text{-}1\,+\,4x\)
\(\displaystyle h(4)\;=\;3\,-\,8x\)
\(\displaystyle h(5)\;=\;\text{-}5\,+\,16x\)
\(\displaystyle h(6)\;=\;11\,-\,32x\)
\(\displaystyle h(7)\,=\,\text{-}21\,+\,64x\)


After the first term, the first constants are: \(\displaystyle 1,\;-1,\;3,\,-5,\;11,\;-21,\;43,\;...\)

These are the partial sums of the series: \(\displaystyle \,1\,-\,2\,+\,4\,-\,8\,+\,16\,-\,32\,+\,\cdots\)

\(\displaystyle \;\;\)In general: \(\displaystyle \,\frac{1\,-\,(-2)^{n-1}}{3}\)


After the first term, the coefficients are: \(\displaystyle \,-2,\;4,\;-8,\;16,\,-32,\;64,\;...\)

\(\displaystyle \;\;\)In general: \(\displaystyle \,(-2)^{n-1}\)


The general term of the recursion is: \(\displaystyle \,h(n)\;=\;\frac{1\,-\,(-2)^{n-1}}{3}\,+\,(-2)^{n-1}x\)