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How to solve the recurrence relation given by the equation below

$$T(n)=T(n-2)+T(n-4)+T(n-6)+...+T(0)$$

It seems to me that $T(n)$ will be exponential but i don't know how to proceed on this problem.

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    What are $T(0)$ and $T(1)$? – 2017-01-03
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    You can simplify the above as $T_n=2T_{n-2}$ – 2017-01-03
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    Your recurrence only works for $n$ even. You should also provide one for $n$ odd. – 2017-01-03

5 Answers 5

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Let $S(k) = T(2 k).$ Then $S(n) = \sum_{k=0}^{n-1} S(k),$ which will grow exponentially with base $2.$

Similarly, define $R(k) = T(2k+1).$ This will satisfy the same recurrence, with the same solution, so your thing grows like $2^{n/2}.$

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    You have no information about the values at odd numbers, there is nothing you can deduce. – 2017-01-03
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    @YvesDaoust Why don't I have information about the values at odd numbers? Because you have reading comprehension issues? – 2017-01-03
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    What do you know about $T(1)$ ? About $T(3)$ ? – 2017-01-03
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    @YvesDaoust The OP says nothing about initial conditions of any kind, so what do you know abut $T(0)?$ Or $T(2)?$ – 2017-01-03
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    The recurrence says $T(2)=T(0)$. There is no equivalent for odd terms. Unless you read a different text than me. – 2017-01-03
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Let $T(0)$=1, then $T(2)=1=2^0$, $T(4)=2=2^1$, $T(6)=4=2^2$, $T(8)=8=2^3$.

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Hint:

$T_2=T_0$

$T_4=T_2+T_0=2T_0$

$T_6=T_4+T_2+T_0=2T_4=2^2T_0$

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$T_n=2T_{n-2}=2^{\frac{n}{2}-1}T_0$

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You can turn the given recurrence in a more traditional form:

$$T(n+2)=\color{green}{T(n)}+\color{blue}{T(n-2)+T(n-4)+T(n-6)+...+T(0)}=\color{green}{T(n)}+\color{blue}{T(n)}.$$

The solution is obviously

$$T(n)=C2^{n/2}$$ for even, positive $n$.

With the initial condition

$$T(2)=T(0),$$

we have $$T(n)=2^{n/2-1}T(0)$$ where $T(0)$ is assumed to be known.

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Hint: $T(n-2)=T(n-4)+T(n-6)+T(n-8)+...T(0)$