## probability – Ways of solving the following recurrence relation system.

Consider the following system of linear recurrence relations.

begin{aligned} p_n &= a cdot p_{n-1} – c_{n-1}\ c_{n-1} &= p_{n-1} – b cdot p_{n-2} + c_{n-2}end{aligned}

with $$p_0 = 1$$ and $$c_1 = 1$$.

I’ve tried to represent $$p_n$$ as a finite linear combination of $$p_{k}, k < n$$, but this doesn’t work for me. Maybe there is any chance to represent the final solution?

Any ideas?

## recurrence relation – Two dimensional recursive function in \$Olog n\$ time complexity

It is well known that a recursive sequence or $$1$$-d sequence can be solved in $$O log n$$ time given that it has the form

$$a_n=sum_{k=1}^{n} C_ka_{n-k}$$

where $$C_k$$ is a constant. Examples would include polynomials like $$n^2$$ or $$n^3$$, exponentials like $$2^n$$, and Fibonacci Numbers defined by $$a_n=a_{n-1}+a_{n-2}, a_0=0, a_1=1$$. Factorials would not be included for example, because they are defined by $$a_n=na_{n-1}$$, and $$C_k=n$$ is not constant.

Let $$a_{n,k}$$ be a two dimensional sequence defined by

$$a_{n,k} = sum_{i=0,j=0, (ine n text{ }∧text{ }jne k)}^{n,k}C_{i,j}a_{i,j}$$

where $$C_{i,j}$$ is constant.

Is it possible to compute $$a_{n,k}$$ in logarithmic time (a.k.a $$O log n$$) or better?

I know one case where this is possible, namely if $$a_{n,k}$$ are coefficients of polynomials in a sequence ($$1$$-d recursive sequence), such that the degrees of each term differ by $$1$$. In this case, the diagonals of the sequence $$a_{n,n+t}$$ for some constant $$t$$, are $$1$$-d constant recursive functions, whose terms can be computed in $$Olog n$$ time. This doesn’t use the initial relation in terms of two indices though as expected.

However, in certain cases, the diagonals do not form $$1$$-d constant recursive functions, namely when the degree difference of consecutive polynomial terms is more than $$1$$.

## asymptotics – How to solve recurrence of a binary tree

I’m trying to solve this recurrence of a function of the height of a binary tree with a recursive tree.But I can’t find any pattern to solve it.

$$n_l$$ is the height of the left subtree, $$n_r$$ is the height of the right subtree and $$n$$ is the height of the tree.

$$T(n) = begin{cases} T(n_l)+T(n_r) + 1 & text{if } n ge 0, \ 1 & text{otherwise.}\ end{cases} \ text{Given that }n_l+n_r+1=n$$

## algorithm analysis – Finding a tight bound for a recurrence relation

algorithm analysis – Finding a tight bound for a recurrence relation – Computer Science Stack Exchange

## recursion – Symbolic Solution of recurrence equation in two variables

I have a recursive equation of the form `R(p,q)=frac{c}{2a-1}(R(p,q-1)+cfrac{2a-b}{2a-1} sum_{j=1}^{p-1} left((1-b)frac{c}{2a-1}right)^{p-j-1} R(j,k-1) )` with `a, b, c` as certain parameters independent of `p` and `q` and the initial condition is `R(i,0)=afrac{c^{p-1}}{(2a-1)^p}`. I have solved it and got the solution of the form `R(p,q)=frac{ab^{q-1}c^{p-1}}{(2a-1)^{p+q}}left(frac{(2a)^q}{b^{q-1}}-sum_{i=1}^{q}(2a-b)^i(1-b)^{p-i}binom{p-1}{i-1}sum_{r=1}^{q-i+1}binom{r+i-2}{i-1}(frac{2a}{b})^{q-r-i+1}right)`. My questions are can we write the solution in a more compact and easy form? Can anybody assisst me in solving it by some other technique?

## asymptotics – Can I use Master Theorem to solve recurrence relations with a constant in the recursive term?

If you are just interested in an upper bound you can notice that $$T(n) le S(n)$$ where $$S(n) = 3 S(n/3) + frac{n}{2}$$ and has solution $$S(n) = O(n log n)$$.

Alternatively there is always induction. You can show that, for $$n ge 2$$, $$T(n) le c n log n$$.

For $$2 le n < 7$$, $$T(n)$$ is a constant and $$n log n ge 1$$. Therefore the claim is true for a sufficiently large (constant) value $$c^*$$ of $$c$$.

For $$n ge 7$$ you have:
$$T(n) = 3Tleft(frac{n}{3} – 2right) + frac{n}{2} le 3 c frac{n}{3} log frac{n}{3} + frac{n}{2} = cn log n – cn log 3 + frac{n}{2},$$

which is at most $$cn log n$$ when $$c n log 3 ge frac{n}{2}$$ or, equivalently, $$c ge frac{1}{2 log 3}$$.

Simply pick $$c = max{c^*, frac{1}{2 log 3} }$$.

## spfx – Sharepoint on prem 19 not getting all recurrence events in calendar

I am getting events with the help of a query and it is working ok. But I get only one item from the recurrence item even tho I have `<ExpandRecurrence>` tag set to true. What am I doing wrong? If my view is set to a Calendar and not AllItems view I am getting multiple items from the recurrence but only for one month. Is there a way to combine the two of them or use AllItems view and get recurrence items as in the Calendar view? I am using sharepoint 19 on prem and sp-pnp-js library.

My code:

``````let data = await pnp.sp.web.lists
.getByTitle(this.props.list)
.renderListDataAsStream({
OverrideViewXml:
`
<QueryOptions>
<ExpandRecurrence>TRUE</ExpandRecurrence>
<Where>
<And>
<Geq>
<FieldRef Name="EventDate"/>
<Value Type="DateTime">` +
firstDay.toLocaleDateString("sl-SI",{year: "numeric"}) + "-" + firstDay.toLocaleDateString("sl-SI",{month: "2-digit"}) + "-" + firstDay.toLocaleDateString("sl-SI",{day: "2-digit"}) +
//"2021-03-01" +
`</Value>
</Geq>
<Leq>
<FieldRef Name="EventDate"/>
<Value Type="DateTime">` +
lastDay.toLocaleDateString("sl-SI",{year: "numeric"}) + "-" + lastDay.toLocaleDateString("sl-SI",{month: "2-digit"}) + "-" + lastDay.toLocaleDateString("sl-SI",{day: "2-digit"}) +
//"2021-04-01" +
`</Value>
</Leq>
</And>
</Where>
</QueryOptions>
`,
})
.then(.....
``````

## time complexity – solving Recurrence relation f T(n)=T(n/2)+log log(n)

``````T(n)=T(n/2)+log log(n)--------1
T(n)=T(n/4)+log log(n/2)+log log(n)-------------2
T(n)=T(n/8)+log log(n/4)+log log(n/2)+log log(n)-----------3
T(n)=T(n/2k)+log log(n/k)+....+log log(n/2)+log log(n)
T(n)=1+log log(n/k)+....+log log(n/2)+log log(n)
n/2k=1 => n=2k
T(n)=1+log log(2K/k)+....+log log(n/2)+log log(n)
T(n)=1+log log(2)+....+log log(n/2)+log log(n)
T(n)=1+log (log(2)*....*log(n/2)+log(n))
``````

I ARRIVED AT THIS POINT BUT I DIDN’T KNOW HOW TO CONTENU ANY HELP PLEASE

## Showing asymptotic lower bound on log of recurrence

I’m trying to prove a lower bound on some computational problem, but in order to do it, I need an $$Omega(nlog(n))$$ lower bound on $$log(T(n))$$, where $$T(n)$$ is a recurrence defined as follows:

$$T(1) = 1$$

$$T(n) = sum_{k=1}^{n-1} T(k)T(n-k)$$

Does this recurrence have a known solution? If not, just giving a lower bound on $$log(T(n))$$ would suffice for me. Optimally, I would want an $$nlog(n)$$ bound, but I don’t know if its possible to achieve. So, any lower bound bigger than a linear one would be appreciated! Thanks in advance for helping me out!

## time complexity – Recurrence equation – \$a,b\$ condition to be \$O(n^c)\$

The reucrrence:
$$T(n)=T(frac{n}{a}) + T(frac{n}{b})+O(n^c)$$
What is condition for $$a,n$$ to $$T(n)=O(n^c)$$?
So with substitution I get
$$T(n)=T(frac{n}{a}) + T(frac{n}{b})+O(n^c)le (frac{n}{a})^c + (frac{n}{b})^c + n^c = n^c ((frac{1}{a})^c + (frac{1}{b})^c + 1)$$
I need $$(frac{1}{a})^c + (frac{1}{b})^c + 1$$ to be less than one if I want to show that it is $$le n^c$$.
So $$frac{1}{a^c} + frac{1}{b^c} le 0$$ but I have troubles with that and I don’t know if my idea of showing this is right.