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.

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