Let me edit that! I feel like I could be using better values instead.
I changed the $(k-1)$ to $k$ instead, in turn changing $diam(G)$ to $k$ as I think it would make things easier. Let me attempt to find the maximum value of $apd(G)$, please correct me if I'm wrong.
We want the sum of all...
Hi, I have attempted and I am stuck at this part.
Fix $ k$. Consider G, with a path $v_{1},v_{2},...v_{k-1}$. Attach each node $u_{1},u_{2},...u_{n-(k-1)} $to $v_{1}$. This means we have attached $(n-k+1) $nodes to $v_{1}$.
$diam(G) = d(u_{1},v_{k-1}) = k-1$.
So the graph looks like this...
I need to prove the above statement. I have a very strong gut feeling that the above equation is not true, and so I need to find a case where the graph diameter is greater than the average pairwise distance.
First off, I would like to clarify about the average pairwise distance, which is given...
Thank you! I understand it now. Since $K_{1}$ is an isolated vertex, there will be no edges anyway. And so we just look at $K_{n-k+1}$ which leads us to C(n-k+1,2).
Thank you for the hint. I think it's cool that you mentioned about SameGame too, I played that game when I was younger and never really thought that you could link it to this topic.
In any case, I followed as per your comment and got this:
Let $n_{q} \ge n_{p} \gt 1$
$K_{n_{q}} + K_{n_{p}} \lt...
Hello! I am having trouble solving the right part of the inequality.
For left part of the inequality $n-k \le m$, here’s how I did it
Let $ n = v_{1} + v_{2} + v_{3}...+v_{k}$, the sum of vertices of each component in G
least number of edges = $(v_{1}-1) + (v_{2}-1) + (v_{3}-1)...+(v_{k}-1)$...
$\left | h + cos(h) \right | \leq C \left | h \right | $can be written as
$\left |h + (1-\frac{h^2}{2!}...) \right | \leq C \left | h \right |$
$\frac{\left |h + (1-\frac{h^2}{2!}...) \right |}{ \left | h \right | } \leq C$
The $\frac{1}{\left | h \right |}$ part in the left hand side...
Now $\sum_{i=0}^{10}(x_{i}+1) L _{10,i}(5) = (x_{0}+1) L _{10,0}(5) + (x_{1}+1) L _{10,1}(5) + ... + (x_{10}+1) L _{10,10}(5)$
Which I can further decompose into
$\frac{(x_{0}+1)(5-x_{1})(5-x_{2})...(5-x_{10})}{(x_{0}-x_{1})(x_{0}-x_{2})...(x_{0}-x_{10})} +...
Question: For the following functions, does the fixed point iteration for finding the fixed point in $\left [ 0,1 \right ]$ converge for all first points $ p_{0} \in \left [ 0,1 \right ]$?
Justify your answer.
a.$ g(x) = e^{\frac{-x}{2}}$
b.$ g(x) = 3x - 1$
Let me attempt for part a first...
$\left | h + cos(h) \right | \leq C \left | h \right |$ And by taylor series, I know that $ h + cos(h) = h + (1-\frac{h^2}{2!}...) $
So solving for the equation, as h tends to 0, lhs tends to infinity while rhs is a constant. This is a contradiction, so h + cos h is not O(h)
I am having trouble solving part 2, for
$ W_{\frac{n(n+1)}{2}} \leq 2^{n} (n-1) + 1 , n \geq 0 $
I know that $W_{m} \leq 2*W_{m-k} + 2^{k} – 1, 0 \leq k \leq m$
Let $m = \frac{n(n+1)}{2}$
So now $W_{\frac{n(n+1)}{2}} \leq 2*W_{\frac{n(n+1)}{2} - k} + 2^{k} - 1, 0 \leq k \leq...
Let \mid h \mid < 1. Which of the following functions are O(h)? Explain.
-4h
h+h^2
\mid h \mid ^{0.5}
h + cos (h)
Based on my notes, f(h) = O(h) only if \mid f \mid ≤ C \mid h \mid , where C is a constant independent of h.
I can only solve for the first function -4h, as I can...
Let S on X be {(a,b), (a,c), (a,a), (b,a), (b,b), (c,a), (c,c)}. S is reflexive and symmetric, and S∘S= {(a,a), (a,b), (a,c), (b,b), (b,a), (b,c), (c,c), (c,a), (c,b)}
So S has 7 elements while S∘S has 9 elements, so S is reflexive and symmetric is not true.