Proving upper/lower bound

$f (n) = Θ(f (n/2))$

The counter example in the solutions was $f(n)=sqrt{n}$.

But then we get for every $nge n_{0}$

$sqrt{n}le c_{0}sqrt{frac{n}{2}} -> nle c_{0}^{2}cdotfrac{n}{2} -> 2nle c_{0}^{2}cdot n -> 2le c_{0}^{2}$

and I don’t see a problem with that, as we can choose $c_{0}=2$. Same with the omega definition. So what am I missing?

inequality – Proving on the equation $(x^2+mx+n)(x^2+px+q)=0$

Find all real numbers k such that if $a,b,c,d in mathbb R$ and $a>b>c>d geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that the following equation has 4 distinct real solutions :

$(x^2+mx+n)(x^2+px+q)=0$

Here all i did :

$(x^2+mx+n)(x^2+px+q)=0$

$Leftrightarrow x^2+mx+n=0$ or $x^2+px+q=0$

so I think the four real solutions, if any, of the equation can only be :

  • $ x= frac{sqrt{m^2-4n} -m}{2} $

  • $ x= frac{-sqrt{m^2-4n} -m}{2} $

  • $ x= frac{sqrt{p^2-4q} -p}{2} $

  • $ x= frac{-sqrt{p^2-4q} -p}{2} $

So I think to solve the problem we just need to find all real numbers k such that if $a,b,c,d in mathbb R$ and $a>b>c>d geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that
$p^2 geq 4q$ and $ m^2 geq 4n$

We can prove that for $k >= 4 $ it is absolutely true. So are there any other satisfying $k $ values ? I’m not entirely sure. Hope to get help from everyone. Thanks very much !

real analysis – Proving $cos x$ has a unique fixed point

I am trying to prove that the function $cos x$ has a unique fixed point. Certainly $-1 leq cos x leq x$, so I’m already restricted to $[-1,1]$. Further, if $x in [-1,0]$, then $cos x > 0$ but $x < 0$, so $cos x neq 0$, so I’m only concerned with $[0,1]$. Every argument for this I have seen uses the contraction principle, though I don’t have access to it yet and can’t use it.

Are there any other ways to approach this?

Proving that a problem is not FPT using reduction

In the Inclusive Vertex Cover problem, For a given graph $G=(V,E)$, each vertex $uin V(G)$ has weight $u_{w} in mathbb{N}$ and value $u_{v}in mathbb{N}$. The value and weight of a set cover $S$ are defined as the sum of weight, value of the vertex in the cover, and its neighbors. Note that a vertex might be counted more than once if it has two neighbors in the cover.

More precisely:

$W = sum_{vin V(G)}(sum_{x in N(v)}x_{w})$

$V = sum_{vin V(G)}(sum_{x in N(v)}x_{v})$

The problem is: Given an instance $(G,k,W^{*},V^{*})$ is there a vertex cover $S$ with size of at most $k$ s.t $W le W^{*}$ and $V ge V^{*}$.

I need to prove that this specific problem can’t have an FPT algorithm with a run time complexity of $f(k)n^{O(1)}$.

The way of doing that is by showing that for a fixed $k$, we can solve some NP-hard problems.

The problem definition is almost identical to the knapsack problem.

I am looking for a reduction from an NP problem to the Inclusive Vertex Cover for a fixed $k$.

I tried taking an instance for the knapsack problem and:

  • Defining different stars for every possible group, but the number of groups is exponential
  • Isolated vertices can’t work since we don’t know the number of vertices on the cover
  • Use more complicated graphs with a predetermined number of items, but it always leads to an exponential number of vertices in the graph

There is a connection to this question and some other questions that were posted on this subject, but I followed this guide that was recommended on how to ask a good homework question. However, I altered the question a bit since it’s much easier to look at the similarity of the IVC problem to the knapsack for me.

Any suggestions? Hints? Ideas?

How can I use induction for proving termination of a string rewriting system?

If we have a string rewriting system within the alphabet {X,Y}* and the rule XY->YX. How can we prove by induction that on every string input the system terminates?

real analysis – Proving Integrals on $C[a,b]$ are Lipschitz under maximum norm

My problem says:

Let $X=C(a,b)$. Define the function $phi:Xtomathbb{R}$ by
$$phi(f)=intlimits_{a}^{b} f(x)dx$$
for each $f$ in $X$

Show that $phi$ is Lipschitz on the metric space $X$, where $X$ has the metric induced by the maximum norm.

So,I’m trying to prove that
$$|intlimits_{a}^{b} f(x)dx – intlimits_{a}^{b} g(x)dx|leq text{c}; max{|f(x)|,|g(x)|}$$ is this correct? And how should I started my proof?

reference request – Proving Uniformity of Infinity (Cantor’s Proof)

I have an idea about the uniformity of infinity between $aleph_0$ and $aleph_1$, motivated by directly disputing Cantor’s proof which I learned of in my discrete mathematics class. I’m going to skip all the formalization of this proof that I’ve written out for brevity’s sake, but I’m rather convinced of it. Surely, I cannot be right, so I’m looking for some criticism on where it is wrong.

The formalized version states that if we prove $(0,1)$ is countably infinite, then $(-infty,infty)$ is. I can explain my reasoning for this if you’d like.

Now this proof relies on Byers’ demonstration I learned about in grade-school that a repeating radix-point number can be represented in two equivalent ways:

$$
x=0.overline1_2text{; definition}\
10_2x=1.overline1_2text{; multiply both sides by radix}\
10_2x=1_2+0.overline1_2text{; decompose into sum substituents}\
10_2x=1_2+xtext{; substitute}\
x=1_2text{; subtract both sides by }x\
$$

The generation of the sequence $(0,1)$ we’ll use to run against Cantor’s diagonal proof is written out with three parts here in a brief, not particularly rigorous manner. Hopefully it’s still rigorous enough to be convincing — if you’re not convinced with the first two steps, just start at step 3, since the first two steps are just contextual. It’ll be done in radix-two, or binary, for consistency with the above demonstration.

Step 1: Enumerate 0 and all positive integers in ascending order — prepend leading zeros for uniformity’s sake. Use each value as the row index.

  1. $…00000$
  2. $…00001$
  3. $…00010$
  4. $…00011$
  5. $…00100$

Step 2: Treat each number like a string of symbols are reverse it.

  1. $00000…$
  2. $10000…$
  3. $01000…$
  4. $11000…$
  5. $00100…$

Step 3: Prepend bicimal point. Recall that trailing zeros are insignificant.

  1. $0.00000…$
  2. $0.10000…$
  3. $0.01000…$
  4. $0.11000…$
  5. $0.00100…$

Run this through Cantor’s diagonal test to generate a unique number (ignoring everything before the bicimal point). Note that the last digit in this sequence is $0.overline1_2$, which is $1$. There are two numbers you can arrive at: $0.overline10$ and $0.overline1$, depending on how you want to view “reaching the end”. For the former, you can say that Cantor’s proof relies on their being an infinitesimal in order to generate a unique number, which means the series is countable in the first place. For the latter, that number is the final number in our sequence by definition: $1$, so it’s not unique.

A caveat of this method is that irrational and transcendental numbers cannot currently be found in the series in a finite number of steps due to their nature of not being perfectly expressible in a radix-point number. I’m not particularly supporting the idea that there really is only one form of infinity, but more disputing Cantor’s proof stating that there is. The call in this is that Cantor’s proof isn’t rigorous, and that proving different infinities requires a definition for the set of irrational and transcendental numbers that is not based on other sets.

Despite the lack of rigorousness in this post, I still would like all criticism, including that of the lack of rigor, and ways to better articulate what I’m saying. Feel free to submit edits or comments and I’ll revise the post.

Most importantly, is there anything in the literature that discusses this idea already?

pushdown automata – Proving a PDA transition with Mathematical Induction

We were told to prove that this transition is valid
$$<q_0, a^n, Z_0> vdash^* <q_0, epsilon, Z^nZ_0>$$

I managed to prove the base case and got the inductive step
$$<q_0, a^{k+1}, Z_0> vdash^* <q_0, epsilon, Z^{k+1}Z_0>$$

But I can’t seem to get the final working, I can’t proceed beyond this point.

In the stack, an input $a$ pushes $Z$ into the stack and $b$ pops from it

logic – Software for proving tautology with steps

I’m looking for a software which can prove a tautology using logical equivalences.
It should be able to show each step. So you can follow the chain of reasoning.

Here is an example:
begin{align}
(pto q)to(neg qtoneg p)&equivneg(neg plor q)lor(qlorneg p)tag{material implication}\(1em)
&equivneg(neg plor q)lor(neg plor q)tag{commutativity}\(1em)
&equiv neg Mlor Mtag{$M:neg plor q$}\(1em)
&equiv mathbf{T}.tag{negation law}
end{align}

regular expressions – Proving $(a + ab)^*a = a(a + ba)^*$

Need to prove $(a + ab)^*a = a(a + ba)^*$ by using regular algebra.

Concatenation does not commute. So repeated use of commutativity will fail. I am getting confused about which identity I should use to prove this. What I can interpret is that I just need to change the side of $a$ but the $^*$ must be kept outside. But I am not able to do this.

Any help will be appreciated.