## functional programming – Proving injectivity for an algorithm computing a function between sets of different types of partitions

I am attempting to solve the following problem:

Let $$A$$ be the set of partitions of $$n$$ with elements $$(a_1, dots, a_s)$$ such that $$a_i > a_{i+1}+a_{i+2}$$ for all $$i < s,$$ taking $$a_{s+1} = 0.$$ Define $$g_n = F_{n+2}-1$$ and let $$B$$ be the set of partitions of $$n$$ as $$b_1 ge dots ge b_s$$ such that $$b_i in {g_n}$$ for all $$i,$$ and if $$b_1 = g_k$$ for some $$k,$$ then $$g_1, dots, g_k$$ all appear as some $$b_i.$$ Prove $$|A|=|B|.$$

Attempt: Let $$e_i$$ be the vector with $$1$$ at position $$i$$ and $$0$$ elsewhere. If $$b_1 = g_k,$$ let $$c=(c_k, dots, c_1)$$ count how many times $$g_i$$ appears. We calculate $$f: B to A$$ as follows:

Let $$c=(c_k,dots,c_1), a=(0,dots,0).$$ While $$c ne 0,$$ let $$d_1 > dots > d_k$$ be the indices such that $$c_{d_i} ne 0.$$ Replace $$c, a$$ with $$c-(e_{d_1}+dots+e_{d_k}), a+(g_{d_1} e_1 + dots + g_{d_k} e_k)$$ respectively. After the while loop ends, let $$f(b)=a.$$

Let $$sum a, sum b, sum c$$ be the sum of the components of $$a, b, c$$ respectively. Since $$sum c$$ decreases after every loop, the algorithm terminates and $$f(b)$$ is well-defined. Since $$c_k g_k + dots + c_1 g_1 + sum a$$ does not change after every iteration, is $$sum b$$ at the start and $$sum a$$ at the end, we have $$sum f(b) = sum b = n,$$ so $$f(b)$$ is also a partition of $$n.$$ Now $$a = (g_k, dots, g_1)$$ after the first loop, which satisfies the condition $$g_i > g_{i-1}+g_{i-2}$$ since $$g_i = F_{n+2}-1 = (F_{n+1}-1)+(F_n-1)+1 > g_{i-1}+g_{i-2}.$$ Furthermore, after every iteration of the loop, the difference $$a_i – (a_{i-1}+a_{i-2})$$ changes by $$0, g_{d_j}-g_{d_{j-1}} > 0,$$ or $$g_{d_j}-(g_{d_{j-1}}+g_{d_{j-2}}) > 0,$$ so we have $$a_i > a_{i-1} + a_{i-2}$$ at the end and hence $$f(b) in A.$$ Thus, $$f: B to A$$ is well-defined.

In order to prove the injectivity of $$f,$$ it suffices to prove each loop iteration as a mapping $$(c,a) to (c’,a’)$$ is injective, which would imply the mapping $$(c,0) to (0,a)$$ that the while loop creates is injective. Indeed, if $$f(b_1) = f(b_2) = a$$ with $$(c_1, 0), (c_2, 0)$$ being sent to $$(0, f(b_1)) = (0,a), (0, f(b_2)) = (0,a)$$ respectively, then we have $$(c_1, 0) = (c_2, 0) Rightarrow c_1 = c_2 Rightarrow b_1 = b_2.$$

Suppose $$d_1 > dots > d_i, f_1 > dots > f_j$$ are the non-zero indices of $$c_1, c_2$$ respectively and $$c_1 – (e_{d_1}+dots+e_{d_i}) = c_2 – (e_{f_1}+dots+e_{f_j}), a_1+g_{d_1}e_1 + dots+ g_{d_i} e_i = a_2 + g_{f_1} e_1 + dots + g_{f_j} e_j.$$ If $$x ge 2$$ is an entry of $$c_1,$$ it decreases by $$1,$$ so the corresponding entry in $$c_2$$ after $$c_2$$ is modified is also $$x-1,$$ which means it must’ve been $$(x-1)+1 = x$$ before since $$x-1>0.$$ Thus, if the values of two positions of $$c_1, c_2$$ differ, one is $$1$$ and the other is $$0.$$ However, if $$c_1 = (1,0), a_1 = (3,1), c_2 = (0,1), a_2 = (4,1),$$ then $$(a_1, c_1), (a_2, c_2)$$ both get sent to $$((5,1), (0,0)).$$ I can rule out this specific example by arguing that one of the pairs is illegal and could not have come from any choice of initial $$c,$$ but I have no idea on how to do this in general.

What should I do next in order to show $$f$$ is injective? Furthermore, since the problem I’m trying to prove is correct, injectivity would imply $$f$$ is secretly a bijection. But I have no clue on how to even start on the surjectivity of $$f,$$ so I just constructed a similar algorithm for $$g: A to B$$ in the hopes of proving $$g$$ is injective too. If I can show $$f$$ is injective I will probably know how to show $$g$$ is.

Here is an example of $$f, g$$ in practice:

Let $$n = 41, b = (12, 7, 7, 4, 4, 2, 2, 2, 1) Rightarrow c = (1, 2, 2, 3, 1).$$

$$((1, 2, 2, 3, 1), (0,0,0,0,0)) to ((0, 1, 1, 2, 0), (12, 7, 4, 2, 1)) to ((0, 0, 0, 1, 0), (19,11,6,2,1)) to ((21,11,6,2,1),(0,0,0,0,0)),$$ so $$f(b) = (21,11,6,2,1).$$

Let $$a = (21, 11, 6, 2, 1).$$

$$((21,11,6,2,1),(0,0,0,0,0)) to ((9,4,2,0,0), (1,1,1,1,1)) to ((2,0,0,0,0),(1,2,2,2,1)) to ((0,0,0,0,0),(1,2,2,3,1)),$$ so $$g(a) = (12, 7, 7, 4, 4, 2, 2, 2, 1).$$

## Help proving one direction of an iff for a self learner in differential equations

I am self learning differential equations from the book “Differential Equations With Hostorical Applications” by George Simmons. The following problems is the one I am having an issue with:

Given the homogeneous equation $$y” + P(x)y’ + Q(x)y =0$$, and change the independent variable from $$x$$ to $$z=z(x)$$. Show that the homogeneous equation can be transformed through this change of variables into an equation with constant coefficients iff $$frac{Q’ + 2PQ}{Q^{3/2}}$$ is constant, in which case $$z = int sqrt{Q(x)}dx$$ will effect the desired result.

As of now I have solved the “only if” direction with the following math:

Let $$z = z(x)$$. We have the following:

$$frac{df}{dx} = frac{df}{dz}frac{dz}{dx} = z'(x) frac{df}{dx} Rightarrow frac{d}{dx} rightarrow z'(x)frac{d}{dz}$$

for the first derivative, and for the second derivative we have:

$$frac{d^{2}f}{dx^{2}} = frac{d}{dx}left(z'(x)frac{df}{dz}right) = frac{d}{dz}left(z'(x)frac{df}{dz}right)z'(x) = (z'(x))^{2}frac{d^{2}f}{dz^{2}} + z'(x)frac{d}{dz}left(z'(x)right) frac{df}{dz} = ldots$$
$$ldots = (z'(x))^{2}frac{d^{2}f}{dz^{2}} +z'(x)frac{d}{dx}left(z'(x)right)frac{dx}{dz} frac{df}{dz}$$

$$frac{dx}{dz} = frac{d}{dz}left(z^{-1}(z) right) = frac{1}{z'(x)}$$
$$Downarrow$$
$$frac{d^{2}f}{dx^{2}} = (z'(x))^{2}frac{d^{2}f}{dz^{2}} + z”(x)frac{df}{dz} Rightarrow frac{d^{2}}{dx^{2}} rightarrow (z'(x))^{2}frac{d^{2}}{dz^{2}} + z”(x)frac{d}{dz}$$

and all together we have the following three equations, which combined gives us the transformed differential equation.

$$y” + P(x)y’ + Q(x)y =0$$

$$frac{d}{dx} rightarrow z'(x)frac{d}{dz}$$

$$frac{d^{2}}{dx^{2}} rightarrow (z'(x))^{2}frac{d^{2}}{dz^{2}} + z”(x)frac{d}{dz}$$

$$y” + left(frac{P(z)}{z'(x)} + frac{z”(x)}{(z'(x))^{2}}right)y’ + frac{Q(z)}{(z'(x))^{2}}y = 0$$

Now suppose that $$frac{Q(z)}{(z'(x))^{2}} = c_{2}$$ and that $$left(frac{P(z)}{z'(x)} + frac{z”(x)}{(z'(x))^{2}}right) = c_{1}$$. Plug $$z'(x) = sqrt{frac{c_{2}}{Q(z)}}$$ and $$z”(x) = frac{Q'(x)}{2sqrt{c_{2}Q(x)}}$$ into the the equation for $$c_{1}$$, and we get that :

$$frac{2PQ + Q’}{Q^{3/2}} = frac{2c_{1}}{sqrt{c_{2}}} = constant$$

I have tried various things for the other direction, but I can’t seem to make any progress.

## expression simplification – Proving Properties of Boolean Algebras \$(x+y) + (ycdot x’) = x+y\$

I’m trying to justify the following simplification:
$$(x + y) + (y cdot x’) = x + y$$

The solution that was provided to me is as follows:

$$= (x + y) + (y cdot x’) – (LHS)\ = x + y cdot (1 + x’)\ = x + y cdot (1)\ = x + y\$$

I’m a little confused on the jump from steps 1 to 2, and what rule was applied and then also the switch from a disjunction to a conjunction.

Is anyone able to provide some insight into the reasoning?

## digital – Proving you were at a certain location on a certain day

If I wanted to show that I’ve visited a place (say Central Park) I can show a photo of myself in Central Park. Let’s say I wanted to show that I’ve visited Central Park for each day of the year in 2019. Then I would need 365 photos of myself in Central Park, one on each day of 2019. A quick Google search reveals a 2006 article saying that timestamps can easily be altered. Is there now a way to take photographs whose timestamps can’t easily be altered?

## turing machines – proving \$E_{TM}\$ is undecidable using the halting language

You are right, assuming $$E_{TM}in R$$ you have Turing machine $$T$$ which decides $$E_{TM}$$ and you can construct with it a Turing machine which decides $$H_{halt}$$:

If we have $$T$$ which decides $$E_{TM}$$ and suppose we want to decide whether $$M$$ halts on $$x$$.
Construct a Turing Machine $$T_{M,x}$$ which irrespective of its input $$y$$ simulates $$M$$ on input $$x$$: if the simulation halts and $$M$$ accepts or rejects $$x$$, then $$T_{M,x}$$ accepts its input, otherwise, it never halts.

You can convince yourself that if $$M$$ halts on $$x$$, then $$L(T_{M,x}) = Sigma^*$$, and if it doesn’t then $$L(T_{M,x}) = phi$$.

Now you can figure out why this acts as a decider for the Halting problem.

## real analysis – sequence characterization of adhering points: am I proving this result correctly?

To let $$X$$ to be a subset of $$textbf {R}$$, and let $$x in textbf {R}$$. Then $$x$$ is an inherent point of $$X$$ exactly when a sequence exists $$(a_ {n}) _ {n = m} ^ { infty}$$, consisting exclusively of elements in $$X$$that are converging too $$x$$.

MY ATTEMPT

Let us prove the implication $$( Leftarrow)$$ first.

Since $$a_ {n} to x$$, for each $$varepsilon> 0$$is there $$N geq m$$ so that
begin {align *} n geq N Longrightarrow | a_ {n} – x | leq varepsilon end {align *}

Since $$a_ {n} in X$$That means no matter what happens $$varepsilon> 0$$ you choose, there is one $$a_ {n} in X$$ so that $$| a_ {n} – x | leq varepsilon$$.

Consequently, $$x$$ is an inherent point of $$X$$.

Conversely, we want to prove that $$( Rightarrow)$$.

If $$x$$ is an inherent point of $$X$$for each $$varepsilon> 0$$Is there a $$a in X$$ so that $$| x – a | leq varepsilon$$.

Especially for everyone $$varepsilon = 1 / n$$there is an element $$a_ {n} in X$$ so that $$| x-a_ {n} | leq 1 / n$$.

If you take the limit, you come to the conclusion $$lim a_ {n} = x$$as requested.

## Probability – How do you generally go about proving a statement?

I struggle to start at all when it comes to proving statements, so I read a book about stochastic processes and in the beginning they offer some basic properties of probabilities, and one of them is

$$P (A ^ c) = 1 – P (A)$$

and next they ask me to prove it as an exercise, as simple as that may seem, I just don't know how to go about proving such a statement myself?

Any guidance?

## Elementary number theory – congruence: proving a set takes all these values?

I came across this question in an Olympics math book:

To let $$gcd (m, n) = 1$$, $$A = {x mid0 le x le m-1, gcd (x, m) = 1 }$$ and $$B = {x mid0 le x le n-1, gcd (x, n) = 1 }$$. If $$C = {na + mb mid a in A, b in B }$$ then prove it $$C$$ takes all values $$equiv 0 le x le mn-1$$ Modulo $$mn$$.

I have successfully proven that $$gcd (mn, na + mb) = 1$$ for all $$a, b.$$ That would imply that $$forall c in C,$$ $$0 le c le mn-1$$ Taken modulo $$mn$$.

However, I have difficulty continuing. My guess is that it has something to do with Euler's phi function, given the number of elements in $$A$$ and $$B$$ are each $$phi (m)$$ and $$phi (n)$$ and $$C$$ must accept $$phi (mn) Rightarrow phi (m) phi (n)$$ Elements.

How do I prove that this is valid for all such values?

## Complexity Theory – Helps to understand questions and how to start proving that language is unrecognizable

I looked at this problem and I don't know how to start. Could someone please lead me in the right direction?

Also help clarify what the question is. I am not 100% clear what L (M) means. Is it the language of machine M? I only saw questions that read "M is a machine that accepts all strings that end with 010", never before L (M) = {…}.

## Proving techniques – How to prove the function of a recursive Big-Oh without using repeated substitution, master phrase or closed form?

I have a function as defined $$V (j, k)$$ with two base cases with $$j, k$$ and the recursive part has an additional variable $$q$$ which it also uses. Also, $$1 leq q leq j – 1$$, The recursive part has the form: $$jk + V (q, k / 2) + V (j – q, k / 2)$$I am not allowed to use repeated substitution and I want to prove this by induction. I can not use the main clause because the recursive part is not in that form. Any ideas how I can solve it with given limitations?