## 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?

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## real analysis – proving a supremum of a crowd

The question:

Find the supremum of the set $${ { sqrt (4) {n ^ 4 + n ^ 3} -n: n in mathbb {N} }}$$
And then it tells us that we need to take large values ​​of n to find an appropriate guess, to show that it is an upper bound, and then to prove that it is the smallest upper bound.

I followed the question and found a suitable guess for s = 1/4, showing that this is a good upper limit. My problem is to prove that there is no lower limit. At this point, my work looks like I'm trying to prove by contradiction:

Suppose h is another upper bound such that h <1/4.
$${ sqrt (4) {n ^ 4 + n ^ 3} -n
$$n ^ 4 + n ^ 3 <(h + n) ^ 4$$

But after the expansion I can only quit $$n ^ 4$$ This gives me a lot of unknown powers and a really complicated solution that I can do by hand

$$n ^ 3

That said, I know that I have taken the wrong path, but I'm not sure which direction to prove it. I adapted a response from another book example, but that was only up to potency 2, so it was much easier to simplify this method.

## turing machines – Here's why it's wrong to stop proving the problem

Instead of solving the halting problem, I will try to solve a less complicated problem in a similar way.
Can we write a function that predicts whether two given numeric inputs are the same?
I will not create such a function, but suppose such a function can exist and we will call it H. Now we have H, a function that works and solves our problem, but we write another function and call it H +, a function that negates the result of our perfectly functioning function.

Pseudocode:

``````def H(p1, p2):
#perfectly working piece of code that will solve our problem
# returns True if p1 == p2, else returns False

def H+(p1, p2):
return not H(p1, p2)
``````

Now if we have the code, we compare p1 = 1, p2 = 2. And if we use the function H +, why not, it's the same function as H, a function that we know works. H + negates only the results of H. The result of H + is true, how can it be? we know that 1 is not equal to 2, so here we have a paradox and prove that we can not write a function to predict whether two numeric values ​​are the same.

Now for the Halting problem,
If I understand that correctly, the Halting problem has been proved in a similar way; There is a machine H that can predict if a problem is solvable. There is a larger machine that uses H, but negates the results named H +. When H + is then fed into H +, a paradox arises. Of course, H + will not work. We assume that machine H delivers the correct result. Why will H still work in the same way after changing and converting to H +?
What happens if we feed H + into H, will we still have the same paradox? I do not think so.

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## Proving \$ ({L_A} ^ * L_B) ^ + = (L_A cup L_B) ^ * L_B \$

I try to prove the following identity:
$$({L_A} ^ * L_B) ^ + = (L_A cup L_B) ^ * L_B$$

This is clearly true, as both sides will agree exactly "any sequence of A and B that ends with B" :

$$B, AB, AAB, ABBABAB, …$$

I tried to think about the quantity properties and even tried to think about the NFA constructed by the two sides of the equation but so far without success.

Does anyone know how to prove it?

## Hurricane Dorian is the second strongest storm in history proving that climate change is real. Will Donald Trump address that?

Accurate records of Atlantic hurricanes began in 1851 before the storm's strength could be surmised.

– and one can say with absolute certainty that hurricanes and the earth are older than that, historically speaking.

Since the reliable recording of tropical cyclone data in the North Atlantic in 1851 , there have been 1,574 systems with at least tropical storm intensity and 912 systems with at least hurricane intensity.

https: //en.wikipedia.org/wiki/List_of_At …

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## Documents proving that I have not used Facebook for social media communication

I was accused of posting things on Facebook, and I have not had any discussions. Do I need a lawyer to obtain legal documents to obtain this information? Many thanks

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## at.algebraic topology – Proving a Kan-like condition for functors to model categories?

I tried to prove this version of the Kan condition for a project that I think about, and I'm pretty stuck. My experience of asking questions about MO in the past was great. I hope that the (higher) category theorists here can help me!

Categories of Functors: To let $$cal {C}$$ be a (closed) model category and let $$mathcal {S}$$ be a finite pose.

The category of functors $$( mathcal {S}, mathcal {C})$$ can be permeated with that Reedy model structure (see (1)). A bunch of posets $$f: mathcal {S} to mathcal {T}$$ induces a pullback functor
$$f ^ *: ( mathcal {T}, mathcal {C}) to ( mathcal {S}, mathcal {C})$$
This functor fits into a jugular extension with the left Kan extension $$f _!$$ (see Barwick (2)).
$$f _!: ( mathcal {S}, mathcal {C}) leftrightarrow ( mathcal {T}, mathcal {C}): f ^ *$$
Besides, if $$iota: mathcal {S} to mathcal {T}$$ is the inclusion of a subheading that is closed downwards (i.e. $$s in mathcal {S}$$ and $$s & # 39; precs$$ implied $$s & # 39; in mathcal {S}$$) then

$$iota ^ *: ( mathcal {T}, mathcal {C}) to ( mathcal {S}, mathcal {C})$$

Maybe for reasons of clarity I should mention that we see $$mathcal {S}$$ as directed category, d. H. a Reedy category in the $$mathcal {S} _ + = mathcal {S}$$ and $$mathcal {S} _-$$ is the trivial subcategory for each object.

Let me set a notation for a specific subcategory of the radio category $$( mathcal {S}, mathcal {C})$$, I have not yet fully defined the definition, the conditions are mainly due to the situation in which I am.

Definition 1: The category $$text {Ch} ( mathcal {S}, mathcal {C})$$ from $$mathcal {S}$$Chains in $$mathcal {C}$$ is the full subcategory of $$( mathcal {S}, mathcal {C})$$ consisting of functors $$x: mathcal {S} to mathcal {C}$$ so that

• (on) $$x$$ is a cofibrant diagram relative to the Reedy model structure.
• (B) $$x_S to x_T$$ is a quasi-isomorphism for everyone $$S to T$$ in the $$mathcal {S}$$,

I may want to improve the above assumptions to prove the result you are looking for (sentence 1 below).

• (A & # 39) $$x$$ is cofibrant and fibrant with respect to the Reedy model structure.
• (b & # 39;) $$x_S to x_T$$ is a trivial cofibration for everyone $$S to T$$ in the $$mathcal {S}$$,

I also like to use the hypothesis $$x$$ is cofibrant in the projective model structure, which is only objectively kofibranz.

Categories of layers: A nice class of posets arises from simple complexes.

Definition 2: To let $$X$$ be a simplicial complex. The Category of layers $$X mathcal {S}$$ is the poset whose objects are simplicity $$S$$ in the $$X$$ and where there is a morphism $$S to T$$ if $$S$$ is included in $$T$$,

Clear the card $$X to X mathcal {S}$$ is functorial. Any map of simple complexes $$f: X to Y$$ induces a map of posets $$f: X mathcal {S} to X mathcal {T}$$,

Main question: The result I have tried to prove is the following horn filling feature.

Sentence 1 (?): To let $$iota: lambda ^ {n, k} to delta ^ n$$ Designate a standard inclusion of a horn $$Lambda ^ {n, k}$$ in the $$n$$-Simplex $$Delta ^ n$$, and let $$iota: lambda ^ {n, k} mathcal {S} to delta n mathcal {S}$$ also designate the induced functor for stratigraphy categories. Then the corresponding pullback functor

$$iota ^ *: text {Ch} ( Delta ^ n mathcal {S}, mathcal {C}) to text {Ch} ( Lambda ^ {n, k} mathcal {S} , mathcal {C})$$
allows a section, i. H. a functor $$sigma: text {Ch} ( Lambda ^ {n, k} mathcal {S}, mathcal {C}) to text {Ch} ( Delta ^ n mathcal {S}, mathcal {C})$$ With $$iota circ sigma = text {Id}$$,

My main question is the following.

Question: Is sentence 1 true? What if I make some of the possible changes to Definition 1 suggested above?

Ideas for the proof: Here is a sketch of the proof I had in mind.

You can extend a functor $$x in text {Ch} ( Lambda ^ {n, k} mathcal {S}, mathcal {C})$$ to a functor that $$bar {x} in ( delta ^ n mathcal {S}, mathcal {C})$$ by filling the two faces $$T_0, T_1$$ from $$Delta ^ n$$ that is missing in $$Lambda ^ {n, k}$$ with the Colimit $$text {colim} (x)$$ and the inclusions $$S to T_i$$ with the colimit maps $$x_S to x_ {T_i}$$, A map $$x to y$$ in the $$text {Ch} ( Lambda ^ {n, k} mathcal {S}, mathcal {C})$$ induces a card $$bar {x} to bar {y}$$ in an obvious way, and this defines a functor $$sigma$$ as in the sentence. We have to show that $$bar {x}$$ has properties (a) and (b) from Definition 1.

To show property (b) from Definition 1, we note that since the nerve of $$Lambda ^ {n, k} mathcal {S}$$ is the barycentric subdivision of $$Lambda ^ {n, k}$$ (contractible) and the structure maps are quasi-isomorphisms, the map $$x_S to text {colim} (x)$$ is a quasi-isomorphism.

Property (a) is the problem: The Colimit $$text {colim} (x)$$ is cofibrant because the colimit is the left Kan extension of the pullback $$Lambda ^ {n, k} mathcal {S} to *$$ and that is a left jaw next door. However, there seems to be no reason for the cofibrancy of the extended diagram. You could try cofibrant replacement, but this will ruin the property $$iota ^ * bar {x} = x$$,

I'm not sure if (a & # 39;) and / or (b & # 39;) help at all and move to the projective model structure (assuming that $$x$$ Seems to ruin the property that the colimit will be cofibrant, which is bad. Anyway, I'm stuck here.

One last remark is, if I only use the (pointwise) left Kan extension $$iota_!$$ then, as far as I can tell, the quasi-isomorphism property (b) of Definition 1 is generally not satisfied.

Many thanks: To read the long question and for any help or advice you may have!

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## Proving \$ binom {n + m} {r} = sum_ {i = 0} ^ {r} binom {n} {i} binom {m} {r – i} \$

To prove $$binom {n + m} {r} = sum_ {i = 0} ^ {r} binom {n} {i} binom {m} {r – i},$$

I have shown that equality applies to all $$n,$$ $$m = 0, 1,$$ and all $$r simply fix it $$n$$ and $$r$$ and insert $$0.1$$ to the $$m.$$ Then I go on $$m$$ (and further $$m$$ just).

But I'm not completely confident because I see two placeholders, $$n$$ and $$m.$$ Is this the case when a double induction is required (first)? $$m$$ and then on $$n$$)

Consider all fixated $$n, r geq 0$$ and the following two cases (I know that only one case is needed to complete this inductive proof).

CASE 1

begin {align} binom {n + 0} {r} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {0} {r – i} \ & = binom {n} {0} binom {0} {r} + binom {n} {1} binom {0} {r-1} + cdots + binom {n} {r} binom {0} {0} \ & = 0 + 0 + cdots + binom {n} {r} \ & = binom {n} {r} end

CASE 2

begin {align} binom {n + 1} {r} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {1} {r – i} \ & = binom {n} {0} binom {0} {r} + binom {n} {1} binom {0} {r-1} + cdots + binom {n} {r-1} binom {1} { r – (r-1)} + binom {n} {r} binom {1} {r – r} \ & = 0 + 0 + cdots + binom {n} {r-1} + binom {n} {r} \ & = binom {n} {r-1} + binom {n} {r} end

INDUCTION

Suppose it is true $$m leq k.$$ Now think $$binom {n + (k + 1)} {r}.$$ Pascal's identity follows

$$binom {n + (k + 1)} {r} = binom {n + k} {r} + binom {n + k} {r-1}$$

And,

begin {align} binom {n + k} {r} + binom {n + k} {r-1} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {k} { r – i} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k} {r – 1 – i} \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k} {r – i} + sum_ {i = 0} ^ {r-1} binom {n } {i} binom {k} {r – 1 – i} \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} bigg[binom{k}{r – i} + binom{k}{r – 1 – i}bigg] \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k + 1} {ri} \ & = sum_ {i = 0} ^ {r} binom {n} {i} binom {k + 1} {ri} end

Therefore, equality applies to $$m = k + 1$$ Since equality is for $$m = 0, 1,$$ and that if equality holds $$m = k,$$ then it applies to $$m = k + 1,$$ It follows that equality holds $$forall m in mathbb {N}.$$

## CNN's Jim Acosta has mocked him for accidentally proving that the boundary walls are working?

Sally, on border security enlightened ???

Forgive me, but unfortunately you have not received a PhD in Border Safety from Harvard. I did my doctorate in Baylor.

However, if I have a plumber problem, I go to a plumber. If I have problems with the car, I go to a car mechanic.

If I ask myself what would be helpful for border security, I go to Border Patrol and ASK YOU.

Well, Sally, why do not you tell me about the training in border security that YOU have to oppose?

,

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## Machine Learning – Square Core Proving

We have the given square kernel.

K (x, y) = (xTy) ^ 2

φ ([x1, x2]) = {x1x1, x1x2, x2x1, x2x2}

Show that K (x, y) = φ (x) T ((y) for arbitrary vectors of length n.

I can show that this works for a two-dimensional vector, but I'm confused, as proven for an n-dimensional vector.