## inequality – When can we deduce \$(mA+nB+pC)>(mX+nY+pZ)\$ from \$(A+B+C)>(X+Y+Z)\$?

I am trying to prove an equality in the form $$(mA+nB+pC)>(mX+nY+pZ)$$.
I can prove the inequality $$(A+B+C)>(X+Y+Z)$$ and I wonder if there is any condition,
under which we can deduce $$(mA+nB+pC)>(mX+nY+pZ)$$ from $$(A+B+C)>(X+Y+Z)$$ ?

And $$A,B,C,X,Y,Z,m,n,p$$ are all positive.

The first thing comes to mind is the Cauchy-Schwarz Inequality:

$$sqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}geleft(mX+nY+pZright)$$

Now it suffices to prove that

$$left(mA+nB+pCright)gesqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}$$
$$iffleft(mA+nB+pCright)^{2}geleft(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)$$

But I can’t find the condition under which we can deduce the above inequality from $$(A+B+C)>(X+Y+Z)$$.

Or using Hölder’s inequality:

$$left(mA+nB+pCright)^{frac{1}{2}}left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{frac{1}{2}}ge A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}$$
$$iffleft(mA+nB+pCright)^{frac{1}{2}}geleft(A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}right)left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{-frac{1}{2}}$$

And so on.

## How can I deduce an app from a notification icon?

I’m concerned about an app on my child’s phone. I saw an app icon in the notification area that looked strange (Andy Android head, slightly tilted as its icon).

How can I find out what app it is from just having seen that icon in the notification bar?

## dnd 5e – What should players roll to “deduce” a solution?

The answer is either Intelligence or Wisdom, but which will depend on what kind of process you are trying to represent. You aren’t describing deduction, but rather induction, which is an important part of understanding which Ability to use.

## Intelligence

The PHB describes intelligence (pg. 177) as:

… mental acuity, accuracy of recall, and the ability to reason.

If you are trying to represent the character’s ability to recall what the lock or key looked like, or to logically determine a relationship between them, then Intelligence is the Ability to use.

Intelligence is almost the perfect answer for the question as you posed it. Deduction is almost definitely an Intelligence based activity without exception. Except that you aren’t describing deduction, but rather induction.

Deduction may look something like, “All simple, aged copper locks can only be opened by simple, aged copper keys. I want to open a simple, aged copper lock, therefore I must require a simple, aged copper key.” However, deduction requires assumptions that are absolutely true. Is it really true that all key-lock combinations share traits like this? Why can’t an iron key open a copper lock? For example, suppose that a replacement key were created at some point. It may use different materials and styles.

So the kind of inference you are looking for is probably not deduction, and may not even be Intelligence based.

## But Really Wisdom

The PHB describes Wisdom (pg. 178) as:

… how attuned you are to the world around you and represents perceptiveness and intuition.

If you want to represent whether a character has noticed something, or is able to reach a conclusion based on intuition rather than reason, then Wisdom is the way to go.

The logical process you are describing is inductive, and a much better match for Wisdom. An inductive process may look something like, “So far, all the locks I have seen are opened by keys made from the same material and in the same style as the lock they open. Therefore, I think that all locks are opened by keys in the same material and style as the lock.”

To me, this sounds like what you are trying to describe. The character is trying to intuit a relationship between locks and keys based on their own experience and observation. Wisdom is a much better fit than intelligence.

Now, if the player still has a problem determining that they should use a simple, aged copper key for their simple, aged copper lock an Intelligence check would be appropriate, because this has become a deductive problem.

## c ++ – How to deduce the type of array in a template?

I write a simple model that returns an array that would be a product of multiplying the values ​​of two arrays sent by parameters.
How can I deduce the type of table I want to return?

For the sake of simplicity, I have specified the size of the two tables because I am only curious to know how to declare this third table.

I could only do it like this:

``````template
S *New_Array(T *tab, S *tab2){

static double tab3(5){};

for (int i = 0; i < 5; i ++){
*(tab3 + i) = *(tab +i) * *(tab2 + i);
}

return tab3;
}
``````

but this type of template is not very useful ...

I thought I might have had to use something like:

``````template
auto New_Array(T *tab, S *tab2) -> decltype(*tab * *tab2){}
``````

but I guess that only gives me a return in the form of (in this case) double, am I right?

In addition, the same problem with the declaration of a new array occurs in main where I want to assign a return value from the function template.
I have therefore tried to use decltype:

``````int tab() = {1, 2, 3, 4, 5};
double tab2() = {1.5, 2.3, 3.6, 7.8, 9.0};

typedef decltype(*tab * *tab2) MYARR;
MYARR tab3(5){};
``````

but that did not work out as I thought and now I'm stuck with my horrible model with no idea of ​​what to do.

Sincere friendships,

## Number Theory – Is it possible to deduce from perfect odd numbers convolution sums involving divisor functions or other arithmetic functions?

Divide and use certain identities of (1) I have deduced the following facts, see also the remarks below. After these introductory paragraphs, to motivate our question, I ask if we can deduce useful information for perfect odd numbers $$n$$ of the literature evaluating the sums of convolutions.

Made. A) if $$n$$ is an odd perfect number of the form $$36 M + 9$$ then
is satisfied with the following identity $$4 sigma left ( frac {2n} {3} right) = 3 sigma left ( frac {2n} {9} right) +2 (n + sigma (n)) tag {1}$$ and in addition also the congruence $$sigma_3 (n) equiv 18 text {mod} 36$$ is holding. B) If our odd perfect number $$n$$ has the form $$12 million + 1$$ then satisfied $$3 sigma left ( frac {n-1} {12} right) + sigma left ( frac {3 (n-1)} {4} right) = 4 sigma left ( frac { sigma (n) -2} {8} right), tag {2}$$
and also congruences $$sigma_3 (n) equiv 6 text {mod} 12$$ and $$sigma_5 (n) equiv 2 text {mod} 12$$ hold.

Notes on previous facts. My belief is that the opposite direction of $$(1)$$and the other equation $$(2)$$ are true, but I can not obtain that these prove that each equation is a characterization of an odd perfect number of the given form. I've tried working by invoking Euler's theorem to get perfect odd numbers $$n = 2 ^ { alpha} 3 ^ { beta} m$$ such as $$(2, m) = (3, m) = 1$$ by case, to be deducted from the declaration $$(1)$$ a contradiction, but if my calculations are correct, it's a failure: case 1 ($$beta = 2$$ and $$alpha = 0$$) returns $$sigma (m) = frac {18} {13} m$$; case 2 ($$beta> 2$$ and $$alpha = 0$$) will be $$(3 ^ { beta +1} -1) sigma (m) = 4 cdot 3 ^ beta m$$; case 3 ($$beta = 2$$ and $$alpha geq 1$$) was $$sigma (m) = frac {2 ^ { alpha + 1} cdot 9m} {13}$$ and finally case 4 ($$beta> 2$$ and $$alpha geq 1$$) returns $$sigma (m) = frac {2 ^ { alpha + 2} cdot 3 ^ { beta}} {3 ^ { beta + 1} -1}$$. And I can not find any counter-examples, I can not find any integers $$n equiv 0 text {mod} 9$$ satisfactory $$(1)$$ and I can not find any integer $$n equiv 1 text {mod} 12$$ and as $$8 mid (-2+ sigma (n))$$satisfactory $$(2)$$. (If my congruences are rights) I have not used a lot of specific information about perfect odd numbers to infer these congruences.

Question. Is it possible to deduce assertions for perfect odd numbers, assuming they exist, using convolution sums with divisor functions $$sigma_k (n)$$ or other arithmetic functions? I speak of congruences, of convolution sums analysis in the case of a specialization related to perfect odd numbers, of useful identities, or of congruences or specializations having applications in the study of perfect odd numbers. Thank you so much.

If you want to follow the approach in my fact, your answer is welcome: try to compute special equations for perfect odd numbers or to study similar congruences for different divider functions. $$sigma_ {k} (n) equiv a text {mod} b.$$ It was my attempt / method of gathering information about odd perfect numbers, extracted from convolution sums involving arithmetic functions.

I add here the references that I used to deduce my facts, I used a specialization of an identity that is shown in the second paragraph of page 255 of (1), I also add the reference for the Touchard's theorem..

## References:

1) James G. Huard, Zhiming M. Ou, Blair K. Spearman and Kenneth S. Williams, Basic evaluation of some convolution sums involving divider functions, Theory of numbers for the millennium II, A. K Peters (2002).

(2) J. Touchard, On prime numbers and perfect numbers, Scripta Math. flight 19pp. 35-39 (1953).

## Aggressive Geometry – How to deduce the next map between the tangent spaces of Zariski is surjective?

Let $$f: X rightarrow Y$$ to be a morphism of diagrams: $$X$$, $$Y$$ are regular diets, let $$Z_1, Z_2$$ to be two regular closed sub-themes of $$X$$, let $$x in X times_Y Z_2$$ such as $$y = f (x) in Z_1 cap Z_2$$.
Then I would like to know how I can deduce that the map
$$T_xX rightarrow (T_yY / T_yZ_2) otimes_ {k (y)} k (x)$$
is surjective knowing that

1) The image of $$T_ {x} (X times_Y Z_1)$$ generates $$T_yY / T_yZ_2$$

2) $$T_x (X times_Y Z_1) rightarrow T_y Z_1 otimes_ {k (y)} k (x)$$ is surjective.

This is supposed to flow from these things without seeing how … I would appreciate any explanation about it. Thank you!

## polynomials – How to deduce Hensel's lemma from the theorem of factorization of master?

In the book p-adic differential equations by Kiran.S.Kedlaya we find the exercise II. (3):

Yes $$F$$ is a complete non-Archimedean field, $$P (X) in mathfrak o_F[X]$$and the reduction of $$P (X)$$ in $$k_F$$ factors like $$overline Q cdot overline R$$with $$overline Q$$, $$overline R$$ co-prime, then there is a single factorization $$P = QR$$ in $$mathfrak o_F[X]$$ with $$Q, R$$ lifting $$overline Q, , overline R$$.

Suspicion says

Since $$overline Q$$, $$overline R$$ are co-prime, there are $$overline S, overline T in k_F[X]$$ with $$overline Q overline S + overline R overline T = 1$$, $$deg ( overline S) < deg ( overline R)$$ and $$deg ( overline T) < deg ( overline Q)$$. Use levees from these to satisfy the conditions of the main factorization theorem (indicated at the end of the question).

Attempt:

We have to choose three elements $$a, b, c in mathfrak o_F[X]$$ and three subgroups of complete additives $$U, V, W$$ satisfying the conditions. Since we want to factorize $$P$$, we will choose $$c = P$$. Then choose arbitrary elevators $$Q, R, S, T$$ and let $$a = Q$$ and $$b = R$$. Since we will have $$ab-c in W$$and we only know that $$ab-c in left$$, we then choose $$W = left subseteq mathfrak o_F[X]$$.

Then leave $$U = left$$ and $$V = left$$. As $$QS + RT equiv1 pmod p$$it follows that $$f (u, v): = av = bu$$ is a surjection of $$U times V$$ at $$W$$.

It remains to show that there is $$lambda> 0$$ such as $$left | f (u, v) right | geq lambda cdot text {max} left { left | a right | left | v right |, left | b right | left | u right | right }$$ and $$left | ab-c right | < lambda ^ 2 left | c right |.$$

But I do not know how to find such $$lambda$$, except of course that $$lambda leq1$$. And I do not see how to apply the degree restrictions to $$overline Q$$ and $$overline R$$.

I've thought about varying the elevators $$Q, R$$ by a few multiples of $$p$$ so we can choose $$lambda = 1$$, but in vain. And I am short of ideas now.

Master factorization theorem:

Let $$R$$ to be a non-archimedean ring. Suppose that there are three elements $$a, b, c R$$ and three additive subgroups $$U, V, W subseteq R$$ such as

1. L & # 39; space $$U, V$$ are complete under the standard and $$U cdot V subseteq W$$.
2. The map $$f (u, v): = av + bu$$ is a surjection of $$U times V$$ at $$W$$.
3. It exists $$lambda> 0$$ such as $$left | f (u, v) right | geq lambda cdot text {max} left { left | a right | left | v right |, left | b right | left | u right | right }, , forall u in U, v in V.$$
4. We have $$ab-c in W$$ and $$left | ab-c right | < lambda ^ 2 left | c right |$$.

Then there is a unique pair $$(x, y) in U times V$$ such as
$$c = (a + x) (b + y), quad left | x right | < lambda left | a right |, left | x right | < lambda left | a right |.$$
We also have
$$left | x right | leq lambda ^ {- 1} left | ab-c right | left | b right | ^ {- 1} quad left | y right | leq lambda ^ {- 1} left | ab-c right | left | a right | ^ {- 1}.$$

Any help or reference is sincerely appreciated. Thank you in advance.

## Algorithms – Using Chebyshev to Deduce an Upper Limit of the Coupon Problem Paste & # 39; s

I am a class and I have trouble solving an exercise.

Let $$X$$ a RV defined as the number of trials required to collect at least one coupon of each coupon type (of which $$n$$). then $$E[X] = nH_n = n ln (n) + O (n)$$ and $$sigma ^ 2_X = n ^ 2 sum_ {i = 1} ^ n frac {1} {i ^ 2} – nH_n$$ with $$lim_ {n to infty} frac { sigma ^ 2_X} {n ^ 2} = frac { pi ^ 2} {6}$$. The exercise asks how Chebyshev inequality can be used to increase the probability that $$X> beta n ln (n)$$ for some people $$beta> 1$$.

I have tried this:
begin {align *} Pr[X>beta n ln(n)] & leq Pr[|X-mu_X| geq (beta – 1)n ln n] \ & leq Pr left[|X-mu_X| geq (beta – 1) left( frac{sqrt 6}{pi} sigma_X right) ln nright] (1) \ & leq frac { pi} {6 ( beta -1) ^ 2 ln ^ 2 n} = O left ( frac {1} { ln ^ 2n} right) end {align *}

where (1) follows because $$n> frac { sigma_X sqrt 6} { pi}$$. However, I believe that the first inequality is incorrect because I reject the $$– O (n)$$ of $$mu_X$$ by deriving an upper limit.

I've also tried something else:
begin {align *} Pr[X>beta n ln(n)] & leq Pr[|X-mu_X| geq beta n ln n – nH_n] \ & leq Pr[|X-mu_X| geq (beta -1)nH_n] \ & = Pr left[|X-mu_X| geq frac{6(beta – 1)H_n}{pi^2n} left(n^2 sum_{i=1}^infty frac{1}{i^2} – n H_n right)right] end {align *}

but I'm not sure where to go from here. Thank you for any help!

## Deduce pdf (X) of E[X] And Var[X]?

"Let X be a random variable with E[X]= 6 and Var[X]= 4 … What is the probability distribution function for X?

Not sure how to solve this one. Thank you.

## combinatoire – How to deduce the number of array subsequences equal to the number of all array combinations?

If I have a size chart `not`. I want the number of all subsequences of it (contains empty).

For example: `[1, 2, 3]` All subsequences are `[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]`So the number of all subsequences is `8`.

Using dynamic programming, we can easily deduce: just choose `1` or not, the result should multiply `2`, So chosen `2` or not, multiply `2`… We will deduce therefore:

• `dp[n] = d[n - 1] * 2`
• `dp[0] = 1`

So the result is $$2 ^ {n}$$

Another approach is the combination. The result should be:

$$sum_ {i = 0} ^ {n} C ^ {i} _ {n}$$

In a nutshell, how to deduce:

$$sum_ {i = 0} ^ {n} C ^ {i} _ {n} = 2 ^ {n}$$