## \$ varphi in C ^ {2} ( mathbb {R}) cap L ^ 1 ( mathbb {R}) \$, so that \$ varphi & # 39; & # 39; leq \$ 0

I have to find a function $$varphi in C ^ {2} ( mathbb {R}) cap L ^ 1 ( mathbb {R})$$ so that
$$varphi & # 39; & # 39; (x) leq 0 qquad forall x in mathbb {R}$$

## ag.algebraic geometry – \$ text {Can we say} varphi (x) = f (x), f (x) in x mathcal {O} _K[[x]]]? \$

I studied the sentence in David Eisenbud's Commutative Algebra book $$7.16$$ regarding cards from Power Series rings. I particularly like the conclusion $$7.17$$The statement reads as follows:

Conclusion 7.17. If $$f in x R ((x))$$ be a power series. If $$varphi$$ is the endomorphism $$varphi: R ((x)) to R ((x)), x mapsto f,$$ Which is it identity on $$R$$ and sends $$x$$ to $$f$$, then $$varphi$$ is isomorphism iff $$f & (39)$$ is a unit in $$R$$.

This is generally for $$g (x) in R ((x))$$,we have $$begin {equation} varphi (g (x)) = f (g (x)), longrightarrow (1) end {equation}$$

Now if we take the ring $$R = mathbb {Z}$$ of whole numbers $$varphi: mathbb {Z} ((x)) to mathbb {Z} ((x)), x mapsto f$$, can be thought $$varphi: mathbb {Z} to mathbb {Z}$$ by $$varphi (n) = f (n) = n, n in mathbb {Z}, f (x) in x mathbb {Z} ((x))$$.
One can also take into account $$varphi: mathbb {Q} to mathbb {Q}$$ by $$varphi (n) = f (n) = n, n in mathbb {Q}, f (x) in x mathbb {Q} ((x))$$.

My question: To let $$K$$ to be a finite extension of $$mathbb {Q}$$ with ring of integers $$mathcal {O} _K$$ and take the algebraic degree $$bar {K}$$ from $$K$$. If we take the isomorphime or the automorphism $$varphi in text {Gal} ( bar {K} / K)$$, $$text {can we say} varphi (x) = f (x), f (x) in x mathcal {O} _K ((x)) ?$$ Can you give me a possible idea?

Approach: If I take $$g (x) = x$$, a term power series in $$(1)$$, we get $$varphi (x) = f (x)$$. Is it like this ?? because $$varphi$$ must be part of it $$text {Gal} ( bar { mathbb {Q}} / mathbb {Q})$$ and also $$varphi: K to K$$ should be identity from the lemma, also because of homomorphism. But I do not know when $$varphi: bar K to K$$.

In any case, this is not a conclusion to the question above.

Please help me with the above question, you can take on further requirements if necessary.

## Solutions for \$ prod_ {n = 1} ^ infty left (1+ frac1 {f (n)} right) = varphi \$

The constant $$varphi = frac {1 + sqrt5} {2}$$ seems to appear everywhere. I wonder what non-trivial functions $$f$$ fulfill
$$prod_ {n = 1} ^ infty left (1+ frac1 {f (n)} right) = varphi$$
I was surprised when I couldn't find any solutions.

## PDF of \$ z = exp (j varphi) \$, where \$ varphi sim mathcal {U}[-a, +a]\$

How do I find the PDF from $$z = exp (j varphi)$$, Where $$varphi sim mathcal {U} (- a, + a)$$. $$d. H.$$an evenly distributed r.v.

My difficulty with this is that the numbers are complex and I don't know how to deal with them.

## Find an injecting group homomorphism \$ varphi: D_4 rightarrow Sym_4 \$. Also: is \$ (D_4, circ) \$ Abelic?

To let $$D_4$$ be the dihedral group (of the place) and $$circ$$ be the composition of the transformations.

Exercise:

1. is $$(D_4, circ)$$ Abelianer?
2. Find an injecting group homomorphism $$varphi: D_4 rightarrow Sym_4$$ and determine if $$varphi$$ is an isomorphism.

Already done:

1. Show that $$(D_4, circ)$$ is a group.
2. Find $$| D_4 |$$,

Can someone help? Thank you in advance!

## Algebraic Geometry – Is \$ varphi: operatorname {Spec} A _ { mathfrak o} to operatorname {Spec} A \$ injective and open to the underlying topological spaces?

To let $$A$$ be a commutative ring with unity, $$mathfrak p$$ a main ideal of $$A$$, then $$varphi: operatorname {Spec} A _ { mathfrak p} to operatorname {Spec} A$$ induced by the canonical homomorphism $$A to A _ { mathfrak p}$$ injective and open to the underlying topological spaces?

## elementary number theory – Show \$ gcd (a, b) = 1 \$ implies \$ varphi (a cdot b) = phi (a) cdot phi (b) \$

This means that if $$gcd (m, n) = 1$$, then $$φ (mn) = φ (m) φ (n)$$, (Evidence: let $$A, B, C$$ are the sets of nonnegative integers that are respectively too and less than coprime $$m, n$$, and $$Mn$$; then there is a bijection between $$A × B$$ and $$C$$, according to the Chinese remainder theorem.)

I also saw this on the wiki page of Euler's Totient function, but I had no idea$$dots$$

My experiments:

After FTA we have:
$$a = p_1 ^ { alpha_1} cdots p_n ^ { alpha_n}$$
$$b = q_1 ^ { beta_1} cdots q_m ^ { beta_m}$$
Since $$gcd (a, b) = 1$$, to have $$p_i neq q_j$$, Where $$i in (1, n), j in (1, m)$$implies:
$$a cdot b = p_1 ^ { alpha_1} cdots p_n ^ { alpha_n} cdot q_1 ^ { beta_1} cdots q_m ^ { beta_m}$$
From that:
begin {align} & ~~~~~~ varphi (a cdot b) \ & = varphi (p_1 ^ { alpha_1} cdots p_n ^ { alpha_n} cdot q_1 ^ { beta_1} cdots q_m ^ { beta_m}) tag * {(1)} \ & = a cdot b (1 frac {1} {p_1}) cdots (1 frac {1} {p_n}) (1 frac {1} {q_1}) cdots (1 frac {1} {q_m}) tag * {(2)} \ & = a (1- frac {1} {p_1}) cdots (1- frac {1} {p_n}) cdot b (1- frac {1} {q_1}) cdots (1- frac {1} {q_m}) tag * {(3)} \ & = varphi (p_1 ^ { alpha_1} cdots p_n ^ { alpha_n}) cdot varphi (q_1 ^ { beta_1} cdots q_m ^ { beta_m}) tag * {(4)} \ & = varphi (a) cdot varphi (b) tag * {(5)} end

Is this proof valid since I have seen the proof of Euler's product formula?$$($$used on step $$(2))$$ It seems like I would use this feature too. If I then use Euler's product formula to prove this property, it seems a bit circular, or are there other approaches $$?$$

## nt.number theory – Reason why the equation \$ operatorname {rad} ( varphi (ab) (a + b)) = 30 \$ has many solutions, where \$ varphi (n) \$ is the total function of the Euler

We denote whole numbers $$m> 1$$ share the product of different primes $$m$$ as $$operatorname {rad} (m) = prod_ { substack {p mid m \ p text {prime}}} p,$$
with the definition $$operatorname {rad} (1) = 1$$ (see you want the Wikipedia Radical of an integer). And we call the Euler's Totient function as $$varphi (m)$$,

I am studying a talk about the abc conjecture of YouTube. What about the solutions for integers? $$a, b> 1$$ and $$gcd (a, b) = 1$$ the equation
$$operatorname {rad} ( varphi (ab) (a + b)) = 30. tag {1}$$

I've done a Pari / GP program that shows this equation $$(1)$$ under the assumption $$gcd (a, b) = 1$$ and the additional requirement that I have added is that our integers are greater than $$1$$has many solutions.

Question. Can you specify a resonance / heuristic for which the equation is easily recognizable? $$(1)$$Does it have many solutions under the given circumstances? Many thanks.

So I ask for the simple consideration that hopes our problem has many solutions, perhaps an infinite number of solutions. I do not know if cheap thinking is easy to get.

Here I add an example of my calculation.

Example. Take, for example $$a = 999$$ and $$b = 601$$ then
$$operatorname {rad} ( varphi (999) varphi (601) cdot1600) = operatorname {rad} (622080000) = operatorname {rad} (2 ^ {12} cdot3 ^ 5 cdot5 ^ 4 ) = 30,$$

## Could I apply the main clause if my \$ N / b \$ \$ varphi (N) \$ is?

To let

$$T (N) = begin {cases} 1 & text {if} N = 1 \ T ( varphi (N)) + lg ( varphi (N)) ^ 3 & text {else} end {cases}$$

from where $$varphi (N)$$ is the deadly function of Euler.

Can I somehow express that? $$varphi (N)$$ as $$N / b$$so that I can apply the main theorem and solve this repetition?

You can assume $$varphi (N) = (p-1) (q-1)$$if it's that much easier. You can also assume if it helps that $$p$$. $$q$$ are safe primes, that is, $$p = 2p & # 39; + 1$$ and $$q = 2q & # 39; + 1$$, (Suppose anything that makes the problem easier, for example, you can replace the function $$lg ^ 3 ( varphi (N))$$ with everyone else who makes the problem easier, but only as a last resort.)

## nt.number theory – Can one make an interesting statement about just perfect numbers from the equation \$ 1 / operatorname {rad} (n) = 1 / 2-2 varphi (n) / sigma (n) \$?

It is well known that the problem with even numbers is to prove or disprove if there are an infinite number of them. A few weeks ago I wrote the following supposition where $$varphi (n)$$ denotes the Euler's dead-ended function, $$sigma (n) = sum_ {1 leq d mid} d$$ the sum of the divisor function and $$operatorname {rad} (n) = prod_ { substack {p midn \ p text {prime}}} p$$
is the product of different primes $$n> 1$$ with the definition $$operatorname {rad} (1) = 1$$see the Wikipedia radical of an integer.

The Euler's totalient function and the sum of the divisor function are found in terms of equivalences to the Riemann hypothesis, and the so-called radical of an integer is the famous arithmetic function found in the formulation of the abc conjecture.

Guess. An integer $$n geq 1$$ is just a perfect number if and only if
$$operatorname {rad} (n) = frac {1} { frac {1} {2} -2 frac { varphi (n)} { sigma (n)}}. tag {1}$$

I quoted this assumption a few days ago in MSE. My intention is to know if it is possible to get a statement about the problem, which also concerns perfect numbers, if there are infinitely many of them, using the equation, or even if one can argue that the Equation seems $$(1)$$ is not useful for this purpose.

Question. Can this equation make an interesting statement about the infinity of even perfect numbers or a fact about their distribution? $$(1)$$ or call up the previous one guess (It's easy to prove that even perfect numbers $$n$$ satisfy, but my proof for the other part of the conjecture has failed)? If you believe that disability is not possible, please explain. Many thanks.

You can refer to statements about even perfect numbers and tools or assumptions from analytic number theory (we can look for these statements and read from the literature). I hope this is a nice exercise for this site, in any case I hope for comments.