## real analysis – Prove there exists a sequence \$sum_{n=1}^{infty} a_n\$ = inf \$A\$

Let $$A$$ be a nonempty set of real numbers with a lower bound. Prove there exists a sequence $$(a_n)_{n=1}^{infty}$$ such that $$a_n in A$$ for all $$n$$ and $$lim_{n to infty} a_n = inf(A)$$.

I try to apply the definition of convergence but struggle to connect that to $$lim_{n to infty} a_n = inf(A)$$. Any helps would be appreciated!

## navier stokes – inf sup condition

I am considering the discrete inf-sup condition for the discrete Stokes problem

$$inf_{q in Q_h} sup_{v in V_h} frac{(q, nabla cdot v)}{| q | | nabla v|} geq beta > 0$$

This means that
$$inf_{q in mathbb{R}^p} sup_{v in mathbb{R}^n} frac{v^T B q}{| q | | v|} geq beta > 0$$

Can a do the following argumentation:

$$inf_{q in mathbb{R}^p} sup_{v in mathbb{R}^n} frac{v^T B q}{| q | | v|} = inf_{q in mathbb{R}^p, |q | = 1 } sup_{v in mathbb{R}^n, | v | = 1} v^T B q = inf_{q in mathbb{R}^p, |q | = 1 } |B q|$$

Is this correct? Or am I missing something?

I would be very grateful about any help.

Best,
frieder

## \$ S: = Big { left (m + n right) ^ { frac {1} {mn}}: ; m, n in mathbb N Big } inf S, sup S \$

If available, you will find $$sup S, inf S$$
$$S: = Big { left (m + n right) ^ { frac {1} {mn}}: ; m, n in mathbb N Big }$$
$$m = 1, n = 1: ; left (m + n right) ^ { frac {1} {mn}} = 2$$

For a firm $$n in mathbb N, m in infty ;$$:

$$displaystyle lim_ {m bis infty} (m + n) ^ { frac {1} {mn}} = lim_ {m bis infty} Bigm (m left (1+ frac { n} {m} right) Big) ^ { frac {1} {m} cdot frac {1} {n}} = lim_ {m to infty} left ( sqrt (m) {m} left (1+ frac {n} {m} right) ^ { frac {1} {m} right) ^ { frac {1} {n}}$$= lim_ {m bis infty} sqrt (n) { sqrt (m) {m}} lim_ {m bis infty} left (1+ frac {n} {m} right) ^ { frac {1} {mn}} = e \$\$

$$f (m): = left (1+ frac {1} {mn} right) ^ { frac {1} {mn}}$$

$$m & # 39;> m implies f (m & # 39;)> f (m)$$
$$m + n geq 2 (mn) ^ { frac {1} {2}} Big / ^ { frac {1} {mn}}$$
$$(m + n) ^ { frac {1} {mn}} geq sqrt { sqrt (mn) {mn}} ; sqrt (mn) {2}> 1$$
Then I concluded: $$f (x) in langle 1, e rangle implies inf S = 1, sup S = e$$
Is that correct?

## complex analysis – \$ f \$ is completely without zeros, then \$ r> 0 \$ m (r) = inf left { left | f (z) right |: left | z right | = r right } \$, \$ m \$ does not increase

$$f$$ is whole function without zeros, then for $$r> 0$$ $$m (r) = inf left { left | f (z) right |: left | z right | = r right }$$. $$m$$ does not increase the function.

I have an idea how to solve this problem, but I encountered a problem.

If $$f$$ is constant, then it's trivial.

If $$f$$ is not constant with $$f (z) neq 0$$ for each $$z in mathbb {C}$$, and $$f$$ is holomorphic, then we define
$$g (r): = frac {1} {m (r)}$$ then
$$= frac {1} { inf left { left | f (z) right |: left | z right | = r right }}$$
$$= sup left { frac {1} {f (z)}: left | z right | = r right }$$

because $$left { frac {1} {f (z)}: left | z right | = r right }$$ is closed and limited $$g$$ takes its maximum values ​​at the limit.

The question is: I assumed that $$g$$ increases with $$r$$and of it $$g$$ is an increasing function, but unfortunately I can't find any right arguments for this statement and I would like if someone could explain this?

## How can I prove that inf {\$ sqrt[n]{a}: n in Bbb {N} \$ *} = 1

My question is the title itself.

inf {$$sqrt [n] {a}: n in Bbb {N}$$*} $$= 1$$ when $$a> 1$$

Even if someone does not want to tell me the answer, a guide would be helpful.
Thanks.

## Algorithms – Create a priority search tree to determine the number of points in the range [-inf, qx] X [qy, qy’] from a series of points sorted by y-coordinates

A priority search tree can be created for a set of points P in time O (n log (n)). However, if the points are sorted by the y-coordinates, it takes O (n). I find algorithms for constructing the tree if the points are not sorted.

I found a way to do this as follows:

1. Construct a BST on the points.
Since the points are sorted, it will take O (n) time

2. Min-heap the BST on the x-coordinates
This will take theta (n) time

The total time complexity will be O (n).

Is this a valid approach to creating a priority search tree in O (n) time?

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## Functional Analysis – Calculate \$ inf_ {a, b, c in mathbb {R}} int _ {- 1} ^ {1} left | x ^ {3} -ax ^ {2} -b xc right | ^ {2} dx \$

calculation $$inf limits_ {a, b, c in mathbb {R}} , int _ {- 1} ^ {1} left | x ^ {3} -ax ^ {2} -b xc right | ^ {2} mathrm dx$$

I'm new to Hilbert space, I see similar questions with the formula: $$langle f, g rangle = int f g , mathrm d mu$$ (real $$L ^ {2}$$), but I have problems transforming the problem into the formula: what is the systematic way to find it $$f$$ and $$g$$? What role does orthogonality play in the question?

## nt.number theory – approximation for \$ inf {r> 0, (n-r, n + r) in mathbb {P} ^ {2} } \$ by minimizing a distance

Under Goldbach's guess $$r_ {0} (n): = inf {r> 0, (n-r, n + r) in mathbb {P} ^ {2} }$$ and $$k_ {0} (n): = pi (n + r_ {0} (n)) – pi (n-r_ {0} (n))$$, The PNT implies that one can assume $$dfrac {2r_ {0} (n) -1} {k_ {0} (n)} sim log n$$,

As for everyone $$n> 1$$ one has $$tau (n) geq 2$$The equality occurs then $$n$$ is prime, a very good approximation (actually a narrow upper bound) of $$k_ {0} (n)$$ is given by the function $$S_ {r_ {0} (n)} (n)$$ from where $$S_ {r} (n): = left ( sum_ {m = nr} ^ {n + r} dfrac {2 ^ m} { tau (m) ^ m} right) – frac {1 } {2}$$,

Can you prove that? $$r_ {0} (n)$$ is the positive integer $$r$$ that minimizes the amount $$vert S_ {r} (n) – dfrac {2r-1} { log n} vert$$? If so, you can get an upper limit for $$r_ {0} (n)$$ in terms of $$n$$?