Calculus – Find $ int frac {1} { sin sqrt {x}} dx $ – Mathematics Stack Exchange

Find the integral:
$$
I = int frac {1} { sin sqrt {x}} dx
$$

Well, here's what I've done so far:
$$
begin {align}
& sqrt {x} = t Rightarrow dt = frac {dx} {2 sqrt {x}}, dx = 2tdt \
& I = 2 int frac {t} { sin t} dt = ?
end {align}
$$

The problem is to calculate the last integral. $ t $ in the meter it makes it rather unsolvable.

Time complexity – Big-O comparison of two algorithms $ sqrt {n} $ and $ 2 ^ { sqrt { log _ {2} n}} $

I have given 2 algorithms with their temporal complexity $ t_a (n) = sqrt {n} $ and $ t_b (n) = 2 ^ { sqrt { log _ {2} n}} $ and I have to show $ t_b (n) = O (t_a (n)) $,

I have created a program to check this statement and it seems that this is for everyone $ c> 0, forall n geq16 $ it applies, but I don't know how to prove it formally because I can't find a simplification for it $ t_b $,

I know I have to prove $ exists c
: forall n geq N: t_b (n) leq c * t_a (n) $
with big-o notation.

A hint / solution would be really great.

complex analysis – branch points of $ f (z) = frac { sqrt {z} log (z)} {(1 + z) ^ 2} $

How do you go about finding the branch points / holomorphic branches of a multifunction that is composed of several other multifunction? Here is an example of what I mean:

To let $ f (z) = ( frac { sqrt {z} log (z)} {(1 + z) ^ 2}) $ be a multifunction. Identify the branch points and find a holomorphic branch.

I have no idea how to go about it, though $ sqrt {z} $ and $ log (z) $ are themselves considered multi-functions. Is anyone able to help me? Thanks a lot!

limits – Discussion on the selection of the optimal method for the search for: $ lim_ {x to 0} frac { sqrt { cos2x} cdot e ^ {2x ^ 2} -1} { ln {(1 + 2x) cdot ln {(1 + 2 arcsin {x})}} $

Does the L & # 39; Hospital rule pay off at all when calculating:
$$ displaystyle lim_ {x to 0} frac { sqrt { cos2x} cdot
e ^ {2x ^ 2} -1} { ln {(1 + 2x) cdot ln {(1 + 2 arcsin {x})}} $$

I posted this question on Quora and didn't expect the answers to be answered when I found the limit myself, but to eliminate all bad options. We did not go through this formally The rule of L & # 39; HospitalI tried to avoid it because it seemed like a cliché.
I also examined the commutativity between continuous functions and boundaries.
Since $ x to $ 0I thought maybe I could replace it $ x $ by $ frac {1} {y} $ when $ y to infty $, But that wasn't helpful either.
I was not sure which terms I could replace with another function because they are not identical. For example $ sqrt { cos {2x}} ; & ; e ^ {2x ^ 2} $,

I have examined an answer from Paramarand Singh (prefer his approach):

Solve without the rule of L & # 39; Hopital: $ lim_ {x to0} frac { sqrt { cosh {(3x ^ 2)}} cdot e ^ {4x ^ 3} -1} {x ^ 2 tan (2x)} $

At some point I realized that my attempts at manipulation were all inadequate.

My question: How do you choose the functional substitutes? And any other suggestions / opinions about whether L & # 39; pays off?

We haven't been through so much, which is troubling, and now we all have to work alone (I'm not complaining, it's great and motivating in most cases, but sometimes it's pretty difficult without proper literature)
Steppan Konoplev Solution:

To let $ f (x), g (x) $ be the numerator or denominator. Since
$$ cos x approx. 1 – frac {x ^ 2} {2}, sqrt {1 + x} approx. 1+ frac {x} {2},
e ^ x approx. 1 + x, $$
we have:$$ f (x) approximately sqrt {1-2x ^ 2} (1 + 2x ^ 2) – 1
approx (1-x ^ 2) (1 + 2x ^ 2) -1 = x ^ 2-2x ^ 4 ; text {around} ; x = 0 $$

My note: $ e ^ {2x ^ 2} approx. 1 + 2x ^ 2 ; $?

On the other hand: $ arcsin x approximately x, ln (1 + x) approximately x ; $so we
to have:$$ g (x) approximately ln (1 + 2x) ^ 2 approximately 4x ^ 2 ; text {around} ; x = 0 $$
Consequently: $$ frac {f (x)} {g (x)} =
frac {x ^ 2 + O (x ^ 4)} {4x ^ 2 + O (x ^ 4)} to frac {1} {4} ; text {as} ; x to 0. $$

I also read there:

The simplified form of the numerator is simple, but the denominator is
chaotic if you write out all the terms in detail. It seems so
would be easier to use the first terms of power expansions.

Complexity Theory – Assume that NP $ = $ DTIME ($ 2 ^ { sqrt {n}} $), prove that DTIME ($ 2 ^ { sqrt {n}} $) = DTIME ($ 2 ^ {n} $)

I've tried using the padding argument to prove something (as it appeared in Arora's book), but I'm not sure how this technique will help me here. I am trying to contradict the time hierarchy theorem.

After accepting it, I want to prove it $ NP = DTIME (2 ^ {n}) $,

The first case $ NP subset DTIME (2 ^ {n}) $is trivial.

In the second case, leave $ L in DTIME (2 ^ {n}) $ and $ M $ be deterministic $ TM $ that can decide, let it $ L_ {pad} = { langle x, 1 ^ {2 ^ { sqrt {| x |}}} rangle: x in L } $, I'm not sure how $ L_ {pad} $I can conclude that $ NP = DTIME (2 ^ {n}) $I have the feeling that something is missing and that my method is not pointing in the right direction or that there is no additional detail.

Number theory – Prove that $ sqrt {p_n} not in mathbb {Q} ( sqrt {p_1}, ldots, sqrt {p_ {n-1}}) $

To let $ p_1, ldots, p_n $ be a sequence of different prime numbers $> 0 $, How can you prove that? $ sqrt {p_n} not in mathbb {Q} ( sqrt {p_1}, ldots, sqrt {p_ {n-1}}) $, I tried to prove that it was the principle of mathematical induction, but I couldn't. I also couldn't complete a proof by contradiction. I am grateful if someone helps me to solve this.

Thank you in advance.

Numerical solution $ f (x) = sqrt {x ^ 2 + 1} -1 $ – mathematical stack exchange

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Ring Theory – Let $ I: = (3,1+ sqrt {5} i) $ and $ R: = mathbb {Z}[sqrt{5}i]$. Show that $ I $ is not maximal.

To let $ I: = (3,1+ sqrt {5} i) $ and $ R: = mathbb {Z} ( sqrt {5} i) $, Show that $ I $ is not maximal. My goal is to show that $ R / I $ is a field. So far I have the following:

begin {align *}
mathbb {Z} ( sqrt {5} i) / (3,1+ sqrt {5} i) & cong ( mathbb {Z} (X) / (x ^ 2 + 5)) / (3 , 1 + X, X ^ 2 + 5) / (X ^ 2 + 5)) \
& = mathbb {Z} (X) / (3,1 + X, X ^ 2 + 5) \
& = mathbb {Z} / 3 mathbb {Z} (X) / (1 + X, X ^ 2 + 2).
end {align *}

I can not see how you reduce $ mathbb {Z} / 3 mathbb {Z} (X) / (1 + X, X ^ 2 + 5) $ further. Ideally, if I could write $ X ^ 2 + 2 $ as a multiple of $ 1 + X $I would be ready.

Calculus – Compute $ lim_ {x to 0 ^ {-}} frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x} $

I tried to rate

begin {equation *}
lim_ {x to 0 ^ {-}} frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x}.
end {equation *}

We have
begin {equation *}
frac { sqrt {x ^ 2 + x + 4} – 2} { ln (1 + x + x ^ 2) – x} = frac {x ^ 2 + x} { ln (1 + x + x ^ 2) – x} cdot frac {1} { sqrt {x ^ 2 + x + 4} + 2} quad text {for all} x in mathbb {R},
end {equation *}

So I think my practice is boiled down to that
begin {equation *}
lim_ {x to 0 ^ {-}} frac {x ^ 2 + x} { ln (1 + x + x ^ 2) – x},
end {equation *}

and this limit is the same $ – infty $ from De L & # 39; Hôpitals set.

Can I rate this limit without De L & # 39; Hôpital's theorem?

Calculus – I'm trying to prove the derivation of $ sqrt {x} $ with geometry.

I try to prove the derivation from $ sqrt {x} $ with geometry.

So far I have created a square with area $ x $ and side lengths $ sqrt {x} $,

The derivative of the function is $ frac {d sqrt {x}} {dx} $ With $ dx $ the enlargement of the area.

I have set up the equations for the change in f,

$ df = 2 ( sqrt {x}) ( text {} d sqrt {x}) text {} + text {} d sqrt {x} text {} d sqrt {x} $

$ df = 2 ( sqrt {x}) ( text {} d sqrt {x}) text {} + text {} dx $

I should end with $ frac {df} {dx} $ = $ frac {1} {2 sqrt {x}} $

What I get in the end is $ frac {df} {dx} $ = $ frac {1} {1-2 sqrt {x}} $

What am I doing wrong?

Thank you very much