## actual analysis – Test \$ sum_ {k-1} ^ { infty} frac {1} {k + a} sqrt { frac {a} {k}}

$$forall a ge 0$$,Try out $$sum_ {k-1} ^ { infty} frac {1} {k + a} sqrt { frac {a} {k}} < pi$$

The book hints at how to convert the sum to integration:$$sum_ {k-1} ^ { infty} frac {1} {(k + a) sqrt {k}} < int_ {0} ^ { infty} frac {dx} {(x + a) sqrt {x}}$$, then I can use integration. But I can't prove it.

## The test explains the algebraic integer in \$ mathbb {Q} ( sqrt {d}) \$

I have a question about the test in the book Elements of Number Theory (Stillwell). The idea is to test what number an algebraic integer can be in $$mathbb {Q} ( sqrt {d})$$.
My question is:
Why is this $$2b$$ under test (red color in the image)?

## asymptotic: why does \$ frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}} \$ refute the lower limit \$ O (n ^ {3- delta}) \$?

I have a simple question:

It is conjectured that the shortest path of all pairs (APSP) does not have $$O (n ^ {3- delta)}$$time algorithm for any $$delta> 0$$ by SETH.

also

there is a result that says that APSP can be resolved in time $$frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}}$$ by Ryan Williams.

But, this improvement does not refute the guesswork.

So what I did is this: I compare between $$lim_ {n -> infty} frac {( frac {n ^ 3} {2 ^ { sqrt { log n}}})} {n ^ {3- delta}} = 0$$ So, $$frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}}$$ is better than the other, so why doesn't that mean we refute the conjecture?

When I have this function: $$frac {n ^ 3} {2 ^ { Omega ( sqrt { log n})}}$$I did not know how to compare it with others, since Big Omega is only a part, how to compare in general with other functions when you have this?

## Why does the solution set of the equation \$ sqrt {x-1} (x ^ 2-1) \$ not include \$ -1 \$?

Why the solution set of the equation? $$sqrt {x-1} (x ^ 2-1)$$ does not include $$-1$$. If I drink $$x = -1$$ I get $$2i times 0$$ which is the same $$0$$ and therefore $$-1$$ the solution should also be considered (along with $$+ 1$$)

## calculation – Evaluate \$ lim_ {n rightarrow infty} sqrt {2 ^ n} int_a ^ b cos ^ n (x + frac { pi} {4}) dx \$

Evaluate
$$lim_ {n rightarrow infty} sqrt {2 ^ n} int_a ^ b cos ^ n (x + frac { pi} {4}) dx$$
Knowing that the integration set $$(a, b)$$ is included in $$(0, frac { pi} {2})$$

Now, my attempt was to first notice that the function that we have to integrate is:$$cos ^ n (x + frac { pi} {4})$$ and I was thinking that I can use this trigonometric identity:$$cos (a + b) = cos (a) cos (b) + sin (a) sin (b)$$, after which I can apply the reduction formula to $$cos ^ n (x)$$. Is this the way to start?

## What is the polar form of \$ z = frac { sqrt (3) + i} { sqrt (3i)} \$?

How to write the complex number $$z = frac { sqrt (3) + i} { sqrt (3i)}$$ in polar form?

## Upper limit \$ int_0 ^ {+ infty} exp left (-a (-b + sqrt {b ^ 2 + cx}) ^ 2 right) dx \$

Is there an upper limit depending on $$to$$, $$b$$ Y $$c$$ (what positive constants) for the next intergral
begin {align} int_0 ^ {+ infty} exp left (-a (-b + sqrt {b ^ 2 + cx}) ^ 2 right) dx. end {align}

## Can a real form (1,1) \$ phi \$ be represented by \$ u sqrt {-1} partial bar { partial} u \$ in the Kahler variety?

Let M be a 2-dimensional dimension (complex dimension) K "{a} ler and $$phi$$ be a real $$(1,1)$$ to form. Is it possible that there is a function? $$u$$ such that $$phi = u sqrt {-1} partial bar { partial} u$$

## calculation – Evaluate \$ lim limits_ {x a 0, x> 0} frac { int limits_0 ^ {sinx} sqrt {tant} dt} { int limits_0 ^ {tanx} sqrt {synt} dt} \$

Solve: $$l = lim limits_ {x to 0, x> 0} frac { int limits_0 ^ {sinx} sqrt {tant} dt} { int limits_0 ^ {tanx} sqrt {synt} dt}$$
So the main problem I have is evaluating $$int limits_0 ^ {tanx} sqrt {synt} dt$$ I was looking to see exactly how to deal with this integral, but it seems that this integral requires college level skills (which I don't possess). This problem is from a high school textbook, so I thought there is some trick that can be applied without having to know about college. The main problem is with this integral because I have managed to solve the first one. Also, it looks like the limit itself will be a problem once you deal with the second integral (what I've tried to do with the second integral is substitute u =$$sqrt {synt}$$ followed by $$sinx = u ^ 2$$ => cosxdx = 2u du => $$dx = frac {2u} { sqrt {1-u ^ 4}} du$$ but I'm still getting stuck).

## real analysis: where am I wrong when integrating \$ int_ {0} ^ {1} { sqrt x} e ^ x dx \$ per substitution?

$$int_ {0} ^ {1} { sqrt x} e ^ x , dx$$

My attempt:

I tried the replacement $${ sqrt x} = t$$ and then I got $$2 int_ {0} ^ {1} te ^ { t ^ 2} , dt$$ So again I replaced $${ t ^ 2} = u$$ and eventually $$2 int_ {0} ^ {1} frac { sqrt u} {2 { sqrt u}} e ^ u , du$$