Expert Answers to Algebra 2 Problems

Why is this function a really good asymptotic for $\exp <!--\; \; -->(x)\sqrt{x}$
$f(x)=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n{x}^n{a}_n=\frac{1}{\mathrm{\Gamma <!--\; \Gamma \; -->}(n+0.5)}$
Why is this entire function a really good asymptotic for $\exp <!--\; \; -->(x)\sqrt{x}$, where for large positive numbers, $f(x)\exp <!--\; \; -->(-<!--\; -\; -->x)\approx <!--\; \approx \; -->\sqrt{x}$?
As |x| gets larger, the error term is asymptotically $f(x)-<!--\; -\; -->\exp <!--\; \; -->(x)\sqrt{x}\approx <!--\; \approx \; -->\frac{1}{x\cdot <!--\; \cdot \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(-<!--\; -\; -->0.5)}$, and the error term for $f(x)\exp <!--\; \; -->(-<!--\; -\; -->x)-<!--\; -\; -->\sqrt{x}\approx <!--\; \approx \; -->\frac{\exp <!--\; \; -->(-<!--\; -\; -->x)}{x\cdot <!--\; \cdot \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(-<!--\; -\; -->0.5)}$. If we treat $f(x)$ as an infinite Laurent series, than it does not converge.
I stumbled upon the result, using numerical approximations, so I can't really explain the equation for the $a_n$ coefficients, other than it appears to be the numerical limit of a pseudo Cauchy integral for the $a_n$ coefficients as the circle for the Cauchy integral path gets larger. I suspect the formula has been seen before, and can be generated by some other technique. By definition, for any entire function $f(x)$, we have for any value of real r:
$a_n=\oint <!--\; \oint \; -->x^{-<!--\; -\; -->n}f(x)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}(r{e}^{-<!--\; -\; -->ix}{)}^{-<!--\; -\; -->n}f(r{e}^{ix}))\mathrm{d}x$
The conjecture is that this is an equivalent definition for $a_n$, where $f(x)\mapsto <!--\; \mapsto \; -->\exp <!--\; \; -->(x)\sqrt{x}$ and $x\mapsto <!--\; \mapsto \; -->r{e}^{ix}$
$a_n=\underset{r\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}(r{e}^{-<!--\; -\; -->ix}{)}^{-<!--\; -\; -->n}\exp <!--\; \; -->(r{e}^{ix})\sqrt{r{e}^{ix}})\mathrm{d}x=\frac{1}{\mathrm{\Gamma <!--\; \Gamma \; -->}(n+0.5)}$

Having trouble understanding why the $r$-th mean tends to the geometric mean as $r$ tends to zero
I am having trouble understanding the proof of Theorem 3 in "Inequalities" by Hardy, Littlewood and Pólya. This theorem states that the $r$-th mean approaches the geometric mean as $r$ approaches zero.
I have seen the following post which makes things a little clearer (albeit using $o(r)$ instead of $O(r^2)$:
Why is the 0th power mean defined to be the geometric mean?
However, I still cannot determine why:
(a): $a^r=1+r.log(a)+o(r)$ as $r$ tends to zero,
and
(b): $\underset{r\to <!--\; \to \; -->0}{lim}(1+rx+o(r){)}^{1/r}=e^x$
I have a pretty solid grasp of limits, as well as the log and exp functions, but I have never really been taught anything substantial on big/little-O notation, in particular as the variable approaches zero. Could somebody point me towards a suitable proof of (a) and (b) above please.

How did Napier rounded his logarithms?
How did Napier round his logarithms? Wikipedia says:
By repeated subtractions Napier calculated $(1-{10}^{-7}{)}^L$ for $L$ ranging from 1 to 100. The result for $L=100$ is approximately $0.99999=1-{10}^{-5}$. Napier then calculated the products of these numbers with ${10}^7(1-{10}^{-5}{)}^L$ from 1 to 50, and did similarly with $0.9998\approx <!--\; \approx \; -->(1-{10}^{-5}{)}^{20}$ and $0.9\approx <!--\; \approx \; -->0.99520$
What was his rule for rounding? May one say that he was computing in the ring
$\mathbb{Z}[x]/(10x-<!--\; -\; -->1,x^8)$?

Solving pre logarithmic problems
How can I solve $2^{6x}$ when $8^x=3$?

Choose the correct term from the list to complete each sentence. complex numbers, complex roots, discriminant, extraneous, Factor Theorem, Integral Root Theorem, lower bound, polynomial function, quadratic equation, Quadratic Formula, radical equation, synthetic division, zero. A ______ is a special polynomial equation with a degree of two.

Solve:
$$3n^2-<!--\; -\; -->15n+18$$

For the following linear equation 5x + 4y = 20, find the x and y intecepts

Find the equation of a hyperbola satisfying the given conditions.
Asymptotes y=4/3x, y=-4/3x, one vertex (3,0)

12.00ft per sec to m per min

Divide polynomial by x + 2 using synthetic division.
$$2x^3+5x^2-<!--\; -\; -->x-<!--\; -\; -->5$$

How many 4 digit pin numbers can you make, consecutive repeats not allowed

Analyze the Complex Function by using the Principal log Branch
I am trying to analyze the function $\sqrt{1-<!--\; -\; -->z^2}$, where the square root function is defined by the principal branch of the log function. I want to locate the the discontinuities.
I know the discontinuities will lie on the negative real axis but I cannot figure out for which values of $z$ this will occur. I first tried rewriting the function as $(1-<!--\; -\; -->e^{2i\mathrm{\theta <!--\; \theta \; -->}})$. But this seems to be getting me nowhere. Any thoughts?

Evaluate the limit of $\ln <!--\; \; -->(\cos <!--\; \; -->2x)/\ln <!--\; \; -->(\cos <!--\; \; -->3x)$ as $x\to <!--\; \to \; -->0$
Evaluate Limits
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{\ln <!--\; \; -->(\cos <!--\; \; -->(2x))}{\ln <!--\; \; -->(\cos <!--\; \; -->(3x))}$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
$\begin{array}{rl}\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{\ln <!--\; \; -->(\cos <!--\; \; -->\left(2x\right))}{\ln <!--\; \; -->(\cos <!--\; \; -->\left(3x\right))}\right)& \stackrel{LHR}{=}\\ & =\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{\ln <!--\; \; -->(\cos <!--\; \; -->\left(2x\right))}{\ln <!--\; \; -->(\cos <!--\; \; -->\left(3x\right))}\right)& \\ & =\underset{x\to <!--\; \to \; -->0}{lim}\frac{{(\ln <!--\; \; -->(\cos <!--\; \; -->\left(2x\right)))}^{\prime }}{{(\ln <!--\; \; -->(\cos <!--\; \; -->\left(3x\right)))}^{\prime }}\\ & =\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{-<!--\; -\; -->\frac{2\sin <!--\; \; -->\left(2x\right)}{\cos <!--\; \; -->\left(2x\right)}}{-<!--\; -\; -->\frac{3\sin <!--\; \; -->\left(3x\right)}{\cos <!--\; \; -->\left(3x\right)}}\right)\\ & =\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{2\sin <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)}{3\cos <!--\; \; -->\left(2x\right)\sin <!--\; \; -->\left(3x\right)}\right)\\ & \stackrel{LHR}{=}\underset{x\to <!--\; \to \; -->0}{lim}\frac{{(2\sin <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right))}^{\prime }}{{(3\cos <!--\; \; -->\left(2x\right)\sin <!--\; \; -->\left(3x\right))}^{\prime }}\\ & =\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{4\cos <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)-<!--\; -\; -->6\sin <!--\; \; -->\left(3x\right)\sin <!--\; \; -->\left(2x\right)}{9\cos <!--\; \; -->\left(3x\right)\cos <!--\; \; -->\left(2x\right)-<!--\; -\; -->6\sin <!--\; \; -->\left(2x\right)\sin <!--\; \; -->\left(3x\right)}\right)\\ & =\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{2(2\cos <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)-<!--\; -\; -->3\sin <!--\; \; -->\left(3x\right)\sin <!--\; \; -->\left(2x\right))}{3(3\cos <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)-<!--\; -\; -->2\sin <!--\; \; -->\left(3x\right)\sin <!--\; \; -->\left(2x\right))}\right)\\ & =\frac{\underset{x\to <!--\; \to \; -->0}{lim}\left(2(2\cos <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)-<!--\; -\; -->3\sin <!--\; \; -->\left(3x\right)\sin <!--\; \; -->\left(2x\right))\right)}{\underset{x\to <!--\; \to \; -->0}{lim}\left(3(3\cos <!--\; \; -->\left(2x\right)\cos <!--\; \; -->\left(3x\right)-<!--\; -\; -->2\sin <!--\; \; -->\left(3x\right)\sin <!--\; \; -->\left(2x\right))\right)}={\displaystyle \frac{4}{9}}\end{array}$
Could we do it in others ways?

What are the asymptotes of ${\displaystyle y=\frac{x^2}{x^2-1}}$ and how do you graph the function?

Is this summation solvable? $S_n=\underset{i=2}{\overset{n}{\sum <!--\; \sum \; -->}}{\log }_i<!--\; \; -->(n)$
Is it possible to solve a summation with a variable base of log?
$S_n=\underset{i=2}{\overset{n}{\sum <!--\; \sum \; -->}}{\log }_i<!--\; \; -->(n)$
Should I use the derivative of ${\log }_i<!--\; \; -->(n)$

How do you graph ${\displaystyle f\left(x\right)=\frac{x^2}{x+1}}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

What are the asymptotes for ${\displaystyle y=-\frac{4}{x+2}}$ and how do you graph the function?

How do you graph ${\displaystyle f\left(x\right)=\frac{x^2-1}{x}}$ using holes, vertical and horizontal asymptotes, x and y intercepts?

How do you find the zeros of the function ${\displaystyle f\left(x\right)=\frac{x^2-x-12}{x^2+2x-35}}$?

How do you graph ${\displaystyle f\left(x\right)=\frac{x^2-x-2}{x^2-2x+1}}$ using holes, vertical and horizontal asymptotes, x and y intercepts?