Expert Answers to Algebra 2 Problems

Smooth curve of genus 1 in P1C×P1C.
Show that
$$\{((x_0:x_1),(y_0:y_1)):(x_0^2+x_1^2)(y_0^2+y_1^2)=x_0{x}_1{y}_0{y}_1\}\subset <!--\; \subset \; -->{\mathbb{P}}_{\mathbb{C}}^1\times <!--\; \times \; -->{\mathbb{P}}_{\mathbb{C}}^1$$is a smooth curve of genus 1.
The author briefly talked about the concept genus in this section without assuming much knowledge in topology. One proposition said that the genus of smooth curve of degree d in ${\mathbb{P}}_{\mathbb{C}}^2$ is $(\genfrac{}{}{0ex}{}{d-<!--\; -\; -->1}{2})$.
Then in one exercise I showed that the arithmetic genus $(-<!--\; -\; -->1{)}^{\dim <!--\; \; -->X}\cdot <!--\; \cdot \; -->({\mathrm{\chi <!--\; \chi \; -->}}_X(0)-<!--\; -\; -->1)$ of a smooth curve in ${\mathbb{P}}_{\mathbb{C}}^2$ agrees with the geometric genus, where ${\mathrm{\chi <!--\; \chi \; -->}}_X$ is the Hilbert polynomial of the projective variety X. The author then stated that this can be generalized to any smooth projective curve over C.
So I think I should use that formula for this exercise. The smoothness can be shown by the Segre embedding. Now the formula says that the genus is
$$1-<!--\; -\; -->{\mathrm{\chi <!--\; \chi \; -->}}_X(0).$$
And ${\mathrm{\chi <!--\; \chi \; -->}}_X(d)=(\genfrac{}{}{0ex}{}{3+d}{3})-<!--\; -\; -->(\genfrac{}{}{0ex}{}{3+d-<!--\; -\; -->4}{3})$. But I cannot get 1. I must have made some mistake when calculating ${\mathrm{\chi <!--\; \chi \; -->}}_X(d)$. But I couldn't figure out where.

Asymptotic complexity of power of logs
I'm trying to simplify $\mathrm{\Theta <!--\; \Theta \; -->}(l{g}^k(n/2))$.
I believe it's $\mathrm{\Theta <!--\; \Theta \; -->}(l{g}^{k}n)$ but i don't know if the following proof is correct... i'd love to receive some input
I tried doing -
$$\mathrm{\Theta <!--\; \Theta \; -->}(l{g}^k(n/2))=\mathrm{\Theta <!--\; \Theta \; -->}(lg(n/2)\ast <!--\; \ast \; -->lg(n/2)\ast <!--\; \ast \; -->lg(n/2)\ast <!--\; \ast \; -->...\ast <!--\; \ast \; -->lg(n/2))=\mathrm{\Theta <!--\; \Theta \; -->}((lgn-<!--\; -\; -->lg2)\ast <!--\; \ast \; -->(lgn-<!--\; -\; -->lg2)\ast <!--\; \ast \; -->(lgn-<!--\; -\; -->lg2)\ast <!--\; \ast \; -->...\ast <!--\; \ast \; -->(lgn-<!--\; -\; -->lg2))=\mathrm{\Theta <!--\; \Theta \; -->}(\mathrm{\Theta <!--\; \Theta \; -->}(lgn{)}^k)=\mathrm{\Theta <!--\; \Theta \; -->}(l{g}^{k}n)$$

$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n\stackrel{?}{=}\sqrt{ab}$
I found this interesting equality, but I could not find a way to prove it. Any (beautiful) idea?
$$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)}^n=\sqrt{ab}$$

Right angled triangle log
If a,b and c(c is the hypotenuse) are sides of a right triangle then prove
$$({\log }_{c+b}<!--\; \; -->a)+({\log }_{c-<!--\; -\; -->b}<!--\; \; -->a)=2({\log }_{c+b}<!--\; \; -->a)\cdot <!--\; \cdot \; -->({\log }_{c-<!--\; -\; -->b}<!--\; \; -->a)$$
The bases are different so can't quite figure out what to do

Find the sum of solutions to:
$2{\log }_4^2<!--\; \; -->(|x+1|)+{\log }_4<!--\; \; -->(|x^2-<!--\; -\; -->1|)+{\log }_{\frac{1}{4}}<!--\; \; -->(|x-<!--\; -\; -->1|)=0$
I'm not sure about what to do with the absolute values, how can I get rid of them?
Should I solve for all various cases depending on the sign of $x+1$ and $x-<!--\; -\; -->1$?

Finding Stationary Points of Natural Log Function
$$f(x)=x-<!--\; -\; -->2\ln <!--\; \; -->(x^2+3)$$
I started by using the chain rule on $x^2+3$ which gives me $\frac{2x}{x^2}+3$
At this point I tried to multiple $\frac{2x}{x^2}+3$ by $x-<!--\; -\; -->2$ - is this correct? How can I then find the stationary point?
I understand that you may not wish to give me the answer on a silver platter. Perhaps somebody could kindly provide me with an example in a similar format and the steps involved in solving it.

Use the Binomial Theorem to find the coefficient of $$x^7$$ in the expansion of $$(x-<!--\; -\; -->2{)}^{11}$$ ?

Modular arithmetic polynomial question
Is it possible to find all polynomials of the form $a{n}^2+bn+c$ where a,b, and c are integers and such that
$$a+b+c\equiv <!--\; \equiv \; -->31(\mathrm{mod}54)$$
$$4a+2b+c\equiv <!--\; \equiv \; -->3(\mathrm{mod}54)$$
$$9a+3b+c\equiv <!--\; \equiv \; -->11(\mathrm{mod}54)$$

Compute the arithmetic genus of the union of the three coordinate axes $Z(x_1{x}_2,x_1{x}_3,x_2{x}_3)\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^3$
Compute the arithmetic genus of the union of the three coordinate axes $Z(x_1{x}_2,x_1{x}_3,x_2{x}_3)\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^3$
The arithmetic genus of $X\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^n$ is $g(X)=(-<!--\; -\; -->1{)}^{dim(X)}(\mathrm{\chi <!--\; \chi \; -->}(0)-<!--\; -\; -->1)$, where $\mathrm{\chi <!--\; \chi \; -->}$ is the Hilbert polynomial of X.
Let $X=Z(x_1{x}_2,x_1{x}_3,x_2{x}_3)\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^3.$
I guess that $dim(X)=1$, since the coordinate $x_0$ is free.
I showed that if X is a hypersurface of degree d in ${\mathbb{P}}^n$, then $g(X)=(\genfrac{}{}{0ex}{}{d-<!--\; -\; -->1}{n})$. I don't think that this can be applied to this problem but it's a general result that I know.
I showed also that if X is a complete intersection of two surfaces of degree a and b in ${\mathbb{P}}^3$, then $g(X)=\frac{1}{2}ab(a+b-<!--\; -\; -->4)+1$.
Is there any solution of this problem that does not involve computing the Hilbert polynomial of X? However, if computing the Hilbert polynomial is simplier, that's ok for me.

An efficient technique to test if an exponential of logs gives an integer
Is there an efficient way of testing if the resulting value of an exponential gives an integer without actually expanding the equation.
For example:
$\log <!--\; \; -->(12)-<!--\; -\; -->\log <!--\; \; -->(4)=1.09861\dots <!--\; \dots \; -->$
and is a transcendental number.
But if we take the exponential, we get an integer:
$e^{\log <!--\; \; -->(12)-<!--\; -\; -->\log <!--\; \; -->(4)}=e^{1.09861\dots <!--\; \dots \; -->}=3$
This can be easily shown by proving that the equation is equivalent to $12/3$
However is there a way to prove this while staying in the logarithm nation... We are assuming that the two numbers $12$ and $4$ are very large...
Can I find say numerical stability bounds to this simple equation:
${2.71828}^{2.48491-<!--\; -\; -->1.38629}=3\dots <!--\; \dots \; -->$
such that if I use $n$ digits, I will get an integer close the the $n$-th digit.
This may look trivial, but let's assume we do not have say $\log <!--\; \; -->(12)$, but only $2.48491\dots <!--\; \dots \; -->$ up to some precision of k-digits. Same thing for $\log <!--\; \; -->(4)$, etc.

Calculating of genus of a curve.
Let C be a curve over ${\mathbb{F}}_q$ in projective plane. So C can be done as zeroes of some gomogeneous polynomial $\in <!--\; \in \; -->{\mathbb{F}}_q[x,y,z]$ with degree n. Whether is there algorithm which is polynomial time in n that calculate arithmetic genus of C?

Primes modulo which a given quadratic equation has roots
Given a quadratic polynomial $a{x}^2+bx+c$, with a, b and c being integers, is there a characterization of all primes p for which the equation
$$a{x}^2+bx+c\equiv <!--\; \equiv \; -->0(\mathrm{mod}p)$$
has solutions?

Simplifying the exponential expression $e^{-<!--\; -\; -->4\ln <!--\; \; -->x+8\ln <!--\; \; -->y+2}$

How to prove that $f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$ approaches $\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$?
Let
$$f(x)=\underset{n=1}{\overset{{10}^x}{\sum <!--\; \sum \; -->}}\frac{1}{n}$$
I noticed that as x approaches $\mathrm{\infty <!--\; \infty \; -->}$,
$f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)\approx <!--\; \approx \; -->2.3025$. After a bit of experimenting, I found that $\mathrm{2.3025...}=\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$
How can I prove that as x approaches $\mathrm{\infty <!--\; \infty \; -->}$, $f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$ approaches $\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$?

Evaluate: $$x^2-<!--\; -\; -->y^2$$ by replacing x with $$\frac{1}{3}$$ and y with $$\frac{1}{2}$$

How to show that $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}=\ln <!--\; \; -->(n{)}^{\ln <!--\; \; -->(n)}$
I have verified that $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}=\ln <!--\; \; -->(n{)}^{\ln <!--\; \; -->(n)}$ by plugging in values for $n$, but do not understand why it is true. I am not aware of any $\log$ rules that can be used to simplify $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}$

Do the equations 4x-3y=5 and 7y+x=-8 form a system of linear equations? Explain

Determine which of the following differential equations are separable. Do not solve the equations. $$y^{\prime }=(\sin <!--\; \; -->x)(\cos <!--\; \; -->y)$$

How many digits will $5^{4^{3^2}}$ produce?
According to Wolfram Alpha, $5^{4^{3^2}}$ evaluates to an integer with $183231$ digits. How does one find out how many digits such a calculation will produce?

I have a following task of determining whether this problem is well conditioned. I tried to solve it but I am stuck because I don't know how to understand the last step.
Is a problem of calculating the root of the polynomial $p(x)=ax+b$ well conditioned for variables a,x and b?
Here is my attempt:
I calculate maximal relative error of result, denoted with U(a,x,b) where a,x,b are arguments with a relative error not greater than arithmetic precision denoted by $\mathrm{\nu <!--\; \nu \; -->}$. Here is equation with distorted data for relative error:
$$\frac{|a(1+{\mathrm{ϵ<!--\; ϵ\; -->}}_1)x(1+{\mathrm{ϵ<!--\; ϵ\; -->}}_2)+b(1+{\mathrm{ϵ<!--\; ϵ\; -->}}_3)-<!--\; -\; -->ax-<!--\; -\; -->b|}{|ax+b|}=U(a,x,b)$$
And
$$\underset{|\mathrm{ϵ<!--\; ϵ\; -->}|\le <!--\; \le \; -->\mathrm{\nu <!--\; \nu \; -->}}{sup}U(a,x,b)=\frac{|ax(2\mathrm{\nu <!--\; \nu \; -->}+{\mathrm{\nu <!--\; \nu \; -->}}^2)+b\mathrm{\nu <!--\; \nu \; -->}|}{|ax+b|}$$
$$cond(a,x,b)=\frac{U(a,x,b)}{\mathrm{\nu <!--\; \nu \; -->}}=\frac{|ax(2+\mathrm{\nu <!--\; \nu \; -->})+b|}{|ax+b|}$$
How this translates to statement that problem is well defined or not? I don't specifically understand this step.