Articles of algebra precalculus

Solve $(x+1)^n=(x-1)^n$, assuming $x$ is a complex number and $n>0$.

How do I solve $(x+1)^n=(x-1)^n$? I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$. How do I solve it using Moivre’s n-th root theorem?

Find the conectives of a composed proposition given a a true table

How can I construct a proposition P with the values given in the table using p,q, and r? \begin{array}{|c3:c|}\hline p & q & r & P \\\hline 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & […]

Roots of $ x^3-3x+1$

How can I find the roots of $x^3-3x+1$ using Cardano’s formula? So far, I found that $$x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}$$ $$x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}$$ $$x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt[3]{\frac{-1}{2}-\frac{i\sqrt{-3}}{2}}$$ I am now trying to express each cubic radicand in their exponential form. Euler’s formula : $e^{i\theta} = \cos \theta + i\sin \theta$. I […]

Number of copper atoms in $1\mbox{cm}^3$ of copper

I’m starting my physics class and I’m really rusty on my conversions and stoichiometry, The mass of a copper atom is $1.37\cdot 10^{-25}$ kg, and the density of copper is $8920 \mbox{kg/m}^3$. I would ask this in physics.se, but the problem is more math based than anything. So far, I’ve concluded that the Volume of […]

What is the domain of $x^x$

I’m trying to figure out the domain of the function $y=x^x$. When I graph it, it appears to be defined on $[0, \infty)$, but then when I plug in individual negative numbers, for some of them I get real numbers, such as $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$. However, for other numbers such as $x=2.2$, the result is imaginary. On […]

If the $m$th term of an Arithmetic Progression is $\frac{1}{n}$ and the $n$th term is…

Problem : If the $m$th term of an A.P is $\frac{1}{n}$ and the $n$th term is $\frac{1}{m}$ then prove that the sum to $mn$ terms is $\frac{mn+1}{2}$ My working : Let $a$ be the first term of the progression and $d$ the common difference then: $$\tag1T_m = \frac{1}{n}= a+(m-1)d$$ $$\tag2 T_n = \frac{1}{m} = a+(n-1)d$$ […]

The inequality $b^n – a^n < (b – a)nb^{n-1}$

I’m trying to figure out why $b^n – a^n < (b – a)nb^{n-1}$. Using just algebra, we can calculate $ (b – a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $ $ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) – (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $ $ […]

General solution to expressions, without calculating exact roots (A generalization of Newton's identities)

Consider the following equations: $$A_1^1=\sum_iy_i=y_1+y_2+\ldots+y_m=a_1$$ $$A_2^1=\sum_{i_1,i_2}y_{i_1}y_{i_2}=a_2\,\,,i_1< i_2$$ $$A_3^1=\sum_{i_1,i_2,i_3}y_{i_1}y_{i_2}y_{i_3}=a_3\,\,,i_1< i_2< i_3$$ $$\vdots$$ $$A_{m-1}^1=\sum_{i_1,\ldots,i_{m-1}}y_{i_1}\ldots y_{i_{m-1}}=a_{m-1}\,\,,i_1< \ldots< i_{m-1}$$ $$A_m^1=y_{1}\ldots y_{{m}}=a_m$$ What is the general solution to the following expressions without computing exact $y_i$’s: $$A_1^n=\sum_iy_i^n=y_1^n+y_2^n+\ldots+y_m^n=?$$ $$A_2^n=\sum_{i_1,i_2}y_{i_1}^ny_{i_2}^n=?\,\,,i_1<i_2$$ $$A_3^n=\sum_{i_1,i_2,i_3}y_{i_1}^ny_{i_2}^ny_{i_3}^n=?\,\,,i_1< i_2< i_3$$ $$\vdots$$ $$A_{m-1}^n=\sum_{i_1,\ldots,i_{m-1}}y_{i_1}^n\ldots y_{i_{m-1}}^n=?\,\,,i_1< \ldots< i_{m-1}$$ $$A_m^n=y_{1}^n\ldots y_{{m}}^n=a_m^n$$ Does anyone know a reference containing the results? As an example, $m=3$,$n=3$: $$(\sum_iy_i)^3=y_1^3+y_2^3+y_3^3+3y_1^2y_2+3y_1^2y_3+3y_2^2y_1+3y_2^2y_3+3y_3^2y_1+3y_3^2y_1+6y_1y_2y_3$$ […]

Can $\frac{a}{b} + \frac{b}{a}$ ever be bounded from above, if $a, b \in \mathbb{R}$?

Good morning everyone! By the arithmetic-geometric mean inequality, we all know that a suitable lower bound for the quantity $$\frac{a}{b} + \frac{b}{a}$$ is $2$. Now my question is: Will this quantity ever be bounded from above, if we allow $a$ and $b$ to be in $\mathbb{R}$? If the answer is NO, under what conditions on […]

Prove by induction that $2^1+2^2+2^3+2^4+ \cdots +2^n=2(2^n-1)$

Alright I have this problem, Prove by induction $2^1+2^2+2^3+2^4+ \cdots +2^n=2(2^n-1)$ Now I’ve done this so far: Base case $n=1$: $$2^1 = 2$$ $$2(2^1-1)=2(2-1)=2(1)=2 .$$ Assume for $k$, prove for $k+1$: $$ \begin{align*} 2^1+2^2+2^3+2^4+ \cdots +2^k+2^{k+1} & =2(2^k-1)+2^{k+1} \\ &=2^{k+1}+2^{k+1}-2 \end{align*} $$ Now the trouble I’m running into is that I don’t know how to […]