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10

Setting $2^x=a,$ We have $$1+a+a^2+a^3(1+a+a^2)=3(1+a+a^2)$$ $$\iff (1+a+a^2)(a^3-2)=0$$ If $x$ is real $2^x>0\implies1+a+a^2>0$ So, we have $2=(2^x)^3\iff2^{3x-1}=1$ Now if $b^m=1$ either $m=0,b\ne0$ or $b=1$ or $b=-1,m$ is even


7

As you want to do calculations by hand, you are merely asking how to build a table of logarithms. There are at least two ways: Computing $\sqrt{10}$, then the square root of it, etc. until the result is small enough to do an approximation. That's described in Feynman lecture. Using series, for example $\ln (1+t)=t-\frac12t^2+\frac13t^3+\cdots$. There is a ...


6

We distribute just as we would distribute $$(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd.$$ $$(2x^3 - x)(\color{blue}{\sqrt x + 2x^{-1}}) = 2x^3(\color{blue}{\sqrt x + 2x^{-1}}) - x(\color{blue}{\sqrt x + 2x^{-1}})\\ =2x^3\sqrt x + 4x^3x^{-1} - x\sqrt x - 2xx^{-1}\\ = 2x^{6/2}x^{1/2} + \frac {4x^3}{x} - x^{2/2}x^{1/2} - \frac {2x}{x}\\ = 2x^{6/2 ...


6

You can factor $x^4+1 = (x^2 + \sqrt{2} x + 1) (x^2 -\sqrt{2}x + 1)$. This allows you to write the integrand as $\frac{a_1 x + b_1}{x^2 + \sqrt{2}x + 1} + \frac{a_2 x + b_2}{x^2- \sqrt{2}x + 1}$. You would then rewrite the denominator in the form of $(x-u)^2 + v$ and rewrite the numerator as $a_i (x-u) + w$, from which you can do a change of variable to ...


5

For quick-and-dirty approximations, I'm fond of the musical logarithms (writeup by Sanjoy Mahajan of work due to I. J. Good, who credits his father). Essentially this boils down to a mathematical fact: $2^{10} \approx 10^3$, and taking 120th roots, $2^{1/12} \approx 10^{1/40}$. and a "musical" fact: many rational numbers with small numerator and ...


4

The order of operations is a convention that give us what operation should we do before. In the given examples there are three operations (given in the priority):operation in parenthesis, power,multiplication and subtraction so we have $$-2^2=-4\leftarrow\text{the power has the priority before the subtraction}$$ $$(-2)\cdot(-2)=4\leftarrow\text{the ...


4

Of course it is not a coincidence. Historically, you first define $b^n$ for $n$ a positive integer, and you go from there. Thing is that with this approach you need quite an insight to define exponentiation for all positive real numbers, and to deduce the existence and properties of the exponential function. You then define the logarithm as its inverse, and ...


4

Consider $z=0 \in \mathbb{C}.$ Then $z^0$ is undefined. Since $z$ can be written in the form $re^{i\theta}$, where $r \in \mathbb{R}$ and $\theta \in (-\pi,\pi]$, we have $$z^0 =\left[re^{i\theta}\right]^0 =r^0e^{i\theta(0)}=r^0e^0=1 \cdot 1=1$$ since $\alpha^0=1 \quad$ for all $\alpha \in \mathbb{R} \setminus\{0\} .$ Alternatively, you can use the ...


3

$$\exp\left(\sum\limits_{n=0}^{+\infty}a_n X^n\right)=\exp\left({\lim\limits_{n\to +\infty}\sum\limits_{k=0}^{n}a_k X^k}\right)=\lim\limits_{n \to +\infty}\exp \left(\sum\limits_{k=0}^{n}a_k X^k\right)=\lim\limits_{n \to \infty}\prod\limits_{k=0}^{n}\exp \left(a_k X^k\right)=\\=\prod\limits_{n=0}^{+\infty}\exp \left(a_n X^n\right) $$ We changed $\lim$ and ...


3

Because $-2^2$ means that you take the number $2$, raise it to the second power ($^2$), and then you consider its additive inverse ($-$). So, $2$ raised to the second power is $4$, whose additive inverse is $-4$. This is because exponentiation has a higher priority and it is the first thing you have to do; hence you compute the power before doing anything ...


3

I think that you are just showing that, whatever $x$ could be, the $n^{th}$ derivative of $x^n$ is a constant $$\frac{d^2}{dx^2}(x^2)=2$$ $$\frac{d^3}{dx^3}(x^3)=6$$ This corresponds to the number of steps required to arrive to your constant term. With $x^4$, one more round would give you $24$ as constant. In fact $$\frac{d^n}{dx^n}(x^n)=n!$$ what you would ...


3

$2^n + 2^n = (1+1)*2^n = (2^1)*(2^n) = 2^{(1+n)}$ The properties used here were $a^1=a$ and $a^{(b+c)} = (a^b)*(a^c)$. $a^n+a^n = a^{(n+1)}$ only if $a = 2$ (or $0$, if $n>0$), but $a*a^n = a^{(n+1)}$ for all $a$ and $n$.


2

The simple answer to the second part of the question is no. For example, $3^n + 3^n \ne 3^{n+1}.$ For any positive integer $m$, however, $$\underbrace{m^n + m^n + \ldots + m^n}_{m \text{ terms}} = m(m^n) = m^{n+1}.$$


2

Note that the left side of the equation can be written as $2^0+2^x+2^{2x}+2^{3x}...2^{5x}$ This is a geometric series, with a=1, n=6, and $r=2^x$ We use the formula: $S_n = \frac {a(1-r^n)}{1-r}$ Substitute the values, you get $S=\frac{(1-2^{6x})}{1-2^x}$ We do the same for the right side $S=3[\frac{1-2^{3x}}{1-2^x}]$ Equate the terms, and ...


2

To give an approximation for at least $4$ digits in general by hand I think it is almost impossible. If you know some results from approximation theory after that you can appreciate logarithm tables. Of course the first idea is the Taylor expansion for few terms. We know that for $|x| \leq 1$ and $x \neq -1$ the series for $\ln(1+x)$ is the following. $$ ...


2

$\sum_{s=0}^{r-1} e^{\frac{-2 \pi i s k}{r}}$ equals $0$ unless $k=0$, in which case it equals $r$. So your expression is equal to $$\frac{1}{r} R_N(x^0) r = R_N(1) = 1$$


2

$0.0625^{-2.25} = \frac{1}{16}^{-\frac{9}{4}} = 16^{\frac{9}{4}} = (2^4)^{\frac{9}{4}} = 2^9 = 512$


2

0.0625^(-2.25) is $\frac{1}{16}$ raised to the $-\frac{9}{4}$. power. Now $\frac{1}{16}$ is $2^{-4}$, so $(2^{-4})^{-\frac{9}{4}}$ is the same as $2^{-4 \dot -\frac{9}{4}}$ which is $2^9$. This site is a good reference for fractional exponentiation: http://www.purplemath.com/modules/exponent5.htm


2

It is true that $-2^2$ is ambiguous (unless you know the convention), because $-(2^2) \ne (-2)^2$. (And, indeed, some calculators or programing languages may do it using their own convention, different from the mathematicians' convention.) The mathematicians' convention is $-2^2 = -(2^2)$. Why? Presumably because we often need to write $-(2^2)$. But if ...


2

Suppose you want to compute ${3.21}^{1/5}$. If you have a logarithm table (say base 10), you only need the logarithms of numbers between 0.1 and 1 stored (alternatively between 1 and 10), as many as is relevant for your precision. Then because $$\log (3.21^{1/5}) = \frac{1}{5}\left(\log(10) + \log(0.321)\right)= \frac{1}{5}\left(1+\log(0.321)\right)$$ ...


2

To compute fifth roots without logarithms, you can use Newton's method. Suppose we want $a^{1/5}$. Define $f(x) = x^5 - a$; we want to find a root of $f$. So let $x_0$ be a reasonable starting point (see Where to start), and define for $n \ge 0$ $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^5-a}{5x_n^4} = \frac{4x_n^5 - a}{5x_n^4}$$ Now ...


2

$$i^{-1}=\frac1i=\frac i{i\cdot i}=\frac i{i^2}=\frac i{-1}=-i\quad\text{QED}$$ $$i^{-3}=\frac1{i^3}=\frac{i}{i^3\cdot i}=\frac{i}{i^4}=\frac{i}1=i\quad\text{QED}$$


2

You have $b^x=e^{x\ln{b}}$, and I presume you have $b^{x+y}=b^xb^y$, so: $$\begin{aligned} b^x &=e^{x\ln{b}}\\ &=e^{\ln{b}+\ln{b}+\dotsb+\ln{b}}\\ &=e^{\ln{b}}e^{\ln{b}}\dotsm e^{\ln{b}}\\ &=b\cdot b \dotsm b \end{aligned}$$


2

Note that $$|x^3 - y^3| = |x^3 + (-y^3)| \leq |x^3| + |(-y^3)| = |x^3| + |y^3|$$ The $\leq$ part comes from the triangle inequality ($|x+y| \leq |x| + |y|$)


2

You know that: $$e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}$$ is a valid equation for every $-\infty<x<\infty$. For every $x\in \mathbb{R}$ we know that $\infty<-x^2\leq 0$ then if we replace $x$ by $-x^2$ in the equation above we get $$e^{-x^2}=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^n x^{2n}}{n!}$$ which is ...


2

The strict inequality is trickier but for $e^C x \geq0 $ to hold with any $x\geq0$ the matrix $e^C$ must have only non-negative entries. This is different from positive or negative definiteness, there is no relation in either direction. For $C$ it is sufficient to have non-negative entries off-diagonal. This is because $e^{tC}=I+tC+o(t)$, so for small ...


2

Isn't this trivial? Proof by contradiction. Suppose $q$ is rational. Then $q^n$ is also rational for all $n>1$. Since the assumption is that they are irrational, the assumption is false and $q$ must be irrational. Don't know if there is a name for this.


2

If you were to write out the number $10^{10^{56}}$, it would be a $1$ followed by $10^{56}$ zeroes. That is, $$10^{10^{56}} = 1\underbrace{000\dots000}_{\large{10^{56}\ \text{zeroes}}}$$ To give you some sense of scale, there are only about $10^{80}$ (i.e. a $1$ followed by $80$ zeroes) atoms in the observable universe. $$10^{80} = ...


1

Let's call $10^{56}$ a grxy. Then, $10^{10^{56}}$ is one with a grxy zeroes after it. That's a very large number, depending on whether you're comparing it to $1$ or $10^{10^{56}}-1$. I'm sure you know what a cell is. The number of cells in the human body is $10^{43}$. Obviously, $43\ll10^{56}$. So it's no wonder that compared to $1$ or $0$, $10^{10^{56}}$ ...



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