16

I my opinion I think that when you know $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i} ~~~~~~~~~~~~~ \cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$ you can derive all the circular trigonometric identities. If you add the "Wick" transformation $ x \to ix$ then you will step into the hyperbolic world, with all the consequent identities. $$\sin(ix) = \frac{e^{-x} - e^{...


10

Yes; in fact they can all be traced to the same point, and it is a point that you brought up in your question. One important thing to note is that in Math there can be many equivalent definitions of concepts that can lead to each other, for instance one can define $\pi$ as the ratio of circumference to diameter; or area to radius squared. Both are equivalent....


8

Any reciprocal of a function that grows slower than linear. E. g. a much slower decay rate than hyperbolic would be $$\frac{1}{\ln(x)}$$ or even $$\frac{1}{\ln(\ln(x))}$$


4

$$f(x) = \frac{1}{x}.$$ Reciprical square root, $$g(x)=\frac{1}{\sqrt{x}} $$ or $x^{-p}$ for any $0<p<1$: $$h_p(x)=\frac{1}{x^p}. $$ These are much slower than $\frac{1}{x}$ as $x \to \infty$, since $\large{\frac{f(x)}{h_p(x)} = \frac{\frac{1}{x}}{\frac{1}{x^p}} = \frac{x^p}{x} =}$ $x^{p-1}\to 0 $ as $ x \to \infty\ $ because $p-1<0. $


2

$$L=\lim_{x \to \infty} x(a^{1/x}-1)$$ let $x=1/t$, then $$L=\lim_{t\to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{1+t\ln a+(t\ln a)^2/2+...-1}{t} =\ln a.$$


2

$\frac{dy}{dx} \propto y \Rightarrow \frac{dy}{dx} = ky \ $ ($k$ is a constant) Now you're provided with 2 arbitrary points $(x_1,y_1)$ and $(x_2,y_2)$. So, $\begin{align}&\int_{y_1}^{y_2}\frac{dy}{y} = k\int_{x_1}^{x_2} dx \\\Rightarrow \ & \ln\frac{y_2}{y_1} = k(x_2- x_1) \text{ or } \boxed{x_1 - x_2 = \frac1k\ln\frac{y_1}{y_2}}\end{align}$ Now, ...


2

HINT: $a^{x_1}=a^{x_2}a^{x_1-x_2}$


2

If this all started because you were squaring the series $\sum_n\tfrac{1}{n!}$, you probably want to see how we show the exponential series satisfies $\exp(x+y)=\exp x\exp y$. It follows by the binomial theorem:$$\begin{align}\exp(x)\exp(y)&=\sum_{k,\,l\ge0}\frac{x^ky^l}{k!l!}\\&=\sum_{k,\,l\ge0}\frac{\binom{k+l}{k}x^ky^l}{(k+l)!}\\&\stackrel{\...


2

Hint: $$\lim_{n\to\infty}\left(1-\dfrac xn+\dfrac x{n^2}\right)^n =\left(\lim_{n\to\infty}\left(1+\dfrac{x(1-n)}{n^2}\right)^{\dfrac{n^2}{x(1-n)}}\right)^{\lim_{n\to\infty}\dfrac{nx(1-n)}{n^2}}$$ Now in the inner limit set $\dfrac{x(1-n)}{n^2}=y\implies y\to0$ and $\lim_{n\to\infty}\dfrac{nx(1-n)}{n^2}=x\lim_{n\to\infty}\left(\dfrac1n-1\right)=?$


1

Notice that $ \left(\forall y\in\left(0,1\right)\right) $ : $$ y\leq-\ln{\left(1-y\right)}\leq\frac{y}{1-y} $$ That can be proven either using the main value theorem,, or studiing some functions. Thus, setting $ y_{n}=\frac{x}{n}-\frac{x}{n^{2}} $, where $ x\in\mathbb{R} $, and $ n\in\mathbb{N}^{*} $ being bigger enough so that $ y_{n}\in\left(0,1\right) $. ...


1

From $x(2a-3)=\log_2(b)$ (let call this equation (*)) If $2a-3\ne0$ then you can solve this equation $x=\frac{\log_2b}{2a-3}$ and the solution is UNIQUE. So $2a-3$ HAVE TO be zero if you want more than one solution. BUT in this case, (*) can be written as $0=\log_2b$. So, the ONLY possibility for your initial equation to have more than one solution is that b$...


1

Here's an answer to a slightly more general limit $ \displaystyle L(m, n) = \lim_{x \to \infty} x^{m} ( a^{\frac{1}{x^{n}}} -1) \tag*{} $ Where $ a>1 $ and $(m, n) $ are natural numbers. $ \displaystyle L(m, n) = \lim_{x \to \infty} x^{m} ( e^{ \frac{\ln(a)}{x^n}} -1) = \lim_{x \to \infty} \sum_{r=1}^{\infty} \frac{x^{m-rn}}{r!}(\ln(a))^{r} \tag*{} $ ...


1

Recall : $\lim_{y \rightarrow 0} \dfrac{e^y-1}{y}=1;$ Set $y:=1/x,$ and consider $y \rightarrow 0^+$. $\lim_{y \rightarrow 0^+}\log a (\dfrac{e^{y\log a-1}}{y\log a}) = \log a. $


1

One of the most important properties of logarithms is that $$ \log a+ \log b = \log ab \, . $$ In fact, if we try to solve the functional equation $$ f(a)+f(b) = f(ab) \tag{*}\label{*} \, , $$ then the solution is $f(x)=c\log x$, where $c$ is an arbitrary constant and $\log$ is the natural logarithm. Now consider the task of finding the area under the curve ...


1

In this answer I define the natural logarithm as $$ \log x =\int_{1}^{x}\frac{1}{t} \, dt \, . $$ and the exponential function as the inverse of the logarithm. We begin with the fact that $$ \frac{x}{n} \geq \log\left(1+\frac{x}{n}\right) \, . $$ which can be deduced geometrically: $\log\left(1+\frac{x}{n}\right)$ is the area of the region bounded by the ...


1

Here is an alternative approach: if we first define the natural logarithm as $$ \log x = \int_{1}^{x}\frac{1}{t} \, dt $$ and the exponential function as the inverse of $\log$, then it is not much work to show that $e^x$ is its own derivative. (I go into more detail about this here.) Then, the limit $$ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n $$ can ...


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