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17

Okay, so you (hopefully) know how to take the derivative of something like $\tan x$, or $$\frac{d}{dx}(\tan x)=\sec^2x$$ well what if I gave you something like $$\frac{d}{dx}(\tan(x^2))$$ Now we have a separate function embedded within our first function. The chain rule says do the derivative like normal (just treat the $x^2$ like an $x$), then multiply ...


6

First, we note that $$\begin{align} \phi(x) &= \lim_{n\to \infty} \frac{x^n+2}{x^n+1}\\\\ &=\begin{cases} 1,\,\,\text{for}\,\,x>1\\\\ \frac32 \,\,\text{for}\,\,x=1\\\\ 2,\,\,\text{for}\,\,0\le x<1\\\\ \end{cases} \end{align}$$ Next, let's form the difference quotients for $f$ at $x=1$. Thus, for $h>0$, we have $$\begin{align} ...


5

Mathematicians often like to compose functions. That is $(f\circ g)(x) = f(g(x))$, as you will often see it. As a more illustrative example of composition we will define two functions: $f:\mathbb{R} \to \mathbb{R}:x \mapsto \frac{1}{x}$ and $g:\mathbb{R}\to\mathbb{R}:x\mapsto x+1$. Using these functions, we get: $(f\circ g)(x) = f(g(x)) = f(x+1) = ...


5

Your example certainly suffices, but I don't think you'll have any luck working with the expression $|p_n(\frac{1}n)-e^n|$ since, for appropriate choice of $p_n$, we can make that $0$ for all $n$ (even if it approximates every other point terribly). The important thing to note is that $p(x)$ is unbounded, whereas all polynomials are bounded. Therefore, for ...


5

This is probably not the type of function you are used to dealing with: $f(x)=(-1)^x=e^{i(2n+1)\pi x}$, for any integer $n$. So this function is actually multi valued, and most of the time, this function takes on complex values. This is an interesting (and often vexing) property of some functions in the complex plane. For more information about the ...


4

Specifically, you have that $$\frac{3x+2}{x^2 - x - 2} = \frac{A}{x-2} + \frac{B}{x+1}$$ The usual technique reveals that $A = \frac{8}{3}$ and $B = \frac{1}{3}$ Hence $$ \begin{align}\frac{x^3}{x^2 - x - 2} &= x + 1 + {\color{blue}{\frac{3x+2}{(x-2)(x+1)}}} \\ \\ &= x + 1 + \frac{8}{3(x-2)} + \frac{1}{3(x+1)} \end{align}$$ So that ...


4

Here is an alternate solution: We introduce $$ I(s) = \int_{0}^{\infty} \left( \frac{1}{e^x - 1} + \frac{1}{2} - \frac{1}{x}\right)^2 \frac{x^s e^x}{(e^x - 1)^2} \, dx. $$ It is easy to check $I = I(0)$ and $I(s)$ is analytic for $\Re(s) > -1$. We can also check that \begin{align*} J(n, s) &:= \int_{0}^{\infty} \frac{x^{s-1}e^x}{(e^x - 1)^{n+1}} \, ...


4

Besides @MichaelBurr 's very smart idea exposed in a comment under the original post, you could also use the change of variable $t = x^2$ which transforms your integral into $\frac 1 2 \int \limits _0 ^\infty \frac {\sin t} t \Bbb d t$, to which you can now immediately apply the Abel-Dirichlet theorem.


3

We have $$\begin{align} \int_0^{1/2}\frac{\log (1+x)}{x}dx&=\int_0^{-1/2}\frac{\log (1-x)}{x}dx\\\\ &=-\text{Li}_2(-1/2) \tag 1 \\\\ &=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2^k\,k^2} \tag 2 \end{align}$$ where $\text{Li}_2$ is the dilogarithm function in $(1)$. We note that $(2)$ is the sum of an alternating series of monotonically ...


3

Test your sum of square $ s = 60.84 + \dots 51.84$. I get it as $s = 1003.20$. Then depending of the standard deviation type you get for the sample std. dev. $$\sigma_s = \sqrt{s/19} = 7.26636$$ or the population std. dev. $$\sigma_p = \sqrt{s/20} = 7.08237$$


3

From the assumption we have $a_{n}<0<b_{n}<c_{n}$. Hence $b_{n}c_{n}% =2n+1-a_{n}\left( b_{n}+c_{n}\right) >2n+1.$ Hence we have $$ a_{n}=-\frac{1}{b_{n}c_{n}}\rightarrow0\text{ and }na_{n}^{2}\rightarrow0. $$ Now from $a_{n}^{3}-\left( 2n-1\right) a_{n}^{2}+\left( 2n+1\right) a_{n}+1=0$ we get $0-0+2\lim na_{n}+1=0$ so $\lim ...


2

We just need to prove that for any $n\geq 2$ we have: $$\frac{1}{n+1}\leq \log\left(1+\frac{1}{n}\right) \leq \frac{1}{n} \tag{1}$$ then apply induction, but $(1)$ is trivial, since $f(x)=\frac{1}{x}$ is a decreasing function on $\mathbb{R}^+$, hence: $$\frac{1}{n+1}\leq\int_{n}^{n+1}\frac{dx}{x}\leq\frac{1}{n}.\tag{2}$$


2

$1$ radian in standard position is in the first quadrant. $10$ radians in standard position is in the third quadrant. How could $\sin1$, $\sin10$, possible have the same value? Maybe your professor is trolling you? Edit: Perhaps, as others suggested, he meant $\sin(10^n\pi)$? My standard Windows calculator, for instance, will claim $\sin(10^{35}\pi) \neq ...


2

your series is $$\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+............$$ rewriting this series we have,$$\frac{1}{2^{1+1}}+\frac{1}{2^{2+1}}+\frac{1}{2^{3+1}}+.........$$ So from here the $n^{th}$ term of this series will be $$\frac{1}{2^{n+1}}$$


2

Let $g(x) = \int_0^x f(t)\, dt - \int_x^1 f(t)\, dt$, we have $g(x) = 0$, for all $x \in [0,1]$ by assumption, hence $g'=0$. On the other hand, by the fundamental theorem of calculus (note that $f$ is continuous, $$ g'(x) = f(x) + f(x) = 2f(x) $$ So $f(x) = 0$ for all $x$.


2

Your solution is not correct. For any rational number $a$ that is not itself an integer, observe that $f(a)=a$, but that for any $\epsilon<\lceil a\rceil-a$, there is no $\delta>0$ with the property that $$|x-a|<\delta\implies|f(x)-f(a)|=|f(x)-a|<\epsilon$$ because there exist irrational numbers $x$ arbitrarily close to $a$ with $f(x)=\lceil ...


2

Information about the chain rule can be found here, it's basically the way of differentiating composite functions, and hence is massively useful in all of differential calculus where most functions are composites of composites of ... etc... of functions, so the chain rule is useful. It basically states that the derivative of a function $$h(x) = f(g(x))$$ is ...


2

When "geometric objects" intersect transversely, or in "general position" ,the dimension of their intersection is the dimension of the ambient space (which I am assuming here is $\mathbb R^3$) minus the sum of their respective codimensions, say "Cod" i.e., $$Dim (S_1 \cap S_2)_{\mathbb R^3}=Dim(\mathbb R^3)-Cod(S_1)-Cd(S_2) $$ Where $S_1, S_2$ are the two ...


2

We have $\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2 x^4\right) = a^2 \cdot 4x^3$. You can see that this is true by using the product rule as follows: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2 x^4\right) = x^4 \dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2\right) + a^2\dfrac{\mathrm{d}}{\mathrm{d}x} \left(x^4\right)$$ Like you said ...


2

Here is my evaluation of the quadratic case. The starting point for the derivation is the algebraic identity $$6ab^2=\left(a+b\right)^3+\left(a-b\right)^3-2a^3.$$ Then, $$\begin{align} \mathcal{I}_{2} &=\int_{0}^{1}\frac{\ln{\left(\frac{1}{t}\right)}\ln^{2}{\left(t+2\right)}}{t+1}\,\mathrm{d}t\\ ...


2

The domain of the function is {$x \in \mathbb{R}, x \neq 1$}, and thus as $x=1$ is not in the domain of $f(x)$ it cannot be a critical point. Logically, if a point is not in the domain of a function $f$, it does not exist for $f$, and thus it cannot be a critical point for $f$.


2

$$a_n=\frac{1}{2^n}$$ $$\implies \lim_{ n\to \infty}a_n=\lim_{ n\to \infty}\frac{1}{2^n}=0$$


2

Your $du$ is wrong. $(\tan x)'=\sec^2 x$ (not the other way around). Instead, write the integrand as $\sec x \tan x \tan^2 x \, (\color{maroon}{\sec x})^{1/2}$ and let $u=\color{maroon}{\sec x}$. Write $\tan^2 x$ in terms of $\color{maroon}{\sec x}$ and note $du$ then is $\sec x\tan x\, dx$.


2

In complex analysis, you learn that $-1=e^{i\pi}$, so $(-1)^x= e^{i\pi x}$, which can be written as $ \cos{x\pi}+ i\sin{x\pi}.$ (The graphs that wolfram alpha gives you are the graph of the real part, $\cos{x\pi}$, and the imaginary part, $\sin{x\pi}$.) There's a way to find the derivative without using the complex logarithm: Using the expression above ...


2

Let $p_n(x)$ be a sequence of polynomials which converges uniformly to some continuous function $p(x)$ over interval $(a,b)$. Since the convergence of $p_n(x)$ is uniform, the convergence of $p_n(x)$ over $(a,b)$ is uniformly Cauchy. By defintion, this means $$\forall \epsilon > 0 \bigg\{ \exists N > 0\bigg[ \forall n, m \ge N, \forall x \in (a,b), ...


2

If your series begins at $\frac{1}{4}$, then it should be: The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$. The series $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}=\frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{1}{2^{n}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$$ is the sum of a geometric sequence with common ratio ...


1

HINT: $$f(x)=\log x$$ The MVT gives $$\log(x+1)-\log x=\frac{1}{c}$$ for $x<c<x+1$. And $\frac{1}{c}<\frac{1}{x}$. Can you finish?


1

The summation is over the green region in the diagram below. To interchange $j$ and $k$, just determine what the region looks like after swapping the axes (reflecting about a $45^\circ$ line). So $j$ runs from $0$ to $n-1$ and $k$ runs from $1$ to $n-j$. $B_n^{f^2}(x) = \sum\limits_{j=0}^{n-1}{\sum\limits_{k=1}^{n-j}{\{...\}}}$


1

You can take that derivative. $\frac{d}{dx}(-1)^{x} = \ln(-1)(-1)^{x} = i \pi (-1)^{x}$. It's the same method for any $a^{x}$. Your issue is that you are now working with complex numbers. $\ln(x)$ is not defined in the reals for x less than or equal to zero. Your calculator can't graph it because it is only showing the values where the expression is ...



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