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8

HINT: As $y^3+1=(y+1)(y^2-y+1),$ $$\dfrac{x^4+1}{x^6+1}=\dfrac{x^4-x^2+1}{x^6+1}+\dfrac{x^2}{x^6+1}=\dfrac1{x^2+1}+\dfrac{x^2}{x^6+1}$$ Set $x^3=u$ for the second part


5

We have that $$ \lim_{x\to a} f(x)=L $$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that if $|x-a|<\delta$, then $|f(x)-L|<\varepsilon$. We prove this is false. Let $\varepsilon =1/2$ and assume it is true. Since for any $\delta>0$, there are infinitely many rational and irrational numbers in the interval ...


4

Suggesting too long for a comment: When $f(n)$ is a positive function with $\sum_1^\infty f(n)$ converging: $$\left(\sum_{n=1}^\infty f(n)\right)^2=\sum_{n=1}^{\infty}\sum_{k=1}^\infty f(n)f(k)=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2$$ with $f(n) = \frac{1}{n^2+n-1}$ Then the sum you are looking for is: ...


3

Hint : $\mathbb Q$ is dense in $\mathbb R$, that means in every neighborhood of some real number $a$, there are infinite many rational and irrational numbers. Now, use the $\epsilon$-$\delta$-formalism to get a contradiction.


3

If degrees are used then $$ \frac d{dx} \sin x = \frac \pi {180} \cos x. $$ If some units other than degrees are used then $$ \frac d{dx} \sin x = \left(\text{some constant}\cdot\cos x\right). $$ If the units are radians, then the "constant" is $1$, so $\dfrac d{dx}\sin x = \cos x$. As for proving this, go back to the proof that $$ \lim_{x\to0} \frac{\sin ...


3

Another way, Solution : \begin{align} \int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx&=2\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx\\[7pt] &=2\int_{0}^{\infty}\frac{x^4\,\mathrm dx}{x^6+1}+2\int_{0}^{\infty}\frac{\mathrm dx}{x^6+1}\tag{$\color{red}{❤}$}\\[7pt] ...


3

Yes, there does. Let $$f(t)=\arctan(\pi t-\frac{\pi}{2})$$ This is the required function. Edit: As to how I came up with this function, it's something that everyone learns I suppose. A priori, you want a function to stretch out the interval to the whole real line, so "send" 1 to $\infty$ and $0$ to $-\infty$, as this is the most intuitive understanding for ...


2

There exists a bijection. Hint: Think about $\tan$ and the fact that there is a bijection between $(a_1,b_1)$ and $(a_2,b_2)$ for any $a_1 < b_1, a_2 < b_2$.


2

If you have gotten that far, you are almost there! To evaluate $\int m\dot{x}\ddot{x}dt$, use $u$-substitution. Let $u = \dot x$, then $du = \ddot x dt$. This gives the integral $\int mu \,du$. Carrying out the integration and substituting $\dot x$ back in gives the result. Don't forget any necessary constant of integration.


2

You can still write $$ \frac{1}{n} \sum_{k=1}^n\sqrt{1-\frac{ k^2}{n^2 } }$$ This is just the "average" height of a semicircle. The diameter is 2 and the area is $\frac{\pi }{2 } $. Then using $A=b \times h$ the average must be $ \frac{\pi}{4} $. The tools to really formalize these dont come until 2nd or 3rd year university. Depending on where you ...


2

You can avoid talking about integrals if you can talk about areas. Then the sum is a sum of the areas of small rectangles that approximate a quarter of the unit disk. That is the real leap. Perhaps this is enough.


2

We may just exploit: $$\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sin\left(\frac{\pi}{t}\right)\,dt=\int_{n}^{n+1}\frac{\sin(\pi u)}{u^2}\,du=\frac{2(-1)^n}{\pi n^2}+O\left(\frac{1}{n^3}\right) $$ that follows from integration by parts.


2

Applying the chain rule to $u(x) = [y(x)]^{1-n}$ we obtain that \begin{align} \frac{d u}{dx}(x)&= \frac{du}{dy}(y(x))\cdot\frac{dy}{dx}(x)\\ &= (1-n)[y(x)]^{-n}\cdot\frac{dy}{dx}(x) \end{align} Futhermore using the Bernoulli equation we have $$ \frac{dy}{dx}=q(x)y^n-p(x)y $$ and \begin{align} \frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\ ...


2

$$\begin{eqnarray*}\int_{\mathbb{R}}\frac{x^4+1}{x^6+1}\,dx&=&2\int_{\mathbb{R}^+}\frac{x^4+1}{x^6+1}\,dx=2\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx+2\int_{1}^{+\infty}\frac{x^4+1}{x^6+1}\,dx\\&=&4\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx=4\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{6k}+x^{6k+4}\right)\,dx\\&=&4\sum_{k\geq ...


2

For each $x\in \mathbb{R}$, we can find sequence $z_k\to x$, such that $z_k\in \mathbb{Q}$ and $y_k\to x$, such that $y_k\in \mathbb{R}\setminus \mathbb{Q}$. We can conclude from $f(z_k)\to 1$ but $f(y_k)\to -1$, that limit doesn't exist for any $x\in \mathbb{R}$.


2

Conveniently, your integration limits are $(0,4)$ so the inside absolute value can be ignored. You are left with the following: $$\int_0^4 |x^2 - 2x| \ dx = \int_0^4x\cdot|x-2|\ dx$$ $$= \int_0^2 x\cdot(2-x)\ dx + \int_2^4x\cdot(x-2) \ dx$$ I leave the rest to you.


2

i am going to offer a more intuitive approach to this problem. let f(x)=sin(x) on closed interval [0,2pi], then max f(x) = 1 at x = pi/2 similarly, let f(x)=cos(x) on closed interval [0,2pi], then max f(x) = 1 at x = 0 and 2pi now consider f(x,y)=sin(x)+cos(y). now is not apparent that max f(x,y) must equal 2 at (pi/2, 0), (pi/2, 2pi). so there are two ...


2

As already pointed by the other answers, we just have to compute: $$ S_1=\sum_{n\geq 1}\frac{1}{n^2+n-1},\qquad S_2=\sum_{n\geq 1}\frac{1}{(n^2+n-1)^2}.\tag{1}$$ and by exploiting the logarithmic derivative of the Weierstrass product for the cosine function we have, for any $\alpha\in\mathbb{R}^+\setminus\mathbb{N}$: $$ \sum_{n\geq ...


1

hint: $f(2) = 4C+1, \displaystyle \lim_{x\to 2^{-}} f(x) = f(2) \to 10-2 = 4C+1$.


1

You want: $$\frac{A(x+3)+B}{(x+3)^2}=\frac{3-2x}{(x+3)^2}$$ You lost the denominator on the right hand side. Much, then, to solve: $$A(x+3)+B = 3-2x$$


1

The function $\frac{x}{\sinh\frac{x}{2}}$ is regular and bounded in a right neighburhood of zero and decays like $x\, e^{-x/2}$ for large $x$, hence the integral over $\mathbb{R}^+$ is converging. By the residue theorem we also have: $$ \int_{0}^{+\infty}\frac{x}{\sinh\frac{x}{2}}\,dx = \color{red}{\pi^2},\tag{1}$$ and by writing the integrand function as a ...


1

After correction, the solution to the initial differential equation is y(x) = +/- sqrt[ 5x / ( 5Cx^5 + 2) ] Regarding the u transformation: from the B.E. we know (1) y' = q y^n - p y. From the u definition we know that (2) u' = (1-n) y^(-n) y'. Substitute y' from (1) into (2) to recover the new ODE for u.


1

For the point $(x+s\cos(\theta),x^2+s\sin(\theta))$ to be on the curve $y=x^2$, we need $$ (x+s\cos(\theta))^2=x^2+s\sin(\theta)\tag{1} $$ Expanding the left side of $(1)$ and cancelling, we get $$ s\cos(\theta)=\tan(\theta)-2x\tag{2} $$ Now $(2)$ must also remain true for $\theta+\frac\pi2$ so that the other corner adjacent to $(x,x^2)$ is on the parabola. ...


1

Radians make many formulas much simpler. For example, the length of an arc of a circle subtended by an angle of $\theta^\circ$ $$s=2\pi r\theta/360^\circ.$$ If $\theta$ is measured in radians: we get $s=r\theta$. Or, the area of a sector is $$A=\pi r^2\theta/360^\circ,\quad\theta\text{ in degrees}$$ but $$A=r^2\theta/2\quad\theta\text{ in radians}.$$ ...


1

For the one-dimensional case it is actually not very difficult. Assume $I\in \Bbb R$ is an open connected subset of real line. Then denote $C(I) = C^0(I)$ the set of all functions $\ f:I\to\Bbb R$ continuous on $I$, i.e. $$ C^0 = \left\{ \ f: I \to \Bbb R \ \Big| \ \ \forall \, x_0 \in I \quad \lim_{x \to x_0} f(x) = f\!\left(x_0\right) \right\} $$ Then ...


1

Let $C^0$ be the set of continuous functions. Then $$C^\infty = \{ f | \:\forall n\:\: \frac{d^nf}{dx^n} \in C^0\}$$


1

There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps. Notice $$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right] = \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right] $$ and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have $$\mathcal{I} ...


1

No. Let $$ f(x)=\begin{cases}x^3\sin\dfrac{1}{x} & \text{if }x\ne0,\\0 & \text{if } x=0.\end{cases} $$ Then $$ f'(x)=\begin{cases}3\,x^2\sin\dfrac{1}{x} -x\cos\dfrac1x& \text{if }x\ne0,\\0 & \text{if } x=0.\end{cases} $$ $f'$ is continuous and $f(x)=o(x^2)$, but $f'$ is not differentiable at $x=0$.


1

Since the function is constant on non-integer, its derivative is $0$ on non-integer values. Since the function is not continuous on integer values, its derivative undefined at the integers.


1

Here are some useful components to developing a solution to the proposed integral. \begin{align}\tag{1} \int_{0}^{\infty} Ei^{4}(-x) \, dx &= -4 \, \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx \\ \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx &= - \ln^{2}2 + \int_{0}^{\infty} x \, e^{-x} \, Ei^{2}(-x) \, dx - 2 \, \int_{0}^{\infty} e^{-x} \, Ei(-x) \, ...



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