New answers tagged

1

As @Mark pointed out, you're problem can be reexpressed as. $$\sum\limits_{n=0}^{\infty}\frac{n!}{2^n \frac{(n+2)!}{2^{n+1}}}$$ $$\sum\limits_{n=0}^{\infty}\frac{2 n!}{(n+2)!}$$ $$\sum\limits_{n=0}^{\infty}\frac{2}{(n+1)(n+2)}$$ $$2 \times \left(\sum\limits_{n=0}^{\infty}\frac{1}{n+1} - \sum\limits_{n=0}^{\infty}\frac{1}{n+2}\right)$$ Making a change of ...


0

Hint: For the integrand $\sqrt{1-49x^2}$, substitute $x=\frac{\sin(u)}{7}$ and $dx=\frac{\cos(u)}{7}du$. Then $$\sqrt{1-49x^2} = \sqrt{1-\sin^2(u)}=\cos(u).$$ Therefore $$\bbox[5px,border:2px solid #C0A000]{\int\sqrt{1-49x^2}\,dx=\frac{1}{7}\int\cos^2(u)\,du}$$ where $u = \sin^{-1}(7x)$. For the integral of $\cos^2(u)$ use the trigonometric identity ...


0

A Puiseux series is formal Laurent power series in $T^{\frac1n}$ for some $n$ (the $n$ may vary with the series). The set of Puiseux series is a field, denoted $k{\ll} T{\gg}$, and it is the algebraic closure of the field of formal power series $k[[T]]$.


1

For a general Lagrangian $L[t,q,q',q'',\ldots]$ the Euler-Lagrange equation reads $$L_q = [L_{q'}]' - [L_{q''}]'' + [L_{q''}]''' - \ldots = \sum_{n\geq 1} (-1)^{n+1}[L_{q^{(n)}}]^{(n)}$$ If the Lagrangian is time-independent. $L_t = 0$, we have $$\frac{dL}{dt} = L_qq' + L_{q'}q'' + L_{q''}q''' + \ldots = L_q q' + \sum_{n\geq 1}L_{q^{(n)}}q^{(n+1)}$$ and ...


0

Changing variables from $x$ to $\theta$ via $x=A(\theta - \sin\theta)$, the integral that represents the time to go from $(x,y)=(0,0)$ to $(x_0,y_0)$ becomes $$ T=\sqrt{\frac 1{2g}}\int_{\theta=0}^{\theta_0}\sqrt{\frac{1+(y')^2}y}\frac{dx}{d\theta}\,d\theta $$ where $\theta_0$ is the value of $\theta$ that corresponds to $x_0$. You've calculated ...


0

Using the substitution you started with: $$\int \frac{v}{2 + v} \, dv = \int \frac{u-2}{u} \, du = \int \left(1 - \frac{2}{u} \right) \, du = u-2\ln|u| + C =v-2\ln|2+v|+D.$$ Here is another method: $$\int \frac{v}{2 + v} \, dv = \int \frac{2+v-2}{2 + v} \, dv = \int \left (1 - \frac{2}{2+v} \right) \, dv = v-2 \ln|2+v| + C.$$


0

Let's rewrite $F(x)$ a little bit. Therefore, let $G$ be the antiderivative of $g$, i.e. $G'(x)=g(x)$ for all $x$. Then, using the fund. theorem of calculus, we can write $$ F(x) = \int_x^{\sin x} \left(\int_0^{\sin t} (1+u^4)^{0.5} \,du\right)dt =: \int_x^{\sin x} g(t)\,dt = G(\sin x) - G(x). $$ Now it is easier to handle: $$F'(x) = G'(\sin x) \cdot \cos ...


0

In general, the $\min$ function takes the minimum value among a list or set of values that was given to it. Something like $\min(x,y)$ means "the minimum value among $x$ and $y$". Informally, people might say "$x$ or $y$, whichever is less". The comma is merely the usual comma that separates the input values passed to a function that takes more than one ...


2

Hint: Consider the function $A \mapsto A^2$ on $2 \times 2$ matrices.


0

$\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ converges for any $\alpha > 1$ to (let us note it) $l_{\alpha}$. Take $\frac{1}{l_{\alpha}}\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$. $ \sum_{n=1}^{\infty} \frac{1}{n\ln^2(n)}$ also converges (even slowlier than the example above). You won't have any nice analytical expression for the limit if you go that way, ...


10

Take any sequence $s_n$ such that $s_n \to 1$ as slowly as you want, with $s_0 = 0$. Let $a_n = s_n - s_{n-1}$. Then the partial sums $\sum_{i=1}^n a_i = s_n$.


2

You can do $\frac 6{\pi^2} \sum_{n=1}^\infty \frac 1{n^2}$ or $\sum_{n=1}^\infty \frac 1{n(n+1)}$


2

You can apply Cauchy's limit theorem $$\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$ with $$a_n = (1+1/n)(1+2/n)…(1+n/n) = \frac{(2n)!}{n! n^n}.$$ Then $$\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!} \\ = \frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}} \\ = 2\frac{(2 + ...


8

Take logarithm: $$\frac1n \sum_{k=1}^n \log \left( 1 + \frac{k}n\right)$$ which converges (using Riemann sums) to: $$\int_0^1 \log(1+x) dx = 2\log 2 - 1$$ Therefore the original limit is: $$e^{2 \log 2 -1} = 4/e$$


2

I found an answer: We will need the following easy to prove (I can expand here if necessary) relations: \begin{eqnarray*} B_{k+1}(x) = \sum_{i=0}^{k+1} B_i \binom{k+1}{i} x^{k+1-i} \quad \quad (1), \end{eqnarray*} and \begin{eqnarray*} \sum_{i=0}^{n-1} (i+m)^{k-1} = \frac{B_{k}(n+m) - B_{k}(m)}{k} \quad \quad (2) \end{eqnarray*} Let us ...


0

It does seem a bit on the ugly side in that the region seems to be now down to one square pyramid with vertex at $(0,0,2)$. I think the $x$-component of $\vec\nabla\times\vec V$ should be $(y+1)e^y$: check it. For part b), I get your answer now that the domain of $w$ has been corrected. For the Jacobian, I get $$J=\left|\det\begin{bmatrix}\frac{\partial ...


4

Without appealing to L'Hospital's Rule, we note that $\log(x)\le x-1$ for $x>0$. Then, we have $$\begin{align} \frac{\log(x)}{x^{1/9}}&\le \frac{x-1}{x^{1/9}}\\\\ &=x^{8/9}-x^{-1/9}\\\\ &\to -\infty \,\,\text{as}\,\,x\to 0^+ \end{align}$$ And we are done! It is interesting to note that L'Hospital's Rule does indeed apply to the limit ...


4

Using your first idea, what about using the fact that $$(1+1)^n = \sum_{k=1}^n \binom{n}{k} \geq \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \geq \frac{(n-3)^4}{24}$$and concluding by the squeeze theorem? (as $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).


2

Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.


7

It is not an indeterminate form: it is: ‘ $\dfrac{-\infty}{0^+}=-\infty\times(+\infty)=-\infty$’.


3

Use the substitution $x=y^9$, where $y>0$.


9

One may just write $$ \lim_{x \to 0^+} \frac{ \ln x}{x^{1/9} } = \lim_{x \to 0^+}\left( \frac1{x^{1/9} } \times \ln x \right)=\frac1{0^+} \times (-\infty)=+\infty \times (-\infty)=-\infty. $$


0

Let $a\in Y$ and take $V=(f(a)-\epsilon,f(a)+\epsilon)$. Since $f$ is continuous, there is an open interval $U=(a-\delta,a+\delta)\subseteq X$ s.t. $f(U)\subseteq V.$ Now, $U'=U\cap Y$ is open in $Y$ (by definition of the subspace topology), contains $a$ and is s.t. $f(U')\subseteq V$, which is what it means for $f|_Y$ to be continuous at $a$. It may be ...


0

Take any open set $V \in \mathbb R$ then $$(f|_Y)^{-1}(V) = Y \,\cap f^{-1}(V) $$ as $f$ is continuous then $f^{-1}(V)$ is open. Thus $(f|_Y)^{-1}(V)$ is open in $X$ showing that $f|_Y$ is continuous.


0

For any $x_0\in Y$, $f|_Y$ is continuous at $x_0$ means $\forall \varepsilon>0\exists\delta>0$ s.t. $\forall x\in B(x_0,\delta)\cap Y$, we have $|f(x)-f(x_0)|<\varepsilon$. You have already known that $\forall \varepsilon>0\exists\delta>0$ s.t. $\forall x\in B(x_0,\delta)\cap X$, $|f(x)-f(x_0)|<\varepsilon$. Note that $Y\subset X$. But you ...


1

A particular solution of the equation given by variation of parameters is $$K(x) =e^{-x}\left(-\frac {\cos2x}2 \int_0^x \sin(2t)f(t)e^t dt+\frac {\sin2x}2 \int_0^x \cos(2t)f(t)e^t dt\right)$$ All the solutions are therefore given by $K(x) + C\cos(2x) + D\sin(2x)$ where $C,D \in \mathbb R$ Note that if $f$ is bounded by some $M>0$, $$|K(x) + C\cos(2x) + ...


1

The absolute value just implies you have symmetry about the y-axis. Assume $x>0$, then $$\frac{d}{dx} x^{3/7} = \frac{3}{7}x^{-4/7} = \frac{3}{7}x^{3/7-1}=\frac{3}{7}\frac{x^{3/7}}{x}(x>0)$$ If $x<0$ then we just need to reverse the sign of the argument so $-x$ is positive: $$\frac{d}{dx} (-x)^{3/7} = \frac{-3}{7}(-x)^{-4/7} = ...


2

Let $g(x)=f(x)-x$ and $m=\inf_{x\in[0,1]} g(x)$ then since $g$ is continuous on a compact so by Weierstrass theorem this minimum is attained i.e. there is $x_0\in[0,1]$ such that $g(x)\ge g(x_0)=m>0$. Take $\epsilon=\frac m2$.


0

$$\int_0^1 \frac{200\sqrt5(1-x^2)-300(1-x)^2}{ \left[5\sqrt5(1+x)^2-15(1-x^2)+2\sqrt5(1-x)^2 \right]^2}dx=(2\phi+1)(\phi+2)$$


4

You are making it much more complicated than necessary. As a first step, compute the expression at the given point. $$\frac{-R_{4}}{R_{3}}\frac{0C_{2}\frac{R_{3}}{R_{3}+R_{4}}\frac{R_{5}+R_{6}}{R_{6}} } {\frac{0^2C_{1}C_{2}R_{1}R_{2}R_{5}}{R_{6}} +0C_{2}R_{2}\frac{R_{3}} {R_{3}+R_{4}} \frac{R_{5}+R_{6}}{R_{6}} +1}=\frac{-R_{4}}{R_{3}}\frac01=0.$$


1

What is a function? A subset $f$ of $A\times B$ such that for each $a\in A$ there is exactly one $b\in B$ for which $(a, b)\in f.$ A function is "its graph." The function $f$ in your first line ,and the function "$f$ restricted to the domain $[-1,2]$" are two different functions. We can say that "a function $g$ is increasing at a point $p$" (when $p\in$ ...


0

Write $f(x)=x$ if $x\geq0$ and $f(x)=2x$ if $x<0$. To make the second conclusion, you would need $f'$ to exist throught the open interval $(-1,1)$, which is obviously not true here.


0

Am giving a try. Did a bit progress in my way and so want to share it. $$I = \int_{-1}^{1} e^x \cdot (f(x) + f(-x))dx$$ $$ = 0 - \int_{-1}^1 e^x \cdot (f'(x) - f'(-x))dx$$ $$ = 2f'(c)(e-\frac{1}{e})$$ Now, $$I = \int_{-1}^1 e^x f(x) + \int_{-1}^{1} e^{-x} f(x)$$ $$ \implies \int_{-1}^{1} e^{-x} f(x)= 2(e-\frac{1}{e})\cdot(f'(c) - f(1) + f(-1))$$ And am ...


0

Note that if $g(x)=e^{-x}$, $f(x)=\ln x$, $$f(x) = \int \left(\frac{1}{x}-\ln x\right)e^{-x}dx=e^{-x}\ln x+C$$ $f(x)=Ce^x+\ln x$. Now, this may prove helpful-$$\int_{0}^{1} \ln x \mathbf{d}x =-\int_{-\infty}^{0}e^x \mathbf{d}x=-1$$ Which follows from the fact that $\ln x $ is the inverse of $e^x$.


0

The Taylor series for $\sin (x)$ is $\sin (x) = \sum_{0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ The series converges uniformly on any bounded subset of $\mathbb R$. You can use M-test with $M_n = \frac{R^{2n+1}}{(2n+1)!}$ for the interval $[-R,R]$ ($R\in \mathbb R$)


0

The coefficients of such power series decay so fast that they ensure that $f(x)$ is an analytic function with radius of convergence $+\infty$. We are so allowed to differentiate $f(x)$ termwise, even twice, and check that $f''(x)=-f(x)$. Since $f(0)=0$ and $f'(0)=1$, $f(x)=\sin x$ for every $x$, because the previous ODE has a unique global solution.


2

No, of course not: Think of the function $e^{x-y}$ which along the line $x=y$ is not only bounded, but even constant $1$ - yet it clearly is not bounded.


1

You can Taylor's formula, which is an exact formula for polynomials: suppose $f$ has degree $n$. Then \begin{align*} f(x)&=f(\alpha)+ f'(\alpha)(x-\alpha)+f''(\alpha)\frac{(x-\alpha)^2}{2!}+\dots +f^{(n)}(\alpha)\frac{(x-\alpha)^n}{n!}\\ &=(x-\alpha)^2\Biggl[f''(\alpha)\frac1{2!}+\dots +f^{(n)}(\alpha)\frac{(x-\alpha)^{n-2}}{n!}\Biggr]. \end{align*} ...


0

Hint for the standard method: The domain of the integrand is $\;0\le x\le 3$. Rewrite it as $\; \dfrac 1{1+\sqrt{\dfrac3x-1}}$ and use the substitution $$\newcommand\d{\,{\mathrm d\mkern1mu}} u=\sqrt{\dfrac3x-1}\iff x=\frac3{u^2+1},\enspace u\ge 0,\qquad \d x=-\frac{6u}{(u^2+1)^2}\d u$$ So one obtains $$\int_0^{+\infty}\frac{6u}{(1+u)(1+u^2)^2}\d u.$$ There ...


3

Substitute with $y=3-x$ gives $$I=-\int_3^0 \frac{\sqrt{3-y}}{\sqrt y +\sqrt{3-y}} dy=\int_0^3 \frac{\sqrt{3-y}}{\sqrt y +\sqrt{3-y}} dy$$ So $$2I=3$$


1

Use property $\int_{a}^{b}f(x)dx$ = $\int_{a}^{b}f(a+b-x)dx$


0

Personally, I'd multiply by $\sqrt{x}-\sqrt{3-x}$ and integrate by parts. And I'd be careful about the domain (separate it in two parts, one $x<\frac{3}{2}$ and the other $x>\frac{3}{2}$) But I am sure there are plenty of other tricks applicable (variable change for instance)


0

$$\int\frac{1}{\sin (x)} dx=\int\frac{sin(x)}{sin^2(x)}dx=\int\frac{sin(x)}{1-cos^2(x)} $$ Use substitution $t=cos(x) \to dt=sin(x)dx\to dx=\frac{dt}{sin(x)}$ $$\int \frac{sin(x)}{1-t^2}*\frac{dt}{sin(x)}=\int \frac{dt}{1-t^2}=arth(t)=arth(cos(x))+C$$


1

Let the roots of $f(x)$ be $a_1,a_2,\cdots,a_n$. Then, $$f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)$$ $$f'(x)=(x-a_1)(x-a_3)\cdots(x-a_n)+(x-a_2)(x-a_3)\cdots(x-a_n)+\cdots$$ Let $a_1$ be a root of $f'(x)$. But $$f'(a_1)=(a_1-a_2)(a_1-a_3)\cdots(a_1-a_n)=0$$ Thus, one of the factors must be zero. Let the factor $(a_1-a_j)=0$ for some $2\le j\le n$. Then, $a_1=a_j$, ...


4

Hint. One may just interchange sum and integration : $$ \int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\sum_{n=0}^{\infty}\int_0^\infty\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx $$ then, by the change of variable $$u=(n+1)^sx^{n+1}, \quad du=(n+1)^{s+1}x^ndx,$$ one gets $$ ...


1

$$ \frac{\partial}{\partial t}f(x,t)=\frac{\partial}{\partial t}\int_{0}^{g(x,t)} e^{-u^2} du=e^{-g^2(x,t)}\frac{\partial}{\partial t}g(x,t)\ , $$ where one uses the fundamental theorem of calculus $\frac{d}{dz}\int_0^z dt\ h(t)=h(z).$ Taking a second derivative (using the product rule) $$ \frac{\partial^2}{\partial t^2}f(x,t)=\frac{\partial}{\partial ...


2

Assuming $y=a x$ ($a\geq0$), the constraint leads to $$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}$$ and $$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Now, computing $$\frac{d}{da}x^2y=-\frac{(2 a-1) \left(3 a+\sqrt{a (3 a+2)+2}+2\right)}{\sqrt{a (3 a+2)+2} \left(a+\sqrt{a (3 a+2)+2}+1\right)^4}k^3$$ which cancels if $a=\frac 12$. All of this makes ...


2

Let $x$ be the quantity we try to find. $$x=\sqrt[6]2$$ $$x^6 = 2$$ $$x^6-2=0$$


0

We are given that $f$ is differentiable on $I$ and $f'(x) > 0$ for all $ x \in I - \{c\}$ where $c$ is some point in $I$. If $f'(c) < 0$ then by intermediate value property of derivatives there will be many points at which $f'(x) < 0$ and this is not allowed hence we must have $f'(c) = 0$. Next let $a, b \in I$ with $a < b$. If $c \notin (a, b)$ ...


1

One easy way to do it to notice that $$\frac{\partial}{\partial s}\int_0^{\infty}\dfrac{e^{-sk}sinkx}{k}=-\int_0^{\infty}e^{-sk}sinkx=-\dfrac{x}{x^{2}+s^{2}}$$ Then you just have to revert the derivative with respect to "s" $$-\int ds \dfrac{x}{x^{2}+s^{2}}=-\arctan(s/x)$$



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