New answers tagged

-1

I honestly didn't read all your post (sorry for that), but there is a way that prove the result very quickly. I suppose that $f$ is $\mathcal C^2$ in a neighborhood of $c$. Using Taylor-Young, there is a $c_x$ s.t. $|c_x-c|<|x-c|$ s.t. $$f(x)=f(c)+f'(c)(x-c)+\frac{f''(c_x )}{2!}(x-c)^2=f(c)+\frac{f''(c_x)}{2!}(x-c)^2.$$ By continuity of $f''$, we can get ...


2

That's not entirely true, it is true for monotone functions, at least up to a bounded constant term found by using the LRAM and RRAM approximations, but this is usually what they mean by the $\approx$ symbol. However, take the function $f(x) = |\sin(\pi x)|$ for example, then $$\sum_{k\in\Bbb Z}f(k) = 0\not\approx \int_{\Bbb R}f(x) \,dx =\infty$$ because ...


2

I'll leave the details to you and go straight to the main integral. Let $$I_n = \int \frac{du}{(1+u^2)^n}.\tag{1}$$ We clearly have $I_1=\arctan u$ and: $$\begin{eqnarray*} I_{n+1} = \int \frac{(1+u^2)-u^2}{(1+u^2)^{n+1}}\,du &=& I_n -\frac{1}{2}\int ...


0

Since $2M_1 t \le 2M_0+M_2t^2$, then the function $f(t)=2M_0-2M_1t+M_2t^2$ is positive, i.e., $f$ does not change sign for all $t \in \mathbb{R}$, which mean that the discriminant which is $4M_1^2 -8 M_0M_1 \le 0$ and this is equivalent to write $M_1 \le \sqrt{2}M_0M_2$.


1

Hint: if the second-degree polynomial $$ p(t) = M_2t^2-2M_1 t+2M_0 $$ is non-negative for every value of $t\in\mathbb{R}$, its discriminant has to be non-positive. Notice that: $$ M_2\cdot p(t) = (M_2 t-M_1)^2 + \color{red}{(2M_0 M_2-M_1^2)}. $$


2

For $\;|x|<1\;$ : $$\frac1{1-x^2}=\sum_{n=0}^\infty x^{2n}\stackrel{\text{diff.}}\implies\frac{2x}{(1-x^2)^2}=\sum_{n=1}^\infty 2n\,x^{2n-1}$$ Now just do a little cosmetics to the above and get your answer.


5

Proofs using geometrical intuition and Euclid axioms were almost the norm in ancient Greek mathematics. However side by side the idea of numbers was also developing for obvious practical needs (i.e. counting and measuring). With advent of algebra it became obvious that methods based on algebraic manipulation of numbers were far more powerful than the ...


1

Let the first point's position be $(t,0)$, and the second $(x(t),y(t))$. We have that the direction of the second point is proportional to the difference between the points: $$(x'(t), y'(t))\propto(t - x(t), -y(t))$$ But, since the speed is constant, we must have that: $$(x'(t), y'(t))=\left(\frac{t-x(t)}{\sqrt{(t - x(t))^2 + y^2(t)}}, ...


1

Your indefinite integral is correct but the integral you are using is definite. The constants are not needed as they cancel out. Note that the definite integral has the formula $\int_{a}^{b}f(x){dx}=(F(b)+c)-(F(a)+c)=F(b)-F(a)$, where $F(x)$ is the indefinite integral. So ...


0

If I have understood correctly, do you want to see why the integral is also equal to $$\frac{\pi/n}{\sin\left(\pi/n\right)} $$ (if I am wrong tell me and I will delete the answer). A way to see it to observe that, taking $1+x^{n}=u^{-1} $ ...


1

Well, then $$A(x) = \left[4t - \frac{1}{2}t^2\right]_0^x$$ $$A(x) = 4x - \frac{1}{2}x^2$$ $$A(x) = \frac{1}{2}\left(8x - x^2\right)$$ $$A(x) = \frac{1}{2}x\left(8 - x\right)$$ The graph of $A(x)$ will be a downward parabola cutting the x-axis at $0$ and $8$. $$A'(x) = \frac{1}{2}\left(8 - 2x\right) = 4 - x$$


1

By definition, $$A(x) = \int_{0}^{x} f(t)~dt$$ where $f(t) = 4 - t$. Evaluating the integral yields \begin{align*} A(x) & = \int_{0}^{x} (4 - t)~dt\\ & = \left(4t - \frac{1}{2}t^2\right) \bigg|_{0}^{x}\\ & = 4x - \frac{1}{2}x^2 - (0 - 0)\\ & = 4x - \frac{1}{2}x^2 \end{align*} Can you take it from there?


2

The measure $\frac{dx}{x}$ assigns the measure $\int_A \frac{1}{x} dx$ to a set $A$ (implicitly considered to be a subset of $[0,\infty)$ so that you guarantee positivity). Note that in this notation Lebesgue measure is sometimes denoted by just $dx$.


3

The series converges: Using $\sin x = x - \frac{x^3}{6} + o(x^4)$ around $0$, we get that when $n\to \infty$ $$ \sin\left(\frac{\sin(3n)}{\sqrt{n}} + \frac{1}{n\ln^2(n)}\right) = \underbrace{\frac{\sin(3n)}{\sqrt{n}}}_{a_n} + \underbrace{\frac{1}{n\ln^2(n)} + O\left(\frac{1}{n^{3/2}}\right)}_{b_n}. $$ The series $\sum_{n=2}^\infty (-1)^n b_n$ converges ...


2

$$\int_{[0,Re^{2i \pi /n}]} \frac{1}{1+z^n} dz = \int_0^R \frac{1}{1+(e^{2i \pi /n}x)^n} d(e^{2i \pi /n}x) = e^{2i \pi /n} \int_0^R \frac{1}{1+x^n} dx$$ the poles of $\frac{1}{1+z^n}$ are at $e^{i\pi (2k+1)/ n}, k = 0 \ldots n-1$. the only one inside the whole contour $\Gamma_R : [0,R] \cup [R, R e^{2 i \pi /n}] \cup [R e^{2 i \pi / n}, 0]$ is the one at ...


0

Substitute $x-1=y$. Now you need to show that for $y\geq 0$ the next inequality holds: $$y+1-\sqrt{(y+1)^2-1}<1$$ $$y<\sqrt{y^2+2y}$$ $$y=\sqrt{y^2}\leq \sqrt{y^2+2y}$$


0

Any function which is Riemann integrable is anti differentiable. Any continuous function is Riemann integrable. Any non decreasing function defined on [a,b] is Riemann integrable on [a,b]. Any non increasing function defined on [a,b] is Riemann integrable on [a,b] These came from an IBL Real Analysis packet by Mahavier.


0

The "horizontal" hyperbola $ \ x^2 \ - \ y^2 \ = \ 1 \ $ has the asymptotes $ \ y \ = \ \pm x \ $ . In the first quadrant then, that "branch" of the hyperbola follows $ \ y \ = \ \sqrt{x^2 - 1} \ $ . At $ \ x \ = \ 1 \ $ , the asymptote passes through $ \ ( 1, \ 1) \ $ , while the hyperbola has a vertex at $ \ (1, \ 0 ) \ $ . For $ \ x \ > \ 1 \ $ , ...


0

$x-\sqrt{x^2-1}<1\implies x-1<\sqrt{x^2-1}\implies x-1<\sqrt{(x+1)(x-1)}$$\implies \sqrt{x-1}<\sqrt{x+1}$ Since these are algebraic operations perserving order, it is easy to see that the implications follow backwards assuming $x>1$.


0

For $x\gt 1$ one has $2x\gt 2$ thus $2x-2\gt 0$. This means $x^2-1\gt x^2-1-(2x-2)$$ But the r.h.s of the previous inequality is $(x-1)^2$ so we have $(x^2-1)\gt (x-1)^2$. Now take the square root and bear in mind that $\sqrt{(x-1)^2}=x-1$ because $x\gt 1$ we get $\sqrt{x^2-1}\gt x-1$ and so $$-\sqrt{x^2-1}\lt -(x-1)$$ Now add $x$ to get the inequality ...


2

WLOG: $x=\csc2y$ with $0\le2y\le\dfrac\pi2$ $$x-\sqrt{x^2-1}=\dfrac{1-\cos2y}{\sin2y}=\tan y$$ As $0\le y\le\dfrac\pi4,\tan y\le1$


3

$f(x)=x-\sqrt{x^2-1}= \frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\frac{x^2-x^2+1}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}$ From this, it should be easy to deduct.


1

$$x-\sqrt{x^2-1}<1$$ $$\iff (x-1) < \sqrt{x^2-1}$$ $$\iff x^2-2x+1 < x^2-1,\ \ \mathrm{for} \ \ x>1$$ $$\iff x>1$$


5

Notice that $(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})=x^2-(x^2-1)=1$, and that $x+\sqrt{x^2-1}>x>1$, and therefore, we have that $x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}<\frac{1}{1}=1$


1

Define $S(x, y) = \frac {f(y) - f(x)} {y - x}$. We have $\alpha \in (a, b), \gamma \in (b, c)$ such that $f'(\alpha) = S(a, b)$ and $f'(\gamma) = S(b, c)$. $S(a, c)$ lies between $S(a, b)$ and $S(b, c)$, so the intermediate value property for derivatives gives us $\beta \in (\alpha, \gamma)$ such that $f'(\beta) = S(a, c)$.


0

What about $$\frac{dy}{dt}=\frac{dy}{dx}\times \frac{dx}{dt}=\frac{\frac{dy}{dx}}{\frac{dt}{dx}}$$ Now $y=\log(\sin(x))$ and $t=\sqrt{\cos(x)}$. I am sure that you can take it from here.


-1

The fact that $\alpha<\beta$ should follow directly from the mean value theorem. You said that you can easily observe that such $\alpha$ and $\beta$ exist. I am assuming that you are doing so by considering $f$ restricted to the intervals $[a,b]$ and $[b,c]$. In which case, we get that $\alpha\in(a,b)$ and $\beta\in(b,c)$. And so $\alpha<b$ and ...


1

Let $x$ and $y$ be functions of $t$. We have $$x'+y'+2x+y=0$$ $$y'+5x+3y=0$$ so $$x'-3x-2y=0$$ $$y'+5x+3y=0$$ Let $$r(t)=\begin{bmatrix} x(t)\\y(t) \end{bmatrix}.$$ $$r'(t)=\begin{bmatrix} 3 && 2 \\ -5 && -3 \\ \end{bmatrix}r(t)$$ Suppose $r(t)=ce^{\lambda t}$ for some nonzero constant vector $c$. The differential equation becomes ...


-1

Is your function convex or concave? For convex function it's proved easily using the second order derivative. But for concave,I don't know


1

You don't need the other $dx$ term. Please see the below solution: $$\int\frac{1}{1 + \log^2 x}\ d(\log x)$$ Let $u = \log x$ so then $du = d(\log x)$ Therefore, $$\int\frac{1}{1 + u^2}\,du = \arctan(u) + \text{constant}.$$ Substituting back in for $u$ you get: $$\arctan(\log x) + \text{constant},$$ which is the antiderivative/primitive we seek.


0

No. For example, for $N$ large, $s=0$, and $x=\frac 12$ your expression becomes $\frac 18 \cdot \frac {\ln (\frac 12+N)}{(\frac 12+N)^2}$, which will exceed any $\frac A{N^2}$ when the log gets big enough.


1

When you have products or quotients, you can make life simpler considering logarithmic differentiation $$z=e^y \cos^2(y)\implies \log(z)=y+2\log\big(\cos(y)\big)$$ Differentiating with respect to $x$ $$\frac 1 z \frac{dz}{dx}=\frac{dy}{dx}-2 \frac{\sin(y)}{\cos(y)}\frac{dy}{dx}=\big(1-2\tan(y)\big)\frac{dy}{dx}$$ Multiplying both sides by $z$, then ...


3

Let's do the job as suggested by @Dr. MV. Integrate by parts with $u=x^{m-1}$ and $dv=x\left(x^2-1\right)^{5}dx$. This means $du=(m-1)x^{m-2}$ and $v={\left(x^2-1\right)^6\over 12}$. And so $$\begin{align}I_m=&=\left[uv\right]_0^1-\int_0^1vdu\\ &=-{m-1\over 12}\int_0^1\left(x^2-1\right)^6x^{m-2}dx\\ ...


2

A simplistic approach $$\frac{dx}{dt}+\frac{dy}{dt}+2x+y=0\tag 1$$ $$\frac{dy}{dt}+5x+3y=0\tag 2$$ From $(2)$ $$x=-\frac 15\left(3y+\frac{dy}{dt}\right)\tag 3$$ $$\frac{dx}{dt}=-\frac 15\left(3\frac{dy}{dt}+\frac{d^2y}{dt^2}\right)\tag 4$$ Use $(3)$ and $(4)$ in $(1)$ and get $$\frac{d^2y}{dt^2}+y=0\tag 5$$ Solve it for $y$ and use the result in $(3)$ to get ...


0

Hint: The tangent line at a point on a circle is perpendicular to the diameter of the circle passing through that point Also, if a line has slope $s$, another line perpendicular to the first has slope $-1/s$


0

Find the slope of the line passing through the center of the circle and the point on the circle. Call it $m$. Then form the equation of a line through the point on the circle but let the slope of this line be the opposite reciprocal of the one you just found, i.e. slope $\frac{-1}{m}$.


0

The ratio test zaps this one right off the bat. $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{(2n+4)}{(n+1)^5}\frac{n^5}1=\lim_{n\rightarrow\infty}\frac{2n+4}{\left(1+\frac1n\right)^5}=\infty$$ So the series diverges. Note how the ratio test created a context where most of the factors of the product canceled out.


2

Think of the function $y=x^2$. We know that the derivative is $2x$. However, $y=x^2$ can also be written as $y=x*x$. If you choose a constant $a$ to be equal to $x$, then $y=ax$. Therefore, the derivative is $a$, and therefore is $x$. That is clearly incorrect. We know it to be $2x$. The error you are making is somewhat similar to the one I just ...


0

I like to write expressions with "..." in terms os $\sum$ and $\prod$. In this case $\begin{array}\\ \dfrac{1\cdot 4\cdot 7\cdots (3n+1)}{n^5} &=\dfrac{\prod_{k=0}^n (3k+1)}{n^5}\\ &\gt\dfrac{\prod_{k=1}^n (3k+1)}{n^5}\\ &\gt\dfrac{\prod_{k=1}^n (3k)}{n^5}\\ &=\dfrac{3^nn!}{n^5}\\ \end{array} $ and this diverges extremely rapidly - so much ...


2

Let $a=0$. Then the series obviously converges to $0$. Now suppose that $a\ne 0$. Then our series is the geometric series $$a^2+a^2r+a^2r^2+a^2r^3+\cdots,$$ where $r=\frac{1}{1+a^2}\lt 1$. Since $|r|\lt 1$, the series converges. It is probably by now a familiar fact that when $|r|\lt 1$ the series $1+r+r^2+r^3+\cdots$ converges to $\frac{1}{1-r}$. So our ...


0

Just a thought: By setting the inequality as an equality and substituting $y=0$, $(x-8)^2+0+14\sin(x-8)+14\cdot 0=30$ I found the two intersections of the surface at $y=0$ at around $x\approx 3.78962$ and $x\approx 13.9171$. I wonder if it's possible to calculate the volume by using the same approach that you used to compute the area. However, the ...


2

hint: $\dfrac{a_n}{a_n+1} \leq a_n$, and $a_n < 1, \forall n \geq n_{0}\implies a_n^2 \leq a_n, n \geq n_{0}$


0

I'll do the $\mathbb{R}$ case. The reasoning can be generalized to $\mathbb{R}^n$. Let $x\in E$ and $\epsilon>0$ small enough, then as $f$ is $C^1$, there exists a neighborhood of $x$ (we can take a sphere $B(x,r)$ of some radius $r>0$) such that $$f'(x')\in[f'(x)-\epsilon,f'(x)+\epsilon].$$ Therefore, for any two points $y,z\in B(x,r)$ we have ...


1

Let $\varepsilon > 0$. Want some $\delta > 0$ such that $0 < |x-b| < \delta$ implies $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$. We start with the inequality $\left| \frac{1}{a + x} - \frac{1}{a + b} \right| < \varepsilon$ and try to manipulate it to give us insight into what $\delta$ should be. We have \begin{align} ...


0

I figured it out on my own. You can arrive at the second by integrating by parts of both terms on the left side and then taking the derivative.


1

$\forall \epsilon>0, \exists \delta>0$ such that $|x - b|< \delta \implies |\frac 1{a+x} - \frac 1{a+b}| < \epsilon$ $|\frac 1{a+x} - \frac 1{a+b}| = |\frac {b-x}{(a+x)(a+b)}| < \frac {\delta}{|(a+x)(a+b)|}$ If you can prove that $|(a+x)(a +b)|> 0$ then you are done. (which is pretty much what you told us, I just need to do certain ...


1

I think we should find a way to keep $|x-b|$ on the LHS of the inequality. For a fixed small $\epsilon>0$, we want to find $\delta>0$ such that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon$ when $|x-b|<\delta$. Note that $|\frac{1}{a+x}-\frac{1}{a+b}|<\epsilon \Leftrightarrow \frac{|x-b|}{|a+x||a+b|}<\epsilon \Leftrightarrow ...


1

Let $u=sin(\frac{x}2)$ Apply the chain rule we get $\frac{d}{dx}sin^2(\frac{x}2)=\frac{d}{du}(u^2)\frac{d}{dx}(sin(\frac{x}2))$ $\frac{d}{du}(u^2)=2u$ $\frac{d}{dx}(sin(\frac{x}2))=\frac12cos(\frac{x}2)$ So $$\frac{d}{du}(u^2)\frac{d}{dx}(sin(\frac{x}2))=2u(\frac12)cos(\frac{x}2)=ucos(\frac{x}2)=sin(\frac{x}2)cos(\frac{x}2)$$


1

Note: $\frac{d}{dx}$ and $'$ denote the derivative. $$y=\sin^2\left(\frac{x}{2}\right)=\left(\sin\frac{x}{2}\right)^2$$ I will refer to $2$ as my power, $\sin$ as my expression, and $x/2$ as my angle. Basically, we are using chain rule twice. Chain rule says we'll get power times expression to power minus one times derivative of expression times ...


4

Write down $$\begin{cases}h(x)=x^2\\g(x)=\sin x\\k(x)=\frac x2\end{cases}\;\;\implies \sin^2\frac x2=h\circ g\circ k(x)=h(g(k(x)))\implies$$ $$\left(\sin^2\frac x2\right)'=h'(g(k(x))\cdot g'(k(x))\cdot k'(x)=2\sin\frac x2\cdot\cos\frac x2\cdot\frac12=\frac12\sin x$$ using the trigonometric identity $$\sin2\alpha=2\sin\alpha\cos\alpha$$



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