# Tag Info

0

HINT: this integral has an elemetary antiderivative and you can use the tan half angle substitution

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I managed to figure this out a little late. In the given functional equation, replace $x$ with $-x$. The equation is now written as: $f(2x^2-1)=-2xf(-x)$ But it is given that $f(2x^2-1)=2xf(x)$ Therefore, $-2xf(-x)=2xf(x)$ $=>$ $-f(-x)=f(x)$ A rather simple functional equation which can be easily differentiated 4 times. The answer is $0$.

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Actually I have found another solution, more satisfactory to my taste (although maybe not 100% complete). Inspired by the answer of this question on integrating the $\text{sinc}$ function, I thought of the following procedure: Write the exponential as a sum of sines and cosines $$\int_0^\infty\text{d}x\frac{e^{i k ... 0 Let u = \frac{x}{1-x}\to x = \frac{u}{1+u} thus we get$$ \int_0^{\infty}\frac{1}{3u^2+3u + 1}\ln u du $$then you have$$ 3u^2+3u+1 = \frac{1}{4}\left[12\left(u+\frac{1}{2}\right)^2+1\right] $$then using the sub t = \sqrt{12}\left(u+\frac{1}{2}\right) you get$$ ...

0

We can also use the substitution $u=\sqrt{1-x}$, then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}$ and $\sqrt{1+x}=\sqrt{2-u^2}$. We will also use $u=\sqrt2\sin(\theta)$ \begin{align} \int\frac{\sqrt{1+x}}{\sqrt{1-x}}\,\mathrm{d}x &=-2\int\sqrt{2-u^2}\,\mathrm{d}u\\ &=-2\int\sqrt2\cos(\theta)\cdot\sqrt2\cos(\theta)\,\mathrm{d}\theta\\ ... 0 The way I understand it is you have the equation x^2+y^2+z^2=4 $$which is equivalent to$$ f(x,y)=z=\pm \sqrt{4-x^2-y^2}, $$therefore$$ \frac{\partial{f}}{\partial{x}}=\pm \frac{x}{\sqrt{4-x^2-y^2}} $$Perhaps more context on where this question comes from could help clarify things. 1 When you have a parametric equation in 2D it is usually defined as \alpha(t)=(\alpha_{1}(t);\alpha_{2}(t)). The velocity vector at time t_{0} would then be the derivative of \alpha at t_{0}, which is,$$ \alpha'(t_{0})= \lim_{t\to t_{0}}\dfrac{\alpha(t)-\alpha(t_{0})}{t-t_{0}} Assuming that \alpha_{1}(t) and \alpha_{2}(t) are both ... 0 If \rho = a + bi (real and imaginary parts), \log denotes the branch whose imaginary part lies in (-\pi/2, 3\pi/2), and \begin{align*} x^{\rho} &= \exp\bigl((a + bi) \log x\bigr) \\ &= \begin{cases} |x|^{a} \exp(ib\log x) & \text{if x > 0,} \\ |x|^{a} e^{-b\pi} \exp\bigl(i(a\pi + b\log |x|)\bigr) & \text{if x < 0,} ... 0 That's the resultant of (2D) speeds in i,j directions and basically its Pythagoras theorem for small parts of velocity in given directions 1 So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here : Lemma 3 : Let X be a càdlàg square integrable martingale and {\xi} be a bounded predictable process. Then, \int\xi\,dX is a square integrable martingale. In your case W is a Brownian motion, so you fit the conditions for the lemma as a ... 1 The derivative is a limit at a single point. When the limit exists for a given point this specific point is said to be differentiable. A differential equation is an equation that generalises all of the differentiable points for a given function. In your case the function f(x) = x^{3} has all of its derivatives generalised by the equation g(x) = 3x^{2}. ... 0 By the chain rule, we havea = \frac {dv} {dt} = \frac {dv} {ds} \frac {ds} {dt} = (6s^2 + 5) v = (6s^2 + 5) (2s^3 + 5s),$$so you're right. 1 Consider f(x) = -x. Then \frac{f(x) - f(x_0)}{x-x_0} = -1 for all x \neq x_0, and so for all sequences x_n converging to x_0 without reaching it, \frac{f(x_n) - f(x_0)}{x_n-x_0} = -1, the lim sup of which is -1 \le 0. 1 Answer : the sum evaluates to \varphi(2000)=800 . why? because : Claim : For any positive integer n we have:$$\varphi(n)=\sum_{i=1}^{n}\gcd(i,n)\cos\left(\frac{2\pi\ i}{n}\right) \tag 1$$where \varphi(n) is Euler's totient function Proof Let \zeta_n := e^{\frac{2\pi i}{n}}, define the function s for every positive integer n by :$$ ...

0

Almost correct. The integral should be $$\int_{-5}^{0}dy\int_{-5}^{-\sqrt{25-y^2}}f(x,y)dx+\int_{-7}^{-5}dy\int_{-5}^{5}f(x,y)dx+\int_{-5}^{0}dy\int_{\sqrt{25-y^2}}^{5}f(x,y)dx$$

0

HINT: $$\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}\left(\cos\left(\pi nx\right)\right)dx= \Re\left[ \int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}e^{i\pi nx}dx\right]$$ Then observe that $$-\frac{x^2}2+i\pi nx=-\left(\frac x{\sqrt2}-\sqrt2i\pi n\right)^2+2\pi^2 n^2$$ and use the well known integral $$\int_{\Bbb R}e^{-s^2}\,ds=\sqrt{\pi}\;\;$$ with a suitable ...

3

Another chance is given by the expansion of $\cos(\pi n x)$ as a Taylor series: $$\cos(\pi n x) = \sum_{m\geq 0}\frac{(-1)^m (\pi n)^{2m} x^{2m}}{(2m)!}\tag{1}$$ together with the fact that: $$\int_{\mathbb{R}} x^{2m} e^{-x^2/2}\,dx = \int_{0}^{+\infty} x^{m-1/2} e^{-x/2}\,dx = 2^{m+1/2}\cdot\Gamma\left(m-\frac{1}{2}\right).\tag{2}$$ These identities ...

7

Here's a funny one: Define the function $f$ as $$f:\mathbb{R}\longrightarrow\mathbb{R},\ a\longmapsto\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\cos(a x)\,\mathrm{d}x.$$ It is well-known that the improper integral defining $f$ is convergent. We can apply the differentiation theorem to differentiate this improper integral (proof left to the reader) and we ...

1

This is not always possible. Take $f_1(x)=x$ and $f_2(x)=x+1$. Then $f_1^{-1}(y)=y$ and $f_2^{-1}(y)=y-1$. Therefore for any $y_0$: $$f_1^{-1}(y_0)-f_2^{-1}(y_0)=y_0-y_0+1=1\neq 0.$$

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Yes, it can be done using two double integrals. Here you are dividing the triangle into two parts by the vertical line $x=2$. The equation of the line joining $(1,2)$ and $(2,1)$ is $x+y=3$, and that of the line joining $(1,2)$ to $(3,3)$ is $x-2y+3=0$. You can set $x=1$ to $x=2$, and then you will have the range of $y$ as $y=3-x$ to $y=\frac{x+3}{2}$. ...

1

You can always parameterize a triangle with $u,v$, where $u,v>0$ and $u+v<1$. This always leads to an integral $$\int_0^1 \int_0^u\ldots dv\,du$$ Convex interpolation inside the triangle with vertices A,B,C is then $$T=Au+Bv+C(1-u-v)$$ Imagine that "u" is the variable that tells you how much you move from C to A, and "v" the same from C to B. ...

3

Maybe you are searching for the differential not the derivative. The differential of a function in a point is a linear map of the displacement (by definition). Please note that linear map means that the transformation law is linear AND the domain and the codomain of the map is a linear space. Indeed the displacement can be taken to be as large as you want ...

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And a simple image that solidifies what everyone has said already :)

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The slope of a single tangent line is a constant: it's the $m$ in the equation $y=mx+b$ for that line. So if the derivative of $f(x)$ is $2x$ (for example) then you already have a different slope at each point of $y=f(x)$, and therefore an entirely different tangent line at each point.

3

The derivative at a precise point $x$ is the slope of the tangent line at this point. But the derivative is a function so the slope is moving while $x$ is moving. Actually the derivative can be viewed as the variation of the slope of the tangent line. For exemple, the slope of the tangent line of the function $x \to x^3$ increases faster than the slope of ...

2

Since your surface is a cylindar of radius $2$, the surface element can be written as $\mathrm{d}S=2\,\mathrm{d}\theta\,\mathrm{d}z$. Then your integral can be written as: $$I=2\iint\limits_{(\theta,z)\in[0,2\pi]\times[0,3]}z^2\,\mathrm{d}\theta\,\mathrm{d}z,$$ and using separation of variables: ...

1

If I well understand (which is not sure), you have two given functions $f_1(x)$ and $f_2(x)$ The inverse function of $y=f_1(x)$ is $x=f_1^{-1}(y)$ The inverse function of $y=f_2(x)$ is $x=f_2^{-1}(y)$ You want to find a root $y_0$ of the equation $f_1^{-1}(y)=f_2^{-1}(y)$ So, you first have to $$\text{solve}\quad f_1(x)=f_2(x)\quad \text{for } x\quad ... 6 No substitutions:$$ \int\left(\frac{1}{\sqrt{1-x^2}}-\frac{-x}{\sqrt{1-x^2}}\right)\,dx =\arcsin x-\sqrt{1-x^2}+c $$You can also do that way; continue with u=\sqrt{t}, so t=u^2 and dt=2u\,du; so you get$$ \int\frac{4u^2}{(u^2+1)^2}\,du= \int 2u\cdot\frac{2u}{(u^2+1)^2}\,du $$Noticing that 2u is the derivative of u^2+1 you can use integration ... 0 Hint: Note that for y=0 we have x=0 so this is the inferior limit for the integral and the axis of rotation is external (it seems that your mistake is here). Your volume is formed by concentric shells of radius 2-x and height y=x^4, so each shell of thickness dx has volume 2\pi(2-x)dx \cdot x^4 and the volume is:$$ V=2\pi\int_0^1(2-x)x^4dx $$... 1 You can find the area under the curve: Pink area A_1 =\int\limits_0^6 \sqrt{16-2x} \mathrm {dx}. You can do this using integration by substitution. You already have the area of the trapezium, so just need to subtract A_1 from 21 to find the required area. Or you can imagine the curve turned the other way and find the area between the curve and the ... 2 Hint: Since you have the upper (y>0) half parabola you can integrate the function$$ y=\sqrt{16-2x} $$so the area is:$$ A_f=\int_0^6\sqrt{16-2x}dx $$that, with the substitution 16-2x=t \rightarrow -2dx=dt becomes:$$ A_f=\int_6^4\sqrt{t}dx $$can you do from this? 0 Let the parametric equation of the tangent be$$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$where \theta is unknown. Plug in the equation of the ellipse to get$$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\\ ...

7

Let's also look at it upside down. You can define analytical (infinitely differentiable) functions with their Taylor series $\sum \frac{a_n}{n!}x^n$. Taylor series are simply all finite and infinite polynomials with coefficient sequences $(a_n)$ that satisfy the series convergence criteria ($a_n$ are the derivatives in the chosen origin point, in my example, ...

2

Note that this is a differential equation. The equivalent Picard integral equation is $$f(x)=f(0)+\int_0^x g(f(t))\,dt$$ From here it is trivial to observe that if $f$ is $C^n$ or better, then the composition $g\circ f$ is also at least $C^n$ and thus the anti-derivative $C^{n+1}$. Which gives that $f$ is also $C^{n+1}$ and so on.

3

This is the same as maximizing $f(x,y) = 2^x + 2^y$ subject to the constraint $g(x,y) = x^2 + y^2 = 1$. Using Lagrange multipliers, we look for points where $g(x,y) = 1$ and $\nabla f(x,y) = ((\log 2)2^x,(\log 2)2^y)$ and $\nabla g(x,y) = (2x,2y)$ are linearly dependent. This is equivalent to $x2^{-x} = y2^{-y}$. But since the function $h(t) = t2^{-t}$ is ...

1

By approximating the partial sums by definite integrals (see here), we have that $$\log(N+2)-\log2=\int_1^{N+1}\frac1{x+1}\mathrm dx\le\sum_{n=1}^N\frac1{n+1}\le\int_0^N\frac1{x+1}\mathrm dx=\log(N+1)$$ for $N\ge1$. Hence, we can bound the partial sums from above with $\log(N+1)$. However, this bound depends on the number of terms in the partial sums and ...

0

In most cases, the solution doesn't exist. As others pointed out, if $z=0\mod 3$, then there's no solution. If $z=\pm 1 \mod 3$, then you have $x$ that is even (plus)/ odd (minus). Even if $z$ has 4k bits, it's still reasonable to try all $x$ (2048 checks are no big deal). 5-minute python code (if no arg given, just goes through first million), printing ...

1

This probably isn't the kind of answer you're looking for, but it's one I like because it doesn't use calculus. The ellipse is the image of the unit circle under the linear mapping $T(x,y) = (ax,by)$. The tangent line to the ellipse is the image under $T$ of the tangent line to the circle. We have $n = T^{-1}(x_0,y_0) = (x_0/a,y_0/b)$, the corresponding ...

0

Use the fact that the gradient of a differentiable function at a point is orthogonal to the level set of that function passing through that point. Here, $$f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2},$$ and $$\vec\nabla f(x_0,y_0)=\frac{2x_0}{a^2}+\frac{2y_0}{b^2}.$$ Hence, for $(x_0,y_0)\in\mathbb{R}^2$ such that $f(x_0,y_0)=1$, an equation of the tangent line ...

2

We will use the following standard limits $$\lim_{x \to 0}\frac{\arctan x}{x} = 1 = \lim_{x \to 0}\frac{\arcsin x}{x},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ We have \begin{align} L &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - \sqrt[3]{1 - \arcsin 3x}} {\sqrt{1 - \arctan 2x} - \sqrt{1 + \arcsin 2x}}\notag\\ &= \lim_{x ...

4

Suppose $f \in C^k$, then $f^{(k+1)} = f'^{(k)} = (g \circ f)^{(k)}$. Since $g \in C^\infty$ and $f \in C^k$, we have that $g \circ f \in C^k$, and thus that $f^{(k+1)} = (g \circ f)^{(k)} \in C^0$, so it follows that $f \in C^{k+1}$. QED Or even more succinctly as John Ma pointed out below: Suppose $f \in C^k$, then since $g \in C^\infty$, we have that ...

4

Notation: For any function $h$ let $h^{(0)}=h$, and let $h^{(n)}$ be the $n$th derivative of $h$ for $n\geq 1.$ Suppose for some $n\geq 0,$ there is a polynomial $P_n$ in $(2 n+2)$ variables such that $$f^{(n+1)}=P_n(g^{(0)}(f),...,g^{(n)}(f),\;f^{(0))},...,f^{(n)}).$$ Then every term occurring in the polynomial on the RHS is differentiable (because $g$ is ...

0

Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$ (polar coordinates), hence $$x^2+y^2=r^2\cos^2(\theta)+r^2\sin^2(\theta)=r^2$$Now,$$\lim_{(x,y)\to(0,0)}\frac{\ln(1-x^2-y^2)}{x^2+y^2}=\lim_{r\to 0}\frac{\ln(1-r^2)}{r^2}\underset{\text{L'Hopital}}{=}\lim_{r\to 0}\frac{-\frac{2r}{1-r^2}}{2r}=-\lim_{r\to 0}\frac{1}{1-r^2}=-1$$

1

Hint: Use the squeeze theorem and the inequality $$\frac{-x}{1-x} \leqslant \ln(1-x) \leqslant -x$$

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HINT: Rationalize the D & N using $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$ Then use $\lim_{u\to0}\dfrac{\arctan u}u=\lim_{y\to0}\dfrac y{\tan y}=1$ setting $\arctan u =y$ Similarly for $\arcsin$

-4

Hint You need to impose some conditions on $x$ to insure convergence

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If $y < x_{0} < z$ then we have (by the assumptions given in question) $$f(y) \leq f(x_{0}) \leq f(z)\tag{1}$$ Now consider the set $$A = \{f(x) \mid x > x_{0}\}\tag{2}$$ then clearly $A$ is bounded below by $f(x_{0})$ and hence $\inf A = a$ exists and $a \geq f(x_{0})$. Similarly if $B = \{f(x) \mid x < x_{0}\}$ then $b = \sup B$ exists and $b ... 1 Let$f^+(x_0)=\inf_{x \in (x_0, \infty)} f(x)$, this is our candidate limit point. By the definition of$\inf$, we can find an for any$\epsilon>0$, there is an$x$s.t.$f(x)<f^+(x_0) + \epsilon$, and since$f(x)$is monotonically increasing, this holds for all$x \in (x_0,x)$. Similar logic shows this in the other direction. If the limit from below ... 1 With your hypotheses, and for the function$f(x)=x$, the Fourier series of sines is equal to$f$everywhere, except at$x=\pi$, where it is$0$(as you can check explicitly, at this point, simply substituting$x=\pi$in the series that you already wrote). More generally, the Fourier series of sines of a function$fwith the hypotheses that you described ... 2 Your substitution is a bit off. Begin with \begin{align*} \int_0^{0.6}{\frac{x^2}{\sqrt{9-25x^2}}}\, dx &= \int_0^{0.6}{\frac{x^2}{\sqrt{25\left(\frac{9}{25}-x^2\right)}}}\, dx \\ &= \frac{1}{5}\int_0^{0.6}{\frac{x^2}{\sqrt{\frac{9}{25}-x^2}}}\, dx. \end{align*} Now let\$\begin{align*} x&=\frac{3}{5}\sin \theta \\ ...

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