# Tag Info

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The cross-product $U\times V$ of vectors $U$ and $V$ can be written in coordinates as the sum $$dy\wedge dz(U,V)\textbf{ i }- dx \wedge dz(U,V)\textbf{ j }+ dx \wedge dy(U,V)\textbf{ k }$$ (note the sign). Since the length of the cross product $U\times V$ is the area of the parallelogram spanned by them, the formula you wrote is quite suggestive. Note that ...

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This is more like the base $b$ notation for a number, but with the base varying per digit. For example: \begin{align}\frac{17}{10} &= 1 + \frac{7}{10} \\&=1 + \frac{1}{2!} + \frac{1}{5} \\ &= 1+\frac{1}{2!} +\frac{1}{3!} + \frac{1}{30}\\ &=1+\frac{1}{2!}+\frac{1}{3!}+\frac{4}{5!} \end{align} If $x$ is a real number, define the ...

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The answer of D. Fischer is shorter and elegant. Let me nevertheless provide my own elementary answer: First note that $$\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^a}-\frac{1}{(2n)^a}\right)=\sum_{n=1}^\infty \left(\frac{(2n)^a-(2n-1)^a}{(2n-1)^a(2n)^a}\right)=\sum_{n=1}^\infty \frac{(1+\frac{1}{2n-1})^a-1}{(2n)^a} \\ ... 2 Having looked at your previous question, I see the issue here. The key point is that we are computing a partial derivative here. \partial is not a number that gets canceled out. The derivative of f(k) = k^{\frac12} is \begin{equation*} f'(k) = \frac12 k^{-\frac12}. \end{equation*}. 0 First note that $\frac{\partial}{\partial x}[x^{n}]=nx^{n-1}$ So here's how the 2 appears $\frac{\partial}{\partial k}[k^{0.5}]= \frac{\partial}{\partial k}[k^{\frac{1}{2}}]=\frac{1}{2}k^{\frac{1}{2}-1}$ $= \frac{1}{2}k^{-\frac{1}{2}}= \frac{1}{2k^{\frac{1}{2}}}=\frac{1}{2\sqrt{k}}$ 2 The series is, for \operatorname{Re} a > 0, the Dirichlet \eta function,$$\eta(a) = \left(1-2^{1-a}\right)\zeta(a).$$Since \eta is an entire function, by continuity, the limit is$$\eta(0) = -\zeta(0) = \frac{1}{2}.$$0 \quad\quad\quad\quad\quad\quad\quad\quad If we look at it like this, with the condition that the ratios of the lengths are the same, we can get the following equality.$$\frac{x-x_{min}}{x_{max}-x_{min}}=\frac{y-y_{min}}{y_{max}-y_{min}}$$We can then rearrange for y & substitute the other values. 2 The 2 shouldn't be there. It's a mistake. 0 The formula you need is$$ valueY = lowerlimitY + (valueX - lowerlimitX)* \frac{upperLimitY - lowerlimitY}{upperLimitX - lowerlimitX} $$There's a geometric reason for this: the point (valueX, valueY) needs to lie on the straight line between the two points (lowerLimitX,lowerLimitY) and (upperLimitX,upperLimitY). That's what the formula above ... 0 You want linear interpolation. m=\frac{y_{1}-y_{}}{x_{1}-x_{0}} then y=y_{0}+m x_{0} 5 Then you need the derivative of$$y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$$Multiply both numerator and denominator by e^{2x}; this gives you$$y=\frac{e^{4x}+1}{e^{4x}-1}$$To make your life simpler, define u=e^{4x} so$$y=\frac{u+1}{u-1}$$and now apply$$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$$I am sure that you can take from here. 3 Using \cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x) and assuming that \cos(18^\circ)\ne0, we get$$ 16\cos^4(18^\circ)-20\cos^2(18^\circ)+5=0 $$therefore$$ \cos(18^\circ)=\sqrt{\frac{5+\sqrt5}8} $$Then, use \cos(3x)=4\cos^3(x)-3\cos(x)=(4\cos^2(x)-3)\cos(x) to get$$ \begin{align} \cos(54^\circ) &=\frac{-1+\sqrt{5}}2\sqrt{\frac{5+\sqrt5}8}\\ ...

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$\cos27^\circ=\cos(72^\circ-45^\circ)=\cos72^\circ\cos45^\circ+\sin72^\circ\sin45^\circ$ where $\cos72^\circ=\frac{\sqrt5-1}4$ and $\cos45^\circ=\frac{\sqrt2}2$.

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Using $\cos2A=2\cos^2A-1$ and as $0<27^\circ<90^\circ,$ $$\cos27^\circ=+\sqrt{\frac{1+\cos54^\circ}2}$$ and $\cos54^\circ$ has been calculated here

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Eric: You are correct that one needs $n\ge 0$ for the proof to work. Indeed, as this plot suggests, there are no zeroes on $x<0$.

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The only case where the formula doesn't work is $n=1$ (because of a $0$ in the denominator). It's perfectly valid to say, e.g., $$\int \tan^{5/2}(x)\; dx = - \int \tan^{1/2}(x)\; dx + \dfrac{\tan^{3/2}(x)}{3/2}+ C$$ (at least on an interval where $\tan(x) > 0$). It might not be too useful unless you can evaluate one of those integrals by other means ...

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You pretty much have all the right pieces. Note that we're not dealing with a biconditional though; we want to show that: $$0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon$$ Here's a cleaned up version of your proof. Given any $\epsilon > 0$, consider $\delta = \min\{1, \epsilon/7\} > 0$. Then observe that if $0 < |x - 3| < ... 1 Using the method of disks($V=π\int_{\alpha}^{\beta} f^2(x) dx$); The function$y=\sqrt{\frac{2-x^2}{2}}$has two distinct real roots$(\alpha,\beta)=(-\sqrt{2},\sqrt{2})$and the solid of revolution of this half-elipsoid will have the same volume with the revolution of the whole elipsoid as the whole elipsoid will just mark the solid of revolution's boundary ... 0$\begin{align} f(x) &= \frac 4 x \\[1ex] f(x+h) &= \frac 4 {x+h} \\[1ex] f(x+h)- f(x) &= \frac 4 {x+h} - \frac 4 x \\[1ex] &= \frac {4x - 4(x+h)}{(x+h)\times x} \\[1ex] &= \frac{-4h}{x(x+h)} \\[1ex] \frac{f(x+h)-f(x)}{h} &= \frac{-4}{x(x+h)} \\[1ex] \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} &= \frac{-4}{x^2} \\[1ex] \therefore ...

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Just write it: $$4/(x+h) - 4/x = 4(x-x-h)/x(x+h) = -4h/x(x+h)$$and divide by $h$.

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We will start by defining $$\log(x) \equiv \int_1^x \frac{dt}{t}$$ and then showing that this is indeed the inverse function of the exponential function $e^{x}$. By definition and for $x>0$, $\log(x)$ is continious and monotonely increasing with $\log(1) = 0$. We first start by showing that $\log(x) + \log(y) = \log(xy)$: $$\log(x) + \log(y) \equiv ... 3 They are the same up to arbitrary integration constant. HINT:$$\log a - c = \log a - \log e^c = \log \frac{a}{e^c}.$$1 As f^{(k)}=0 for k>5, the Taylor series is a finite sum and =f. 2 Hint: For h\neq0, we have ~\displaystyle\int_1^xt^{~h-1}~dt=\bigg[\frac{t^h}h\bigg]_1^x=\frac{x^h-1}h.~ For h=0, we have ~\displaystyle\int_1^x\frac{dt}t= =\displaystyle\lim_{h\to0}\frac{x^h-1}h.~ At the same time, ~\Big(a^x\Big)'=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^x}h=a^x~\lim_{h\to0}\dfrac{a^h-1}h.~ Do you notice anything ... 1 Hint \,\ (x\!-\!a)^2\mid P(x)\,\color{#C00}\Rightarrow\, \color{#0a0}{x\!-\!a}\mid P'(x)\,\Rightarrow\, 0 = P'(a) = \frac{a^2}2+a+1,\, contra \,a\in R\ \  QED {\rm\color{#c00}{Indeed}}\,\ P = (x\!-\!a)^2 Q\,\Rightarrow\, P'\! = (\color{#0a0}{x\!-\!a})^2 Q' + 2(\color{#0a0}{x\!-\!a}) Q\,\Rightarrow\,P'(a) = 0 Remark \  It generalizes as follows. Let ... 0 Cop-out answer Define \ln(x):=\int_1^x \frac{1}{t}dt. Then the result follows immediately. Serious answer$$ \int \color{green}{\frac{1}{x}}dx = \int \color{green}{\sum_{k=0}^\infty (-1)^k (x-1)^k} dx \\= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \tag{pulling out the constant }\\ = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C = \ln ...

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Here is the proof I wrote with Ian's guiding. Ian, thank you for pointing me in the right direction for what I was looking for; $ln(x)$ is defined to be the inverse function of $e^x$. The inverse function theorem states that $(f^{-1})'(b)=\frac{1}{(f'(a))}$, $b=f(a)$. Let $f(x)=\ln(x), f^{-1}(f^{-1}(x))=f(x)$, thus, $f^{-1}(x)=e^x$. $(f^{-1})'=(e^x)'=e^x$. ...

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