# Tag Info

12

HINT I would say: $\quad\displaystyle \lim_{x\to0}\frac{\sqrt{1-\cos(x)}}{x}=\left(\lim_{x\to0}\frac{1-\cos(x)}{x^2}\right)^{1/2}=\cdots$

6

Hint: $$1-\cos x=2 \sin^2 \frac{x}{2}.$$

6

$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$

4

When you do it with integration by parts, you have to go in the "same direction" both times. For instance, if you initially differentiate $e^{2 \theta}$, then you need to differentiate $e^{2 \theta}$ again; if you integrate it, you will wind up back where you started. If you do this, you should find something of the form $$\int e^{2 \theta} \cos(3 \theta) d ... 4 HINT: you Need the power and the chain rule$$2\,{1\arctan \left( 3\,{\frac {x}{2\,{x}^{2}-8}} \right) \left( 3\, \left( 2\,{x}^{2}-8 \right) ^{-1}-12\,{\frac {{x}^{2}}{ \left( 2\,{x} ^{2}-8 \right) ^{2}}} \right) \left( 9\,{\frac {{x}^{2}}{ \left( 2\,{ x}^{2}-8 \right) ^{2}}}+1 \right) ^{-1}} $$the first derivative of$$\frac{3x}{2x^2-8}$$is given by ... 4 This is word for word the exact statement of the fundamental theorem of calculus. The fundamental theorem of calculus states that if f is continuous on [a,b] then the function F(x)=\int_a^x f(t) dt is differentiable and F'(x)=f(x) 4 Consider h>0 then there is n so that  \frac{1}{(n+1)\pi +\pi/2} < h\le \frac{1}{n\pi + \pi/2}. In this interval (called I_n), \cos(1/t) is positive (resp. negative) if n is odd (resp. even). Also$$\int_0^h \cos\left(\frac 1t\right) dt = \sum_{k=n+1}^\infty \int_{I_k} \cos\left(\frac 1t\right) dt + \int_{\frac{1}{(n+1)\pi + \pi/2}}^h ...

4

Although you have already answered your own question, another way to see it is to consider the $C^2$ function $h(x) := f(x) - g(x)$. This function must have a local minimum at the point $x_0$, since that point is at least as small as all other nearby points. But since it is a minimum, we must have $h'(x_0) = 0$, and $h''(x_0) \ge 0$, which implies $f''(x_0) ... 4 You should always be careful when dividing, because you'll run into trouble if you try and divide by 0. We have that for all$x$, $$f(x+1)(f(x) + 1) = -1$$ Notice that$f$can never be zero, then, because that would mean$0$on the left-hand side was$-1$on the right-hand side. This justifies your division step, and proves that$0$is never in the range ... 4 Assume that the range of$f$is all of$\Bbb R$. Then there are$a,b\in\Bbb R$with$f(a)=0$and$f(b)=-1$. But then $$1=f(a)f(a-1)+f(a)+1=0$$ and $$1=f(b+1)f(b)+f(b+1)+1=0,$$ both of which are absurd. 3 This is not a bug in Mathematica, it is a feature; see Examples$\to$Scope in the Mathematica documentation for the command$\texttt{Limit[]}. As that documentation shows, one can compute left- and right-handed limits of real-valued functions in Mathematica respectively with the commands Limit[1 / x, x -> 0, direction = -1] and Limit[1 / x, x -> ... 3 Partial integration \begin{align} \int \cos \bigg(\frac{1}{t} \bigg) dt &= \int t^2 \frac{1}{t^2}\cos \bigg(\frac{1}{t} \bigg)dt \\ &= -t^2 \sin \bigg(\frac{1}{t} \bigg) + \int 2 t \sin \bigg(\frac{1}{t} \bigg) dt \\ \end{align} 3 Another hint : Multiply the top and the bottom by\sqrt{1+\cos(x)}$. 2 You may observe that $$\sqrt{1-\cos x}=\sqrt{2\sin^2 (x/2)}=\sqrt{2}\:|\sin(x/2)|.$$ 2 Let$g=\frac1h$and$s=\frac1t. Then integration by parts gives \begin{align} \lim_{h\to0}\frac1h\int_0^h{\cos\!\left(\frac1t\right)\mathrm{d}t} &=\lim_{g\to\infty}g\int_g^\infty{\frac{\cos(s)}{s^2}\,\mathrm{d}s}\\ &=\lim_{g\to\infty}g\left[-\frac{\sin(g)}{g^2}+2\int_g^\infty\frac{\sin(s)}{s^3}\,\mathrm{d}s\right]\\ ... 2 As indicated by @levap in the comments, you are likely confusing the (right-hand side, RHS) derivative at 0 with the (RHS) limit of the derivative. I will just add details to clarify. Assuming you mean f(0)=0. The derivative at 0 is by definition \displaystyle\lim\limits_{h\to0}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0}\frac{h^2\sin(\frac1h)-0}h = ... 2 Beside the trivial solution f=c_1, as Paul Evans commented, the only solution of the differential equation\left(\frac{df}{dx}\right)^2=\frac{d^{2}f}{dx^{2}}$$is$$f=c_2-\log \left(c_1+x\right)$$This is obtained setting first p=\frac{df}{dx} which reduces the equation to p^2=\frac{dp}{dx} which is separable and easy to solve. Once p is obtained, ... 2 \cos x = 1 -\frac12x^2+\frac{1}{24}x^4+\cdots e^{-\frac12x^2} = 1 -\frac12x^2+\frac{1}{8}x^4+\cdots So if \varepsilon is small and |x| < \varepsilon, they differ by at most \approx \frac{1}{12}\varepsilon^4. 2 Just for some new ideas! I would reccomend a completely different method. This method uses the Gudermannian \text{gd} function. So you would substitute x=\text{gd}(a);\text{d}x=\text{sech}\space a\text{d}a That transforms the integral into:$$\int \frac{\tanh a}{\tanh a+\text{sech}\space a}(\text{sech}\space a)\mathrm da$$Through some hyperbolic trig ... 2 We don't need any calculus to find the minimum surface area - AM-GM works fine. Solution 1. AM-GM We have the surface area as xy+2yz+2zx with the constraint of xyz=216.$$xy+2yz+2zx = xy+\frac{432}{x}+\frac{432}{y} \ge 3 \sqrt[3]{xy \cdot \frac{432}{x} \cdot \frac{432}{y}} = 3\sqrt[3]{432^2} = 108 \sqrt[3]{4}$$The equality holds at x=y=6\sqrt[3]{2} ... 1 Suppose k<x_n\leq k+1. We want to show that there exist x_m>k+1, so we can plug m back in to find a x_{m'}>k+2 and so on. Consider x_{n+(k+1)^2}. Suppose for sake of contradiction it is \leq k+1. Then using the fact that \{x\}_{i=1}^\infty is strictly increasing, it follows that for all n\leq l \leq n+(k+1)^2, k<x_l\leq k+1. As ... 1 Substituting x=r\cos\theta and y=r\sin\theta,$$\int_0^{2\pi}\int_0^3\frac{5r}{\sqrt{5^2-r^2}}drd\theta=\int_0^{2\pi}\left[5\sqrt{5^2-r^2}\right]_3^0d\theta=10\pi$$Using the already known formula for spherical cap (https://en.wikipedia.org/wiki/Spherical_cap) and r=5,h=1,$$A=2\pi rh=10\pi$$Those two results match. 1 a) f(x) = 0 if x < 1; f(x) = 1 if x \ge 1. f(x) is continuous on (-\infty,1). f(x) is continuous on [1, \infty). But f is not continuous on (-\infty,1) \cup [1, \infty) = \mathbb R. b) If the closures of C and D are disjoint then any point of C will not be a limit point of D and no point of D will be a limit point of C. Let x \in ... 1 You cannot express it because the number is transcendental. You could express it as a limit of a sum (@AJ Stas comment) 1 Given \epsilon > 0 and x, we want to show that there exists a \delta>0 such that |F(x)-F(y)|<\epsilon whenever |x-y|<\delta. Since [-x,x] is compact, the minimum is attained at some point(s) t_0\in[-x,x]. Suppose first that \pm x are not points of minimum, i.e. the minimum is attained strictly inside. Let \delta be such that ... 1 Well firstly, sin(x) \approx x when x is very small. You can see this by taylor expansion of sin(x) as you did in the question, and noticing that x^n, n > 1 goes to 0 very fast. 4 degrees is quite close to zero, so you could conclude that$$sin(4^\circ) = sin(\frac{\pi}{45}) \approx \frac{\pi}{45} \approx 0.069813$$And we are done if we ... 1 \int \dfrac{s^{2} + \sqrt{s}}{s^{2}}ds=\int (1+ s^{-3/2})ds 1$$\frac{\cos(4\pi a)-\cos(4\pi b)}{8\pi}=0\Longleftrightarrow\cos(4\pi a)-\cos(4\pi b)=0\Longleftrightarrow-\cos(4\pi b)=-\cos(4\pi a)\Longleftrightarrow\cos(4\pi b)=\cos(4\pi a)\Longleftrightarrow4\pi b=4\pi a+2\pi n_1\Longleftrightarrow\space\space\vee\space\space 4\pi b=2\pi n_1-4\pi a\Longleftrightarrow$$... 1 the roots are real, they are the eigenvalues of the symmetric real matrix$$ \left( \begin{array}{ccc} 0 & -a & -b \\ -a & 0 & -c \\ -b & -c & 0 \end{array} \right) $$You may bound the eigenvalues by various operator norms, either the induced L_1 or L^\infty norm tells you that$$ |t_j| \leq \max \{ |a| + |b|, |a| + |c|, |b| + ... 1 Hint: convince yourself that it is enough to investigate the convergence of\int_{-1}^1 \frac{dx}{x^{5/3}}\$. Then observe that the integrand is odd...

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