# Tag Info

3

This is a fairly technical question, and therefore will admit a technical answer. If $x,y$ are restricted to be real (or rational) numbers, then $\sqrt{-x}$ is undefined for $x>0$. Once we have an undefined quantity, we cannot proceed further, even multiplying it by zero. Hence with this restriction Desmos is correct. However, if $x,y$ are allowed to ...

2

Either order is fine. The issue with Desmos is that it is restricting the domain (artificially) to non-positive numbers because $\sqrt{-x}$ is complex for other numbers. In reality, having complex numbers is fine, and indeed, the next step, which is to multiply by zero makes the number real again. Wolfram alpha's plot is more correct, while Desmos' plot is ...

1

It's because WolframAlpha assumes polar coordinates and not spherical. Using Mathematica I get this: $$\text{Simplify}\left[\text{Laplacian}\left[\frac{\exp (i k r)}{r},\{r,\theta ,\phi \},\text{Spherical}\right]\right]=-\frac{k^2 e^{i k r}}{r}$$ whereas in polar coordinates I get this: $$\text{Simplify}\left[\text{Laplacian}\left[\frac{\exp (i k ... 2 When I work it out I get exactly what Wolfram Alpha gets. It looks to me like you simply have the formula for the Laplacian in polar coordinates totally wrong! Check out https://en.wikipedia.org/wiki/Laplace_operator#Two_dimensions . 0 You could achieve simultaneous circles in two of the three planes but not in all three. E.g.$$ x = \cos t \\ y = \sin t \\ z = \cos t $$which is a circle in the x,y-plane and the y, z-plane, but not in the x, z plane. 0 Over the real numbers \log(x) is defined only for x > 0. So for example \log(x^2) is 0 at -1, while 2\log(x) is undefined at -1. On the other hand, \log(x^2) = 2\log(x) for x > 0, and WolframAlpha does tell you so. 4 For positive real x,$$2\ln(x) = \ln(x^2).$$As others have pointed out, this does not hold for all reals. The domain of f(x)=\ln(x^2) is (-\infty, 0)\cup (0,\infty), while the domain of g(x)=2\ln(x) is (0,\infty). Functions that have different domains are different functions. 1 No, it isn't. When you say f(x) = g(x), where x is not a precise value, what is implied is that you are saying f \equiv g; that's, f and g are identical as mappings. This is clearly false here, because \ln(\cdot) is defined only on (0, \infty), whereas \ln(\cdot^2) is defined on (-\infty, \infty) \setminus \{0\}. In brief, your input is ... 2 Let x=-1. Then,$$2\ln(-1) \Rightarrow \;\; \text{undef}$$and$$\ln[(-1)^2] = \ln(1) = 0$$So not always. 9 You can show them that they are adding the digits of multiples of 9 between 18 and 90 and then just list them. 18 \implies 1+8=9 27\implies 2+7=9 36\implies 3+6=9 45\implies 4+5=9 54\implies 5+4=9 63\implies 6+3=9 72\implies 7+2=9 81\implies 8+1=9 90\implies 9+0=9 While this approach is not sophisticated, it might be ... 9 For any integer k, 3(3k+3)=9(k+1) is a multiple of 9, and one can prove that such a number has digital root 9. 1 As anakhronizein wrote product sin (k), k=0..10 The trick lies in the main page, where you find on the right side the bar "Example". If you click it, you find pretty much everything. For example. the answer to your question can be found by then choosing "Mathematics", and finally "Calculus & Analysis". I hope it helps for the future! 1 For example: product (1)/n, n=1..10 Computation here. 2 It is a plot of \Gamma(x+1), known to equal x! at integers. 0 The problem is rounding and precision. Assuming that we are using radians$$ \sin(0.0000000001)\doteq0.0000000000999999999999999999998333333333 $$and$$ \tan(0.0000000001)\doteq0.0000000001000000000000000000003333333333 $$both of which, when rounded to 20 significant places are$$ 0.000000000100000000000000000000 $$Thus, assuming the calculator has no ... 0 tan(x) and sin(x) are equal numbers in a calculator for small enough x because the calculator does not represent enough digits. Hence the numerator turns to 0. In the Windows calculator, your example works fine, but: sin(0.0000000000000001) = tan(0.0000000000000001) = 1.7453292519943295769236907684886e-18 The difference is in the mantissa (the first ... 38 As a rule of thumb, try express everything in term of either \sin(x) or \cos(x) to see whether there is any obvious cancellation. For this case, we have$$\frac{\tan(x) - \sin(x)}{\sin(x)^3} = \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin (x)^3} = \frac{1-\cos(x)}{\cos(x)(1-\cos(x)^2)} = \frac{1}{\cos(x)(1+\cos(x))}$$you don't need any calculator to ... 2 Tip of the day: You can focus in on a region by typing into WolframAlpha: Plot[y^(1/y) == x, {x, 0, 5}] Plot[y^(1/y) == x, {x, 0, 1.5}] Plot[y^(1/y) == x, {x, 1, 1.5}, {y, 1, 20}] ... and so forth. 2 This is not an answer but it is too long for a comment. If you are concerned by the plot of y as a function of x such that \space y^{1/y}=x, it means that$$y=-\dfrac{W(-\log (x))}{\log (x)}$$where appears Lambert function. In the real domain, x must be positive and function W(a) is only defines if a \geq -\frac 1e which makes 0 < x \leq ... 3 The Wolfram Alpha one doesn't seem to start at  x=0  (look at the numbers at the bottom). 3 If you can use the standard limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}=\frac{1}{3},\ \text{and}\ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6},\ \ \ \ \text{and}\ \ \ \ \ \lim_{x\rightarrow 0}\frac{x}{\sin x}=1 \end{equation*} then, \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\tan x-\sin x}{\sin ^{3}x} ... 8 The problem is not just that \tan x - \sin x is approaching zero rapidly; the real problem is that as x approaches zero, \tan x - \sin x approaches zero much more rapidly than either \tan x or \sin x, because (as shown by Ross Millikan) \tan x - \sin x \approx \frac12 x^3 but \tan x \approx x + \frac13 x^3 and \sin x \approx x - \frac16 x^3. ... 13 One way to do it by hand is to use Taylor Series. For x\to 0, \tan x = x+\frac 13x^3+o(x^3), \sin x = x - \frac 16x^3+o(x^3) So$$\frac {\tan x - \sin x}{\sin^3 x}= \frac {\frac12x^3+o(x^3)}{x^3}=\frac12+o(1)\to\frac12

27

$0.0000000001$ is too small of a number: the calculator got such a small answer for the top that it assumed it was zero (since the values subtracted in the numerator were rounded to the same value). The bottom was non-zero, so there was no division by zero error. Zero divided by anything non-zero is zero. If you're going to use the calculator method, I ...

1

That is because Wolfram alpha uses the formula for the DFT as $X(k)=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}x(n)e^{-j2\pi k/N}$ -- Samrat Mukhopadhyay The terms like $0.i$ appear because the coefficients are in general complex numbers, but in some of them the imaginary part is zero within machine precision. (Theoretically, it is exactly zero, but the computation ...

3

Let $u=t-x$ , Then $x=t-u$ $\dfrac{dx}{dt}=1-\dfrac{du}{dt}$ $\therefore1-\dfrac{du}{dt}=\dfrac{A(1-t+u)}{t-t^2}-\dfrac{B(t-u)-C(t-u)^2}{(t-t^2)u}$ $u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{Au^2}{t-t^2}+\dfrac{Cu^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$ $u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{(A+C)u^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$ ...

2

I tried to use Mathematica to solve your problem and it solved pretty fast. Here is what I've done.

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