For questions related to usage of Wolfram Alpha.

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14
votes
6answers
30k views

cubic root of negative numbers

Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$. But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, ...
26
votes
4answers
2k views

A limit wrong using Wolfram Alpha

I want to calculate the following limit: $$\displaystyle{\lim_{x \to 0} \cfrac{\displaystyle{\int_1^{x^2+1} \cfrac{e^{-t}}{t} \; dt}}{3x^2}}$$ For that, I use L'Hopital and the Fundamental Theorem ...
2
votes
1answer
468 views

Computing 2d radially symmetric Fourier transforms (with Wolfram Alpha)

Let's assume that I "know" $\mathcal{F}\{\operatorname{circ}(r)\}(\rho)=\frac{J_1(2\pi\rho)}{\rho}$ $\mathcal{F}\{(1-r^2)\operatorname{circ}(r)\}(\rho)=\frac{J_2(2\pi\rho)}{\pi\rho^2}$ because I ...
12
votes
4answers
678 views

The last few digits of $0^0$ are $\ldots0000000001$ (according to WolframAlpha).

Can anybody explain how this comes about?
0
votes
3answers
248 views

Parametrization of curve

I need to find parametrization of $(1+x)y^2=(y+2x)x$. I tried to put all $x$-s on one side, and all $y$-s on other but that can't be done. Could someone please help? Thanks.
1
vote
0answers
97 views

A WolframAlpha error?

Consider the equation: $$ \dfrac{1}{\sqrt[3]{(x+3)^2}}-\dfrac{1}{\sqrt[3]{x^2}}=0 $$ that has the solution $x=\dfrac{-3}{2}$ as can be easely verified. But WolframAlpha gives no solutions (here). ...
1
vote
1answer
829 views

Plotting the intersection of multiple surfaces with WolframAlpha

I want to plot the intersection of two surfaces like in this post. But if I enter the much simplified expression ContourPlot3D[{x^2 + y^2 + z^2 - 4=0, xy=1}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}] ...
6
votes
3answers
132 views

How does Wolfram Alpha come up with the substitution $x = 2\sin u$? Integration/Analysis

I have to integrate $$ \int_0^2 \sqrt{4-x^2} \, dx $$ I looked at the Wolfram Alpha step by step solution, and first thing it does is it substitutes $x = 2\sin(u)\text{ and } \,dx = 2\cos(u)\,du$ ...
4
votes
3answers
298 views

Inverse Function (and WolframAlpha gives different Result)

I wanted to calculate the inverse function of $$ f(x) = \frac{1}{x} + \frac{1}{x-1} $$ Quite simple I thought, put $$ y = \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)} $$ rearrange and solve $$ ...
3
votes
2answers
133 views

Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$

I calculated this integral : $$\int{x \cdot \sin^2(x) dx}$$ By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a ...
3
votes
1answer
3k views

Set of modular equations in Wolfram|Alpha

Am I able to solve set of modular equations in Wolfram|Alpha, like for example: $ \left\{\begin{matrix} x \equiv 2 \pmod{3} \\ x \equiv 3 \pmod{5} \end{matrix}\right. $ ?
2
votes
2answers
133 views

WolframAlpha seems to suggest that $log(z) = log(-z)$

Wolfram-Alpha returns this snippet at the end of the "step by step solution": I'm confused by the negation of the expression inside the logarithm. The definition of complex logarithm is: $$log(z) ...
2
votes
1answer
27k views

Taylor series expansion of arctan(x) around the point 0

This is actually a technical question about why I'm getting different results when trying to do the same thing using Maxima and WolframAlpha. When I enter expand arctan(x) in WolframAlpha I ...
1
vote
1answer
283 views

Why is Wolfram Alpha wrong?

I calculated $$\tan 75^o - [\cot 13^o\cdot \cot 23^o \cdot \tan 31^o \cdot \tan 35^o\cdot \tan41^o]$$ and I got a nonzero answer: ...
1
vote
1answer
516 views

Numerical values of the Jacobi elliptic function sn: Wolfram Alpha vs. Maple vs. C++?

I have a problem to check the validity of an algorithm I've implemented in C++ to compute the Jacobi elliptic function $\mathrm{sn}(u, k)$ (inspired and improved from Numerical Recipes 3rd edition). I ...
0
votes
1answer
120 views

Help me with this mock-long-division $\frac{-x+2}{x^{2}+x-2}=\frac{-4}{3(x+2)}+\frac{1}{3(x-1)}$

$$\frac{-x+2}{x^{2}+x-2}=\frac{-4}{3(x+2)}+\frac{1}{3(x-1)}$$ Wolphram Alpha states that one can do this with long division, I cannot immediately realize it. Could someone show the trick to simplify ...