# Tag Info

2

Since the function is not defined for $x=0$, it's not really meaningful to have a single constant of integration for the whole thing. The most general function $F$ (not defined at $0$) for which, at each point $x\ne0$, $F'(x)=\frac{1}{\sqrt{|x|}}$, is $$F(x)=\begin{cases} -2\sqrt{-x}+c_1 & \text{if x<0}\\ 2\sqrt{x}+c_2 & \text{if x>0} ... 9 Using \operatorname{sgn}(x) is just a (half-dirty) trick to put the two cases into one. Put in -1 vs. +1 for \operatorname{sgn}(x) and your eyes will be open. 0 Another person asked the same question. It's just a quirk of wolfram. It's something akin to their system saying that there are multiple points at zero. 3 OK I tried it. Now what? Is this different from yours? 3 This is equivalent to American Mathematical Monthly Problem 11148 published in April 2005. Let u=x-1. Then rewrite the integral as$$\int_{0}^{\infty}{\frac{u^8-4u^6+9u^4-5u^2+1}{u^{12}-10u^{10}+37u^8-42u^6+26u^4-8u^2+1}}du. The value of this integral is equal to the value of the original integral. The method to solve this equivalent integral can be ...

-1

Mathematica has a lot of weirdness surrounding its use of floor. For instance, the integral of $[x]$ yields $x[x]&. Such an answer is sorta correct but it lends credence to the fact that its implementation of it is highly buggy and incomplete with regards to the non-intuitive uses of floor. It's highly likely that the number it gave is something akin to ... 0 You could have negative values for K in your solutions. There are two branches to this DE. If you had a suitable initial condition to the DE then wolfram (I don't have maple) find it, e.g.: http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%28y%5E2-1%29%2F%282y%29%2C+y%280%29%3D-1%2F2 As mentioned in the comments a complex value of$C$could give rise to a ... 0 You are right, except that you should be more careful with the constant solutions$y=\pm 1$. (In the final answer, they correspond to$K=0$, but you have defined$K=\pm e^C$, which can never be zero, so you have to "repair" the solution by including those cases "by hand".) -1 Solve[ ..., Element[..., Integers]] 4 This is a good example for why one should not blindly trust Wolfram Alpha when it comes it multidimensional limits. Such limits are hard to do for a computer as the result can depend on the path taken towards the limit-point and there are infinitely many paths to approach any given point. When the limit does not exist the standard solution method is to find ... 0 It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for$x^{2/3}$it is treating the$x$as a complex variable, which can cause all kinds of shenanigans, whereas$\sqrt[3]{x^2}$has$x$as a purely real variable, since there is no way that$x^2$can be negative. Thus in the first case you must consider cube roots ... 2 Paste this into Wolfram Alpha: Simplify (D[#, t, t] - 3D[#, t] - 4#)& @ (-t^2 - 3t/2 - 11/8 ) The above has the form of$O(y)$where operator$O$is defined as$d^2/dt^2 - 3d/dt -4$which operates on (via the @-sign) the polynomial$-t^2 - 3t/2 - 11/8 \$. To define a function or an operator, use # as the variable and terminate the definition with ...

1

It might be worth pointing out that you can use WolframAlpha to get the correct result by simply specifying greater precision. You can specify 1.5 to 50 digits of precision like so: 1.550. So your query would be: 1e48 mod 1.550 You can even specify exact precision: 1e48 mod 3/2

Top 50 recent answers are included