Recent content by W3bbo

(Apologies for the lack of LaTeX formatting - I usually do my typesetting with MathType, but I only have Microsoft Equation Editor on this computer which doesn't have LaTeX export). This isn't a homework question, but is problem I've discovered I'm facing after diving into another problem, but...

Easier, yeah; but due to requirements elsewhere in the program I need to be able to derive a permutation from a number.

I'm writing a program that generates every possible valid Mastermind code. That itself is easy. There are 6 colors in 4 possible positions. Cycling through them all is done like so: for(int i = 0 to 1295 ) { // 1295 == 6^4 - 1, there are 1296 possible permutations of colors color1 = (...

Ah, how careless of me. And we're done: \displaylines{ I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx \cr \sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right) \cr I = \int {2\sin \left( x \right)\cos ^2 \left( x \right)dx} \cr u = \cos...

Thanks all. This does my working look alright? \displaylines{ I = \int {\sin \left( {2x} \right)\cos \left( x \right)} dx \cr \sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right) \cr I' = 2\sin \left( x \right)\cos ^2 \left( x \right) \cr u = \cos \left(...

[SOLVED] Some antiderivatives I've got a few antiderivatives to find, I've found most of them and they check out fine with my CAS, but three of them I'm having difficulties with. The first: Homework Statement I = \int {{{\sec ^2 \left( x \right)} \over {\left( {1 + \tan \left( x \right)}...

Whoa, that's voodoo. How did you figure that out?

Indeed, that solved it. Thank you. I have one more question to ask: I need to show that: {\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\mathop{\rm Ln}\nolimits} \left( {1 + \sqrt 2 } \right) Here's my working so...

The original question was "Find x, using the definitions of the hyperbolics in terms of exponentials" ...from this expression: 2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right) Here's my working so far, reducing down to the final polynomial...

No, that's my problem. :)

[SOLVED] 3rd degree exponential polynomial Homework Statement Derived from the original question: "Reduce to find x" 2e^{3x} - e^{2x} - 2e^x = 1 The Attempt at a Solution Inspection fails, since the answer is transcendental. I have the answer from my CAS, but I can't figure...

It's part of a larger exercise on a differential equation. Just a simple example for an implicit function.

Ah, that solves it then. Thanks. As an aside, how can I make this: {{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}} into this: {{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}} They're meant to be identical, but I can't think how. Thanks!

Yes. \displaylines{ \sqrt 2 \left( {1 + \sqrt 2 } \right) = \left( {\sqrt 2 \times 1} \right) + \left( {\sqrt 2 \times \sqrt 2 } \right) \cr = \sqrt 2 + 2 \cr}

Starting with simple fractions, it's known that: {{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}} So when b == d: {{{a \over b}} \over {{c \over b}}} = {a \over c} But what if in the case of: {{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}...