<p>Here's a cool one:</p>
<p>( R* = R without 0 )</p>
<p>f(xy) = f(x) + f(y) for every x,y in R<em>. Proove that if f is continuous for x=1 then it's continuous in R</em>.</p>
<p>Integral of x^x for positive real numbers...</p>
<p>But....</p>
<p>Is not it better to seek solutions of some practical problems? ;)</p>
<p>I can provide some:
- How to find a way to immortality within next 40 years and with less than 20*10^6 $?
- How to increase intelligence and memory capacity in man?
- How to convince congress, UN, EU etc that it is morally good?
- How to synthesize enough air on mars to set up colonies there [and found second MIT - Mars Institute of Technology. Scholarship does not cover travel expanses. hehe ]? </p>
<hr>
<p>And reverting to the main subject: I had problem with this function. Maybe because I have never had integrals :D. I wonder what is the solution. I got one for integers, and then I was thinking about some smart technics of binding the points, and approximate solution from it, since it will have some constant trend after 1...</p>
<p>If You would like to be more advanced take: x^x^x .
It is probably some simple question, and they teach it on calcoulus classes, right?</p>
<p>f(x) = ln(|x|)</p>
<p>:) ill have to think about how to prove that its continuous</p>
<p>natural log is continuous everywhere except at 0. So f(x) <> ln(|x|).</p>
<p>EDIT: sorry, just saw the question said "R* = R without 0"</p>
<p>the solution is more general, think limits:</p>
<p>f is continuous in x0 when:
<a href="http://users.tellas.gr/%7Erandomgr/pics/index_1.gif%5B/url%5D">http://users.tellas.gr/~randomgr/pics/index_1.gif</a></p>
<p>Simple limit for those doing calc ab:</p>
<p>lim 2x^2 + 3
x->0 x^3+5x</p>
<p>Haha...lurkers on this board are going to think they've stumbled across MIT's/Caltech's by accident. This thread cracks me up.</p>
<p>nah mit aint that geeky... caltech, sure.</p>
<p>I know! There was a great task, maybe it is not calcoulus at all, but take a look at this anyway:</p>
<p>1 ; 2R ; [pi]R^2 ; 4/3[pi]R^3 ...</p>
<p>What comes next :)?
I like this one very much. But I will not post answer, since I got two and do not know which one is better :/.</p>
<p>mit has awesome pranks.</p>
<p>anyone know of any princeton pranks?</p>
<p>hey hey they're called 'hacks' rather than pranks. not sure about princeton.</p>
<p>jpsi, shouldn't it be: 1, 2πR, πR², (4/3)πR³.....</p>
<p>you missed a pi in the second term, and I know what this is... now I have to find the "hypervolume" for a "hypersphere" in 4 dimensions, isn't it? :cool:</p>
<p>randomgr: </p>
<p>define g(x)=f(e^x) (of course e^x>0)... then g(x+y)=g(x)+g(y), forall x,y\in R* </p>
<p>*she<a href="g%20:">/b</a>) is a Cauchy function, and being continuous in 1, she's continuous on R... and thus so is f...</p>
<p>i can prove that too!! prove that g(x)=a<em>{\alpha}x for all x\in \alphaQ, where \alpha is an irrational, and all \alpha's are not linear combinations of other \alpha's... in other words, g is linear on Q's (do it first on natural numbers, then for 1/n then for p/q)... if g is continuous in 1, then all a</em>{\alpha}'s are equal...</p>
<p>oh joy.. that's it :)</p>
<p>Not calculus, but still cool (in that geeky way)</p>
<p>Yeah. And now guess this function equation. That's indeed fun. But where's utility? ;)</p>
<p>x^x=5, this one you should ponder for a while</p>
<p>also
sin(-1)x= (1/(sqrt2))(In((sqrt2+x)/(sqr2-x)) this came from the sqrtroot algorithm, and integration by parts.</p>
<p>i could show anyone a really cool method for integrating inverse trig functions. without that integration by parts.</p>
<p>pi^2r^4/2 is the volume of a hypersphere</p>
<p>ForeverZero </p>
<p>how is integral of e^(-x^2) from 0 to infinity multivariable?</p>
<p>no point in learning calculus if all u know is integrals and derivatives... application is more important...</p>
<p>It's multivariable in that that's the only way (that I know of) to solve it explicitly. You can't integrate it in its original form. You have to square it with respect to both x and some other variable (aka "y") and do a polar substitution, which I posted yesterday.</p>