@Sawarnik Okay, I will think about it...I will need to go for dinner...and I have to go for class tomorrow morning, so will see you tomorrow...thanks for the problem...bye
HI, I have to prove that for all integer $t^{2n}-1 \ge n(t^2-1)$. I tried induction but I have to prove (the last step) : $t^{2n+2}-1 \ ge (n+1)(t^2-1)$, and I don't see I can use $t^{2n}-1 >n(t^2-1)$. Someone has an idea please ?
Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
@MickLH Good, then a quick and easy q for you. Prove that if $f$ is differentiable on $[a,b]$, then there exists $x_1,x_2,...x_n$ such that $f'(x_1)+f'(x_2)+f'(x_3)+...f'(x_n)=0$.
@Hawk Heh... I answered and then got downvoted. Hilarious. >8(
@Hawk Granted Antonio Vargas's answer is valid for complex roots, but my answer is valid for real roots (as you seemed to be looking at in your question)
I have to prove that for all integer $t^{2n}-1 \ge n(t^2-1)$. I tried induction but I have to prove (the last step) : $t^{2n+2}-1 \ ge (n+1)(t^2-1)$, and I don't see I can use $t^{2n}-1 >n(t^2-1)$. Someone has an idea please ?
lol might not be able to do that, at least I don't see a way, I didn't mean it like I literally flew here on a spaceship from outer space
Unless you want to get metaphorical, I guess
@Sawarnik Here, being an emotional thing, the closest I believe I will come to proving anything is this: secure.szifler.com/music/dwob_4.mp3 I just recorded it lol feel my emotions and make whatever judgement you want
@skullpatrol Imagine you have 8 identical squares numbered from 1 to 8 and then you sequentially lay them down in the biq square. You need to find out which is the first square you laid down.
Hey folks. Can someone advice on how to solve $15e^{5x}=5x+1$? If I take $\ln$ of both sides, I can get rid of the $e$, but am stuck with $x$ inside the $\ln$.
@Balarka: i have no idea what the EL field is. I think I can assume that the HS teacher who gave this to my friend's son was mistaken when he said "no calculator".
@mick they haven't learned numerical methods yet (i think) and no reasonable teacher would expect someone to solve that by hand. possibly the kid copied instructions down wrong, too.
@chris if 7 and 8 are side-by-side, then you can put 8 down, then 7, then the rest of the pages in the correct order. or you can put page 7 down, then 8 (they're next to each other) , then the rest of the sequence.
@chris if page 6 is under 7, but over 8, then page 7 and 8 are not side-by-side. 7 is "over" 8 (even if it doesn't overlap 8).
To show the rationals are not compact then, I would have to show that an open cover of the rationals does not have a finite subcover. What would be the best step into this direction
Whoof, I barely managed the last through very heavy guns. This is mad : $$\displaystyle \int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx=\frac{\Gamma \left(\frac{1}{8}\right) \Gamma \left(\frac{3}{8}\right)\sqrt{1+\sqrt{2}} }{8 \sqrt[4]{2} \sqrt{\pi }} \left\{\frac{\sqrt{2}-1}{4} \log \left(2+2 \sqrt{2}\right)-\frac{2-\sqrt{2}}{4} \pi \right\}$$
@Chris'ssis @BalarkaSen Get a load of this $$\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}dx=\frac12\gamma+\frac{11}5\ln2-\frac52\ln\sqrt{5}+\frac12\ln\pi-\frac12\ln \phi $$
Also, I found this the other day, it's my new favorite expression for the golden ratio: $$\phi = e^{{\it ArcCsch}\left(2\right)}$$ if you want to inject some fun looking complexity into many mundane things
@DanielFischer Link? At any rate, anyone who has a bit of brain can see the level of your answers are usually over-average. If someone cannot, I wouldn't worry about him/her. =P
@DanielFischer It's simple to construct examples of $L^2$ distributions that are $C^k$ and don't come from $C^k$ functions. (Integrate a discontinuous $L^2$ function $k$ times, say, to get a $C^{k-1}$ function that gives a $C^k$ distribution.) Are there smooth distributions that don't come from smooth functions?
@robjohn There I misunderstood 2 questions or so and that's why I think I chose the wrong answers. The test also measures the out-of-the-box thinking. (of course, not accurately, but it's a kind of measurement there)
@DanielFischer I'm talking about distributions on $\Bbb R$. I can induce a distribution $f^*(\phi) = \int f(x) \phi(x) dx$. I say a distribution is $L^2$ if it's representable in this manner by an $L^2$ function $f$; I say it's $C^k$ if its first $k$ distributional derivatives are also $L^2$; I say it's smooth if distributional derivatives of all orders are $L^2$.
@Chris'ssis I don't think this answer had anything wrong. I acknowledged and upvoted the better answer, but I think my answer was good for the OP since he originally wanted real methods (then someone else changed tags).
That might not be another in a string of downvotes I've gotten recently
Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
