Recreational Problems (SM level)

[dcc]

for more rigour i would sandwhich the motherfucker:

The order hierarchy is rigorous, but there is no point reproving all of those results here.  The idea I've used is that , which is enough to establish that this limit goes to zero

Refrain from spamming, if you will TrueTears.  This isn't the place.

[TrueTears]

Must there be a purpose?

(what was the purpose of your answer?)

The purpose of my answer was to illustrate this question's lack of a purpose.

How was it spamming when I was replying to this?

(btw I loved your double ninja delete)

[Over9000]

Is it wrong for people to ask questions, lol??

[kamil9876]

for more rigour i would sandwhich the motherfucker:

The order hierarchy is rigorous, but there is no point reproving all of those results here.  The idea I've used is that , which is enough to establish that this limit goes to zero

Refrain from spamming, if you will TrueTears.  This isn't the place.

O ok. I had this in mind: then and hence the sequence contains smaller terms than hence approaches . That's like my noob way that I've been using for some time now.

What is that function btw?

[humph]

for more rigour i would sandwhich the motherfucker:

The order hierarchy is rigorous, but there is no point reproving all of those results here.  The idea I've used is that , which is enough to establish that this limit goes to zero

Refrain from spamming, if you will TrueTears.  This isn't the place.

O ok. I had this in mind: then and hence the sequence contains smaller terms than hence approaches . That's like my noob way that I've been using for some time now.

What is that function btw?

http://en.wikipedia.org/wiki/Big_O_notation

Very useful tool for talking about the behaviour of a function as it approaches a limit (you use it all the time in areas like analytic number theory or the theory of algorithms).

[Ahmad]

I'm guessing TT had Stirling's formula (or a slightly weaker version) in mind, this post is an interesting read.

[brightsky]

Not sure if these problems from the original set have been solved yet, but I'll have a crack at some of the 'easier' ones:

13. If made sense, what would x equal?

Since , then which means .

16. Solve the equation .

Factorise:

By null factor law: , , or

17. Solve .

By the quadratic formula,

15. Solve .

Probably not the best way, but by inspection, 9 is a factor.

Rearranging,

Since 9 is a factor, by long division, we have:



Solving ,

So solutions are: , , or

18. (n-radicals).

or

[evaever]

(x-7)(x-3)(x+5)(x+1)=1680
(x+5)(x+1)(x-3)(x-7)=7.5.3.2.2.2.2
Notice that on the left the first factor > second factor by 4, the second factor > third factor by 4 , etc.
Arrange the factors on the right to get 4 factors: (7.2)(5.2)(3.2)2=14.10.6.2
Another possible arrangement: -2(3.-2)(5.-2)(7.-2)=-2.-6.-10.-14
So x=9 or x=-7

[pi]

Could someone explain the trick used by brightsky for questions 13 and 18 please? I've never seen it in a textbook and it seems really useful. Thanks in advance.

[evaever]

13.    sqrt(x+sqrt(x+...))=2,  square both sides, x+sqrt(x+sqrt(x+...))=4, x+2=4, x=2

18.    same as 13,
Or for any n radicals, the statement is true, consider n=1, so sqrt(3x)=x, 3x=x^2, x^2-3x=0, x=0 or x=3.

[evaever]

Now try this

1+2/(1+2/(1+2/(1+2/........../(1+2/x)..........)))=x, find x.

[enwiabe]

[brightsky]

Notice

Integrate by parts:



Since

Therefore

Also,

Therefore the integral becomes:





Since

So:





Notice that the can be manipulated to the LHS:



So now the only problem we have is: , which isn't really methodical (easiest way is probably by trick):



Do u-substitution with u = csc(x) - cot(x) and it will result in ln|csc(x) - cot(x)|.

Divide both sides of the original equation by 2 and we have the answer we need.

[/0]

Another method:



Then u-substitution will finish it.

u-substitution works for all odd powers of and too:

[enpassant]

evaluate the definite integral of sin^2011 (x) /(sin^2011 (x) + cos^2011 (x)) from 0 to pi/2 wrt x.