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## #1 Re: Help Me ! » complex plane » 2013-02-08 20:44:52

Hi cooljackiec,
a) do you know a locus of points that dist 4 from zero?
b) " " " " that dist 3 from y=2?

## #2 Re: Help Me ! » Is this a group ? » 2013-02-03 23:05:13

Still Learning wrote:

Yes,but ~ is associative when you only use 0 and n,does that count?

Ok, I thought that {0,n} was {0,1,...,n}

## #3 Re: Help Me ! » Is this a group ? » 2013-02-03 21:31:12

~ is not associative: (x~y)~z!=x~(y~z). For example, take x=5, y=3, z=1. You have:

(5~3)~1 = ||5-3|-1| = 1 != 3 = |5-|3-1|| = 5~(3~1)

so ({0,n} ,~) is not a group, if I understood what you meant.

## #4 Re: This is Cool » tau=2*pi » 2013-01-24 01:36:05

MathsIsFun wrote:

Maybe we could come up with an argument for pi/2. It is a "right" angle, so it must be right!

## #5 Re: Help Me ! » Little derivative problem (prove or give a counterexample) » 2013-01-16 08:02:59

However, we observe that the derivative (where it does exist) goes to zero... maybe moving (1) to the hypothesis... Anyway, that was a good example, thank you

## #6 Re: Help Me ! » Little derivative problem (prove or give a counterexample) » 2013-01-16 07:43:00

uhmm i'm trying to figure out how do you see that f goes to a y0... let's see:

so the c(n) part converges and the x part of course does. Is there a easiest way to see that?

## #7 Help Me ! » Little derivative problem (prove or give a counterexample) » 2013-01-16 06:49:29

Fistfiz
Replies: 3

Hi guys, i'm trying to figure out if this is true or not, can you help me?

Conjecture: Let f:[m,+∞)->R be a continuous and monotonous function with a horizontal asymptote y0 (as x->+∞). Then:
1) f is derivable.
2) f'->0 as x->∞.

I ask for f being monotonous because the only counterexamples, to the non-improved conjecture, that came to my mind are things like f(x)=sin(x^2)/x.

## #8 Re: Help Me ! » Wronksian determinant in 2nd order linear DE » 2013-01-13 02:06:45

Thank you, this remark:
"Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent."
opened my eyes

## #9 Re: Help Me ! » Wronksian determinant in 2nd order linear DE » 2013-01-12 02:06:52

Of course but what i meant was: can I avoid to include W(t0)!=0 for some t0 in my hypotesis? In other words, if I have two linearly independent solutions u and v, can I automatically say W(u,v)!=0 for all t?

## #10 Help Me ! » Wronksian determinant in 2nd order linear DE » 2013-01-11 22:06:44

Fistfiz
Replies: 4

Hi guys,
I'm studying some demonstrations about 2nd order differential equations of the form:
y''+2by'+ay=f(t)
where a,b are constants.
Suppose that u,v are linearly independent solutions. Now, in several demonstrations, it's needed that the Wronksian determinant of u,v it's different from zero.
I see from Abel's identity (http://en.wikipedia.org/wiki/Abel's_identity) that if this is true for some t0 value, then it's true for all t. Provided this, can I always say that the Wronksian of u,v is always non-zero??

You're welcome.

## #12 Re: This is Cool » Teach Pythagorean Theorem in 10 seconds » 2013-01-08 03:14:44

Hi there! These are very simple and smart proofs. I suggest you also to give a look to:

http://farside.ph.utexas.edu/euclid/Elements.pdf
Book1, Prop.48, Prop.49

the second one (if i remember good) is the Inverse theorem (i.e. if a triangle is such that c²=a²+b², then it's rectangle).

## #13 Re: Dark Discussions at Cafe Infinity » Why in this 1=0? » 2012-12-17 07:38:43

21122012 wrote:
Fistfiz wrote:

Hi Bob,

if I may, it seems to me that the (logical) error is deeper:
because

is just a symbol to denote the class of antiderivatives; so, saying class=number makes me think 21122012 is totally missing the meaning of it all.

Here a problem here in what:

Calculus doesn't distinguish an arithmetic increment from a geometrical increment! ! !

Calculus - bad science! ! !

I'm sorry, but I really don't find the connection you see beetwen this and the main topic...
However, don't you feel a little ashamed by saying "Calculus - bad science! ! !"??
I may be wrong, because i read your first post and didn't either understand what you're talking about... but, honestly, seems to me (and not only to me, as I see) you don't know what an indefinite integral is.

## #14 Re: Dark Discussions at Cafe Infinity » Why in this 1=0? » 2012-12-17 05:46:38

I had a quick glance and it is very obscure to me; i'll try to read it later, thank you.

## #15 Re: Dark Discussions at Cafe Infinity » Why in this 1=0? » 2012-12-17 00:23:13

Hi Bob,

if I may, it seems to me that the (logical) error is deeper:
because

is just a symbol to denote the class of antiderivatives; so, saying class=number makes me think 21122012 is totally missing the meaning of it all.

## #16 Re: Dark Discussions at Cafe Infinity » Why in this 1=0? » 2012-12-16 08:38:50

bob bundy wrote:

hi Fistfiz

If you accept the premise that 1 = 0, then you don't need calculus to get 2 = 1 (just add 1 to each side).

Alternatively suspect that 1 isn't 0 after all.

Bob

cool!

seems like:

waaaaa

## #18 This is Cool » tau=2*pi » 2012-12-16 02:54:25

Fistfiz
Replies: 17

http://tauday.com/tau-manifesto

To be brief, the "tauists", as they call themselves, argue that the constant pi should be replaced with tau=2*pi, which is much more natural.

I have to say that, to me, the manifesto was really convincing; as i see it, the use of tau instead of pi brings clarity and coherence with other formulas (for example area of a circular sector).

What do you think about it?

## #19 Re: Help Me ! » Mathematic of the XXI century » 2012-11-16 22:11:40

21122012 wrote:

is error!

Hi endoftheworld,

I may be wrong, but I think stating that:

∫f(x)dx=F(x)+C

is an error, is itself a (logical) error. Because the indefinite integral is DEFINED AS the solution to the problem:

F'(x)=f(x)

It is the antiderivative, and what you gave is just the definition... does it make any sense to ask if is a definition right or wrong?

Saying that ∫f(x)dx=F(x)+C is an error seems to me like saying that it's wrong to put the ' to indicate the derivative...

I could have totally missed the point, maybe for example your paper says the defining the indefinite integral this way leads to some contradiction; in case i hope you can explain us.

## #20 Re: Help Me ! » limit » 2012-11-06 08:28:14

noelevans wrote:

How does this look? :0)
i*180                    i*(180/n)                                  i0
(-1)^(1/n) = (1*e       )^(1/n) = 1*e              so this approaches 1*e   = 1 as n goes to infinity.

(The angles are in degrees.)

I have to admit that at first sight this looked funny; but after being (maybe) less superficial i'm seeing a meaning behind this:
look it geometrically (i write polar coordinates for complex numbers)...

the (first) square root for -1 is    (1,pi/2)       (midnight)
the (first) 3rd root for  -1        (1,pi/3)          (one o'clock)
the (first) 4th root for -1 is    (1,pi/4)            (half past one)
.....
.....                                                             (...some time passes...)
.....
the (first) nth root for -1 tends to (1,0)         (almost three o' clock)

so it seems to me that your limit is what the first nth root of (-1) tends to.

EDIT: I want to add something:

where k=0,1,2...,n-1. In particular, the integer part of (n+1)/2 (which is n/2 if n is even and (n+1)/2 if odd) belongs to the list of k's;
If we accept your and my proceeding then we get:

(where i put n/2 or n+1/2 as k)

so one of us (or eventually both ) must be wrong.

## #21 Re: Help Me ! » limit » 2012-11-06 07:49:10

anonimnystefy wrote:

What do you mean by a succession from N to C?

You see that, for example

## #22 Re: Help Me ! » limit » 2012-11-06 04:55:47

hi mitu, I would say it does not exist if your succession is from N to R;
here's a short proof of non-existence:
if LIM[a(n)]=L, then for all a(n(k)) LIM[a(n(k))]=L
you see that LIM[a(2k)]!=L since a(2k) is not defined for each k. But maybe someone would argue that for each n in dom(a(n)) a(n)=-1, so LIMa(n)=-1... i see it just as a formal problem, maybe someone can be more precise.

While writing my post i realized that if your succession is from N to C it is not even a function, so i don't know if it has any meaning to talk about limit...

## #23 Help Me ! » detA conjecture » 2012-10-31 03:01:37

Fistfiz
Replies: 0

It is just a conjecture that i've made so i'm asking if you can prove it or give a counterexample.

My conjecture :
let A(n) be the square matrix (n+1)x(n+1) with the generic element be

(<--- it's i^(j-1); i post a pic:

then det(A)!=0

I've tested it to n=9...
moreover, det(A) seems to diverge to +inf with n and i've not found a negative value.

hope someone finds this interesting, goodbye!

## #24 Re: Help Me ! » sup(A+B)<?=supA + supB » 2012-10-15 06:54:23

solved, correct proposition is:

## #25 Help Me ! » sup(A+B)<?=supA + supB » 2012-10-14 23:23:52

Fistfiz
Replies: 1

Hi guys,
I have a doubt about a question on my book:
it says:
"let

, and

show that

"

...my question is: can you provide an example where

?

thank you