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a) do you know a locus of points that dist 4 from zero?

b) " " " " that dist 3 from y=2?

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}

~ 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.

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

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?

**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.

Thanks in advance.

"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

**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.

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).

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.

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

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.

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

**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?

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.

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.

anonimnystefy wrote:

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

You see that, for example

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...

**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

[Link fixed by admin]

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!

solved, correct proposition is:

**Fistfiz**- Replies: 1

Hi guys,

I have a doubt about a question on my book:

it says:

"let

show that

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

?thank you