Hi,

I have difficulty in taking fourier transform of a function f(t).

The function f(t) is not integrable. Through analytical analysis, f(t)
behaves asymptotically the same as a*log(t)/t

for t-> +infinity; and f(t) behaves asymptotically the same as b*log(-
t)/t

for t-> -infinity; where "a" and "b" are some constants.

Is there a way to work around this difficulty and get some sort of
fourier transform of this function f(t), possibly

involving extended functions or functions which can only be evaluated
numerically or distribution functions...

Any thoughts?

Thanks a lot!

Hello,

MeijerG[{{-4, -7/2, 1/2}, {}}, {{0, 0, 1/2}, {}}, 1]

?
Best wishes,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC

...meets all the axiomatic requirements of a vector space over ring as
opposed to a field, then why aren't modules simply called "vector
spaces over a ring". OTOH, if "over a field" is understood in the
term "vector space", while module includes the understanding "over a
ring", why are so many authors punctilious about the tautological
"vector space over a field"? As opposed to what?

COHO

(Committee to Obviate Historical Obfuscation)

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Hi,

Let {u_1,...,u_M} be M unit vectors in C^K that are mutually
orthogonal.
How many free parameters does this set of vectors have?

I came up with the answer M*(2*K - (M+1)/2), but in a journal article
I am reading, they come up with a different answer M*(2*K-(M-1)). I
can't see a flaw in my argument, so I thought I'd run it past the ng,
to see if I had made an error someplace.

The total number of params we have starting off is 2*M*K (M complex
vectors of length K each).

The condition that each vector have unit length reduces the number of
free params it has by one per vector. This can be seen by expressing
the components of each vector in polar form.

The condition that all M vectors be pairwise orthogonal reduces the
number of free parameters by M*(M-1)/2.
The argument goes as follows. Consider the condition that (u_M)^H u_j
= 0 for j=1,...,M-1. We...

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in the past i have called myself a constructivist , but at the same time expressed i liked cantors function and the diagonal method and the CH.

and in a way i fought against cantor by pointing out unfalsifiability , mathematicians like kronecker etc.

yet i do accept the axiom of infinity.

i accept that the cardinality of the reals is bigger than that of the naturals.

and i gave a proof that C^n = C without cantor or ZF(C).

the reason for all of this ?

in one word : subcountable.

2^aleph_n can be mapped to aleph_n with using subcountable.

thus it seems both have same cardinality.

but as N and C have shown (using cantors diagonal argument or other ) this is not true.

we can map C -> N 1 to 1 using subcountable,

or as N as a subset of C.

but what matters is if N maps to C.

it doesnt !

in a similar way 2^C can be shown to have cardinality greater than C , in fact 2^C can be seen as f(R) -> [0,1].

this is intresting , it means
card f(R_1,R_2,R_3) -> [0,1] < card g(R_1) -> [0,1]

for a fixed function f ( g is not fixed , all g )

this can be proven by using R^2 < 2^R combining R^2 = R and 2^C > C ( assuming CH ) where the first can be proven by my binary method and the second by cantors diagonal argument.

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Are these the only two examples up to equivalence?
x^8 + 1600*x + 2800
x^8 + 750000*x + 2187500