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SuicideFuel Math thread problem (official)

Integrate ln(x + 1)/(x^2 + 1) from 0 to 1
 
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(pi/8)*ln(2)

one can use integration under the integral sign -- e.g., consider ln(ax + 1) / (1 + x^2)

The other way of doing it is trig sub:

x = tan(u)

=> integrate ln(tan(u) + 1) from 0 to pi / 4

=> integrate ln(sin(u) + cos(u)) - ln(cos(u)) from 0 to pi / 4

sin(u) + cos(u) = sqrt(2) * sin(u + pi / 4)

=> rewrite above integral as (ln(sqrt(2)) + ln(sin(u)) from pi / 4 to pi / 2) - (ln(cos(u)) from 0 to pi / 4)

=> integrate (ln(sqrt(2)) + ln(sin(u)) from pi / 4 to pi / 2) - (ln(sin(u)) from pi / 4 to pi / 2)

=> ln(2) * (pi / 8)

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Let S be a size n set of tuples s_i = {a_i, b_i, c_i}. All s_i are distinct, but all a_i, b_i, and c_i are not necessarily distinct. Let n_a be the number of distinct tuples {b_i, c_i}, n_b be the number of distinct tuples {a_i, c_i}, and n_c be the number of distinct tuples {a_i, b_i}. Prove that n^2 <= (n_a * n_b * n_c).
 
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Do you mean that, e.g., a_i = a_k for some i != k or that, e.g., a_i = b_k for some i & k or both?

The sets A, B, C are completely separate, a_i could equal a_k for some i != k, or they could all be different.

S is a subset of the Cartesian product of A, B, and C
 
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Integrate ∫(cos x)/cos(x-a) dx
 
Integrate ∫(cos x)/cos(x-a) dx
Substitute u = x − a, use cos(u + a) = cos(u)cos(a) − sin(u)sin(a) and the fact that the antiderivative of tan(u) is ln|sec(u)| + C. Technically we can choose a different constant of integration for every connected component of the domain of tan(u). Lastly we should rewrite the result in terms of x of course.
 
technically they are then multisets
Also, as writ, it suggests that s_1 = {a_1, b_1, c_1} and s_2 = {a_2, b_2, c_2} and {a_1, b_2, b_3} ∉ S. I take it that's not intended?

To clarify, A, B, C are finite sets, S is a subset of the Cartesian product A * B * C, n is the cardinality of S, n_a is the number of distinct pairs {b_i, c_i} found in S, n_b is the number of distinct pairs {a_i, c_i} found in S, and n_c is the number of distinct pairs {a_i, b_i} found in S.
 
Integrate ∫(cos x)/cos(x-a) dx
x-a =t
dx =dt
Int cos(t+a)/cos(t) dt
Int (Cost*cosa-sint*sina)/Cost dt
Int ((cosa-sint(a)*tan(t))
t*cos(a) - ln|(sec(t))| * sina
(x-a)*Cos(a) +sin(a)* ln|cos(x-a)| +C
 
Given nonnegative reals a_1, ... , a_n and b_1, ... , b_n prove:

GM(a_1, ... , a_n) + GM(b_1, ... , b_n) <= GM(a_1 + b_1, ... , a_n + b_n)

GM = geometric mean
 
Given nonnegative reals a_1, ... , a_n and b_1, ... , b_n prove:

GM(a_1, ... , a_n) + GM(b_1, ... , b_n) <= GM(a_1 + b_1, ... , a_n + b_n)

GM = geometric mean
using the subadditivity of the nth root this is trivial
 
find a set of orthonormal vectors from the independent vectors a_{1}=(1,-1,0) , a_{2}=(0,1,-1) and a_{3}=(1,0,-1) How many non-zero orthonormal vectors are obtained?
 
71x187x7916931 to the power of 6 divded by 16
 
How many Jews does it take to make the holocaust victim numbers to 7 million
 
find a set of orthonormal vectors from the independent vectors a_{1}=(1,-1,0) , a_{2}=(0,1,-1) and a_{3}=(1,0,-1) How many non-zero orthonormal vectors are obtained?
Use gram-schimdt process
 
Prove that the convolution of any two square integrable functions on R is a continuous function vanishing at infinity.

Bonus: Can every continuous function vanishing at infinity on R be expressed as a convolution of two square integrable functions?
 
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Prove that the convolution of any two square integrable functions on R is a continuous function vanishing at infinity.
Use approximations to reduce to the (trivial) case where one of the functions is an inducator function with finite support.
 
Bonus: Can every continuous function vanishing at infinity on R be expressed as a convolution of two square integrable functions?
I dunno nigga. Doubt it. Do you think you can pass through Fourier transforms to establish properties of a convolution of two square integrable functions that don't hold true for an arbitrary function in C0(R)?
 
Use approximations to reduce to the (trivial) case where one of the functions is an inducator function with finite support.
I take it you mean compact support in lieu of finite support. I'm sure something along these lines works, but I don't think this is nearly as trivial as you portray it. E.g., what breaks when I try to do this when one of the functions is integrable and the other one is essentially bounded?
 
Do you think you can pass through Fourier transforms to establish properties of a convolution of two square integrable functions that don't hold true for an arbitrary function in C0(R)?
I know you can
 
I take it you mean compact support in lieu of finite support. I'm sure something along these lines works, but I don't think this is nearly as trivial as you portray it. E.g., what breaks when I try to do this when one of the functions is integrable and the other one is essentially bounded?
Right, the nontrivial part lies in the technicalities of how you 'approximate' your function (say f) with indicator functions (say f_n), and establishing that such an approximation exists. A pointwise approximation isn't sufficient for instance. You want the existence of a uniform approximation in the sense that the L2-norm of (f - f_n) approaches 0.

If one function is integrable and the other is bounded, well, everything breaks no? Even the 'trivial' case of where the integrable function is a compactly supported indicator function doesn't work. Just take the bounded function to be 1 everywhere.
 
Right, the nontrivial part lies in the technicalities of how you 'approximate' your function (say f) with indicator functions (say f_n), and establishing that such an approximation exists. A pointwise approximation isn't sufficient for instance. You want the existence of a uniform approximation in the sense that the L2-norm of (f - f_n) approaches 0.
Indeed. You also need that f_n * g → f * g in a pretty strong sense (which necessitates that f_n → f in a pretty strong sense) to be able to deduce the vanishing at infinity of f * g from f_n * g so you could say that establishing the continuity of f ↦ f * g w.r.t. the right topologies is nontrivial as well.
If one function is integrable and the other is bounded, well, everything breaks no? Even the 'trivial' case of where the integrable function is a compactly supported indicator function doesn't work. Just take the bounded function to be 1 everywhere.
I meant "what part of your argument breaks when I try to apply it to the case where one is integrable and the other is essentially bounded". I think we've established the answer thereto en passant, so no need to answer it anymore.
 
find a set of orthonormal vectors from the independent vectors a_{1}=(1,-1,0) , a_{2}=(0,1,-1) and a_{3}=(1,0,-1) How many non-zero orthonormal vectors are obtained?
3?
 
2. Notice that a_1 + a_2 = a_3.

PS @im done there are some problems with how you phrased the question
  • "non-zero orthonormal vectors" is tautological as orthoNORMAL vectors have a norm of 1 and therefore cannot be zero
  • since a_1 + a_2 = a_3 they are only PAIRWISE independent
  • the phrase "a set of orthonormal vectors from ..." is somewhat ambiguous; better would be "an orthonormal basis of the span of ..."
  • your question can be far more concisely phrased as "find the dimension of the span of a_1, a_2 and a_3"; concise phraseology is generally preferred
this is intended to be constructive feedback
 
2. Notice that a_1 + a_2 = a_3.

PS @im done there are some problems with how you phrased the question
  • "non-zero orthonormal vectors" is tautological as orthoNORMAL vectors have a norm of 1 and therefore cannot be zero
  • since a_1 + a_2 = a_3 they are only PAIRWISE independent
  • the phrase "a set of orthonormal vectors from ..." is somewhat ambiguous; better would be "an orthonormal basis of the span of ..."
  • your question can be far more concisely phrased as "find the dimension of the span of a_1, a_2 and a_3"; concise phraseology is generally preferred
this is intended to be constructive feedback
Thanks for the correction, can't believe I overlooked that
 

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