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I would like to show a diagram an essentially equivalent
evaluation of the angle.
The angle between two planes is the angle between two
intersecting lines respectively in the planes perpendicular
to the line of intersection of the planes. In the figure, OA
is the line of intersection of planes OAB and OAD. By
symmetry, the perpendiculars dropped from B and D
on OA intersect, say at point E.
Consider
. BE is the altitudeI think we do not need to assume
to be prime.Hi, I did not read Yaglom's solution but I think there
shoud be a formal proof. I filled in more than 10 rows
and columns for the matrix and think I see a pattern,
which (I think) could be proved if written carefully.
1. For each
To find the (401,666) entry, bobby gave the simple computation rule and here is an illustration:
a. notice both 666, 401 are less than 1024=
I think we don't need to apply the Euclidean Algorithm in
Just as Jane did. Given
. Ifso
is either itself or 1. The former is rejectedYes
is a maximal ideal. There are infinitely many maximal ideals in ,You might want to first consider the boundary of S:
In general, if
, andIn the question,
The inequality
Or we could see geometrically this
satisfiesHi, I think you just need to check the definition
of eigenvector.
So we want here
with ,Hence any basis of
, in particular the canonical basisSorry, we should multiply
I think we could still somewhat apply an argument of symmetry,
and it might even be related to vectors. I write it just for the
sake of philosophical argument (but otherwise meaningless).
A simply proof is very unlikely but an intuitive, acceptable argument
is here depending on how we define the center of mass.
Aside, I think we are all sure vector algebra could show the
3 medians are concurrent: e.g.
If we define the CM as the point that balance a physical uniform triangular plate,
then the argument of symmetry is already a good one for me:
the median AD (D is the midpoint of BC) balances the plate, so does the median
BE (E is the midpoint of AC). Consider the (vector) moment rxF of the plate (at CM)
about the intersection M of AD and BC. Since AD and BC do not overlap, we can resolve
r into components along AD and BC, the same on rxF which become a sum of zero vectors.
So r must be the zero vector else there exist positions r not paralell to F (and r non-zero).
If we define the CM most fundamentally as a (2-D) integral (limit of Riemann sums),
Sorry for the long wind. Thanks for reading.
This is basically the (algebraic) technique of
rationalizing the numerator: multiply both
numerator and denominator by
Hi ziggs,
I would like to point out something called compactification
(of non-compact sets). You could google search and look
for the most general topological definition. A subset S of
a set T is compact with respect to a topology on T if every
open cover (collection of open sets whose union contains S)
of S has a finite subcover (finite sub-collection whose union
already contains S).
Now something concrete. We learned at an early age we cannot
divide by 0! That is: 1/0 is NOT defined. Well we could say 1/0
is not a real number as we know. The extended real number
system [-oo, +oo] (only the topological part) is a 2-point
compactification of (-oo,+oo), by adding 2 points, -oo and +oo.
We can define 1/0 to be either -oo or +oo to get a function f
from [-oo,+oo] to itself, with -oo, +oo going to 0.
However f is not continuous.
For continuity of f(x)=1/x, we would like the 1-point compactification,
or further identifying -oo and +oo as a number oo. Define 1/0=oo
and 1/oo=0. Then f is a continuous bijection (homeomorphism) from
(-oo,+oo)U{oo} onto itself. This 1-point compactification of (-oo,+oo)
can be viewed as a circle (but its topology is not the subspace topology of
the real plane). The 1-point compactification of the real plane can be
viewed as a spherical surface (but its topology is not the subspace
topology of the 3-D real space).
I think Ricky and Mathsyperson basically explain the idea.
I would just like to point to an algorithmic perpective.
Equivalence classes are disjoint and elements
in an equivalence class are pair-wise related.
A list of relations (related pairs) and non-relations (unrelated
pairs) could be
1. consistent and uniquely determines all the equivalence classes, or
2. consistent but does not uniquely determine all the equivalence
classes (it might determines some of them), or
3. inconsistent, which unfortunately might only be found out at
a late stage.
For me, I might want to consider all the relations first and apply
transitivity to get a collection of largest possible subsets of pair-wise
related elements. Then I would apply the non-relations:
a. on each subset above to check for consistencies
b. on each pair of subsets above for possibility that they are
part of the same equivalence class.
Sorry for making a simple exercise too complicated.
Thanks for reading.
I think we could use inequalities just to show the divergence:
I think the assertion to be proved is:
there is a nonzero x in R^n, such that x.v = 0 for all v in V.
Actually, the statement is true for any finite-dimenisional
inner product space U (say dim U = n).
Since V is not the entire U and 0 is in V, there is nonzero x in U\V.
Let {v_1, v_2, ..., v_(n-1)} be a normalized orthogonal base of V.
Consider w = x - (x.v_1)v_1 - (x.v_2)v_2 - ... - (x.v(n-1))v_(n-1).
Then w is nonzero, else x is in V.
For any v_i, i = 1, 2, ..., n-1,
w.v_i = x.v_i - (x.v_i)(v_i.v_i) = 0
since v_i.v_i = 1, and v_j.v_i = 0 for i different from j.
For any v in V, v = a_1(v_1) + ... + a_(n-1)(v_(n-1))
for some scalar (R here) a_1, ..., a_n-1.
So w.v = a_1(w.v_1) + ... + a_(n-1)(w.v_(n-1)) = 0.
I got a sum of 512 for the second part of Qu.2.
I think we could look at both parts of Qu.2 the following way
to arrive at dofference equations.
Let A_N = the # of jump sequences from step 1 to step N.
Then A_1 = 1 (no jump at all) and A_2 = 1.
Consider A_N, N > 2.
Part 1: The last jump can be 1 step or 2 steps (but not both).
So A_N = A_(N-1) + A_(N-2) for N > 2.
{A_N} is a Fibonacci sequence,
i.e. A_N = a(alpha^N)+b(beta^N)
where alpha = (1+sqrt(5))/2, beta = (1-sqrt(5))/2
and a, b are determined from A_1 = 1, A_2 =1.
Part 2: The last jump can be any number of 1, 2, .., N-1 steps.
So A_N = A_(N-1) + A_(N-2) + ... + A_1
and by induction A_N = 2^(N-2). In the question, N = 11.
This answer is from an algebraic point of view, i.e. setting up
and solving an equation.
A geometric point of view is to consider which intermediate steps
the rabbit lands. A set of intermediate steps corresponds uniquely
to a sequence of jumps. It starts at step 1 and ends at step N.
A set of intermediate steps is a subset of {2, 3, ..., N-1}.
So there are 2^(N-2) sets of intermediate steps, so is the number
of jump sequences.
You could see it this way:
denote nCk = (n choose k)
The Binomial Theorem says
(1+x)^n = summation (k=0 to k=n) nCk(x^k)
Put x=-1: 0 = summation (k=0 to k=n) (-1)^k(nCk)
So sum of odd terms = sum of even terms.
Put x=1 to see both sums are 2^(n-1).
With this we could see the probability questionthis way:
Prob(sum of n dice is odd)
= 1/(2^n) (# of combinations in which an odd number of dice is odd)
= 1/(2^n) (nC1 + nC3 + nC5 + ...)
= 1/(2^n) (2^(n-1))
= 1/2
Similar for the even case.
Thanks.
I think the answer is surely it can. Sorry I cannot think of a simpler construction
than the following. I will explain the idea after it.
We know the natural log base, e = 2.718281828459045... is irrational,
e = 1 + 1 + summation (n=2 to n=infinity) (1/n!)
and for n >=2, 1/n! = 1/(2.3.4...n) <= 1/(2.2..2) (n-1 times) = 1/(2^(n-1)).
So the series of non-negative rational numbers
summation (n=2 to n=infinity) (1/n!) = e -2
summation (n=2 to n=infinity) 1/(2^(n-1)) - (1/n!) = 1-(e-2) = 3-e
Consider the sequence a_n of "merging" the above and that of e:
a_0 = 1
a_1 = 1
a_n = 1/n! if n is even and n > 1
a_n = - (1/(2^(n-1)) - (1/n!)) if n is odd and n > 1
Then
summation (n=0 to n=infinity) abs(a_n)
= 1+1+ (n=2 to n=infinity) ((1/n!) + 1/(2^(n-1)) - (1/n!))
= 1+1+1
=3 is rational
but
summation (n=0 to n=infinity) a_n
= 1+1 + (n=2 to n=infinity) ((1/n!) - (1/(2^(n-1)) - (1/n!)))
= e - (3 -e)
= 2e -3 is irrational
The general idea is now that summation (n=0 to n=infinity) a_n converges,
we would like to find an example for which both
A = summation (n=0 to n=infinity) a_n with a_n >=0 and
B = summation (n=0 to n=infinity) a_n with a_n < 0 converge.
Then
summation (n=0 to n=infinity) abs(a_n) = A - B
summation (n=0 to n=infinity) a_n = A + B
Pick any A, B such that A+B is rational but A-B is not, say
A = e = 2.718281828459045
B = 3-e = 0.28171...
as above.
The set of rational numbers being "dense" in the set of real numbers, meaning
there are sequences of rational numbers arbitrarily close to any given irrational
number and vice versa, we could construct series of rational numbers (even >=0)
converging to A and B respectively. Then "merge" the series to get {a_n}.
For example,
e = 2 + .7 + .01 + .008 + .0002 + ... ( I use decimals for convenience)
3-e = .2 + .08 + .001 + .0007 + ...
3 = 2 + .7 + .2 + .01 + .08 +.008 + .001 + .0002 + .0007 + ...
2e-3= 2 + .7 - .2 + .01 - .08 +.008 - .001 + .0002 - .0007 + ... (answer)
Thanks.
If a series converges, the limit of the nth term is 0:
let S_n = sum (i=0 to i=n) a_i, then a_n = S_n - S_(n-1).
Take limit n -> infinity as both limits on RHS exist, and equal too.
But now the nth term of this series approaches +1 or -1, i.e.
its limit even does not exist. So the answer is no.
I think from y_n to x_n, you make a mistake @ 2^(n-1) (2^n -1) which = (4^n - 2^n)/2.
But I think you have already make a mistake @ y_n:
y_n - 2^n(y_0) = (y_n -2y_(n-1)) + 2(y_(n-1)) - 2y_(n-2)) + ... +2^(n-1)(y_1 - 2y_0)
= 3z_(n-1) + ... + 3z_1 + 3z_0
= 3z_0(sum 4^k, k=0 to k=n-1)
= (4^n - 1)z_0
So y_n = 2^n(y_0) + (4^n - 1)z_0.
Hi everyone,
I am sorry I cannot type special characters just with my keyboard.
#1. Try this "diagonal method of counting":
a_11 -> a_12 -> a_21 -> a_13 -> a_22 -> a_31 -> a_14 -> ...
With this indexing, a_ij appears in count # i+((i+j-1)(i+j-2)/2).
In case any of these sets are indeed finite, then we could count any fixed element, say a_11
when the set is already exhausted since if f:X->Y is surjective and X is countable, then so is Y.
#2. We could try induction starting with n=1.
Prob_1(sum is odd) = Prob_1(sum is even) = 3/6 = 1/2.
Assume for n dice, Prob_n(sum is odd) = Prob_n(sum is even) = 1/2 for n >= 1.
Prob_(n+1) (sum is odd) = Prob_(n+1)(sum of first n dice is odd AND (n+1)st die is even)
+ Prob_(n+1)(sum of first n dice is even AND (n+1)st die is odd)
= Prob_n(sum of first n dice is odd)XProb_1((n+1)st die is even)
+ Prob_n(sum of first n dice is even)XProb_1((n+1)st die is odd)
throwing dice being independent events, and now apply induction assumption
= (1/2)X(1/2) + (1/2)X(1/2)
= 1/2
Similarly, Prob_(n+1) (sum is even) = 1/2.
So the assertion is true for all n >= 1 by induction.
#3. I think we just need to look into the meaning of "n choose k". Let's label n objects
A_1, A_2, ... A_n. Let's choose k objects out of these n. If k = 0 or k = n, there is 1 way
(even for choosing nothing). Assume k > 1. A combination of k objects is either but not both
(i) k-1 objects from A_1, A_2, ..., A_(n-1) AND A_n or
(ii) k objects from A_1, A_2, ..., A_n-1 (and A_n not chosen)
By induction ON n+k, namely the sum of # of objects and # to be chosen,
there are (n-1 choose k-1) combinations of type (i) and (n-1 choose k) combinations of type (ii).
We finish with Pascal's formula.
Thanks.
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