Poppy wearer
Father of RCAF pilot & 2 beautiful girls
🇨🇦
SW; Physics; Math; History
Sep 1, 2022 • 6 tweets • 3 min read
@Daily_Epsilon Let G be any group of order 1009 and m any element of G other then the identity e.
By Lagrange the order of m (the power of m that equals e) is either 1 or 1009, since 1009 is prime; but only e is of order 1.
So order is 1009 for all m; and G is the cyclic group of order 1009.
@Daily_Epsilon To expand on the second paragraph:
Lagrange's Theorem. states that for subgroup H of a group G: has order (number of elements) which divides order of G.
It's readily shown that the set of powers of any element m of G satisfies the group axioms; and thus forms a subgroup of G.
Jan 12, 2022 • 6 tweets • 4 min read
@JohnsonFido@Daily_Epsilon Let f(x,y) be the polynomial expression; and note that
f(y,-x) = f( -y,x) = f(x,y)
so that we have 4-fold rotational symmetry.
Secondly note the coefficient symmetry in f(x,y) with same signs for even powers of x, y and opposite signs for odd powers of x, y.
1/@JohnsonFido@Daily_Epsilon This is a tell that roots occur in pairs of the form (r - 1/r).
A proof for the Vedic Divisibility Rules discovered by Hindu mathematicians.
For any natural number n = 10a + b with 0 ≤ b ≤ 9;
divisor p coprime to 10; and
M(p) = m as described below
Thm: n is divisible by p exactly when (a+m∙b) is.
M(p) is found from p as:
ones digit of p = 1: m = ⅒⋅(1 + 9⋅p)
ones digit of p = 3: m = ⅒⋅(1 + 3⋅p)
ones digit of p = 7: m = ⅒⋅(1 + 7⋅p)
ones digit of p = 9: m = ⅒⋅(1 + 1⋅p)
Mar 11, 2021 • 10 tweets • 5 min read
@GWOMaths Spoiler - hint and solution follows
@GWOMaths Define Sₖ(c) as the sum of k consecutive integers with offset c thus:
Sₖ(c) = (c+1) + (c+2) + ... + (c+k-1) + (c+k)
= k⋅c + (1 + 2 + ... + (k-1) + k )
= k⋅c + ½k(k+1).
I believe this is only relevant for d ≅ 1 (mod 4) even though that condition doesn't appear in the calculation.
Jan 17, 2021 • 9 tweets • 4 min read
@johnrobb We have abandoned competence,
in the futile search for universal expertise.
"What?" I hear you say.
Clearly definitions and explanations are in order.
1/@johnrobb Expertise, certainly as regards this discussion,
but perhaps also universally, should be regarded
as the knowledge of a very great deal - about very
very little.
2/
Jan 15, 2021 • 14 tweets • 5 min read
@GWOMaths One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.
1/@GWOMaths Observing symmetry in nature, such as noting the similarity between the symmetries of a snowflake and a hexagon, is readily comprehensible. What does it mean then to study "the nature of symmetry"?
2/
Dec 2, 2020 • 4 tweets • 2 min read
@pnjaban None of the above:
Field-trained former military paramedic.
I don't want anyone not trained specifically in emergency medicine anywhere near; and none of those civilian trained E.R. either, cause they're trained in CYA first, and I want someone who's serious about saving my life
@pnjaban When my Dad suffered acute hemechromatosis at ~34,
for which the ONLY treatment is *bleeding*,
and the only options "a whole lot" and "a whole lot more",
the attending physician was too embarrassed to prescribe the correct treatment.
Dec 2, 2020 • 4 tweets • 1 min read
From Albert H. Beiler's
Recreations in the Theory of Numbers:
Calculate in your head:
47² = ?
96² = ?
113² = ?
179² = ?
@GWOMaths Hint follows
@GWOMaths Rochambeau is just a fancy name for Rock-Paper-Scissors.
Nov 6, 2020 • 9 tweets • 4 min read
@LarrySchweikart My take:
SCOTUS must not be seen to play favourites - yet has an obligation to ensure that corruption is voided.
So wherever widespread and continuous failure to observe due process (as per State law) is seen, void all ballots from that county / counting centre.
1/@LarrySchweikart If the number of voided counties exceeds more than 1 or 2, then void the entire election for the State.
In the case of House and Senate representatives, require special elections.
2/
Mar 20, 2020 • 5 tweets • 3 min read
@GWOMaths Define
y ≡ x - 16
to allow factoring the term 2¹⁶ completely.
- Generating Function f(x) for a single standard die is
x.(1+x+x²+x³+x⁴+x⁵)
= x+x²+x³+x⁴+x⁵+x⁶
= Σ xⁿ . C(n,1)
1/@nklym143@GWOMaths - For two Std. die f(x)² is
= 1.x²+1.x³+1.x⁴+ 1.x⁵+ 1.x⁶+ 1.x⁷
+ 1.x³+1.x⁴+1.x⁵+ 1.x⁶+ 1.x⁷+ 1.x⁸
+ 1.x⁴+1.x⁵+1.x⁶+ 1.x⁷+ 1.x⁸+ 1.x⁹
+ 1.x⁵+1.x⁶+1.x⁷+ 1.x⁸+ 1.x⁹+1.x¹⁰
+ 1.x⁶+1.x⁷+1.x⁸+ 1.x⁹+1.x¹⁰+1.x¹¹
+ 1.x⁷+1.x⁸+1.x⁹+1.x¹⁰+1.x¹¹+1.x¹²
2/
@GWOMaths The game Nim is played by two players alternately removing items from several piles of items. Absent physical materials, with pen and paper a game can be setup as rows of vertical lines, players stroking them out to "remove" them.
@GWOMaths Each turn a player removes 1 or more items from precisely one pile, or row.
The winner is the player who does not (or, as a variant, does) remove the last item.
Consider the setup of two piles of 2 items:
xx
xx
The player to move can be forced to lose by his opponent.
@GWOMaths@MathsTim Recall from Feb. 1 this about generating functions for the odd and distinct partitions of n.
Then the generating function for unrestricted partitions of n, P(n), is given by
SUM(P(n) . q^n) = PROD(1 / (1 - q^i) )
where n ∈ {0, ... } and i ∈ {1, ...}.
@GWOMaths@MathsTim Then P(7) will be the coefficient of q^7 from the expansion of the RHS, noting that
1 / (1 - q^i)
is just
SUM(1 + q^i + q^2.i + q^3.i + q^4.i + ...)
and that we only require in each expansion powers of q ≤7.
Feb 1, 2020 • 8 tweets • 5 min read
@GWOMaths@MathsTim Lemma 1:
From every distinct partition of n an odd partition of n can be uniquely associated.
@GWOMaths@MathsTim Proof:
For the integer n, any distinct partition can be written as
SUM_i(Gi . 2^Ki)
where every Gi is odd and Ki >= 0
because every positive integer is uniquely expressable as an odd positive integer times a power of 2.
Cases n=3,4,5 fail because a friendship graph that is an n-gon fails with friendship # = 2 for all nodes.
1/@GWOMaths@EstherOfReilly Case n=6:
If the node with highest Friendship # has Friendship # >= 3 then that immediately either defines a triad of mutual friendship or a triad of mutual strangers from the first three friend nodes.
and equivalently
2x^2 - 5x -10 - 23n = 0 for some small n
To be factored requires that for our n, find two integers with difference
5
and product
2.(-10 -23n)
@GWOMaths@MathsTim Suppose n=0:
Require two integers with difference
5
and product
-20.
None found
Suppose n=1:
Require two integers with difference
5
and product
-66.
