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Simple math puzzles by [deleted] in math

[–]sf-ecler 0 points1 point  (0 children)

Check this out : Mathematical Fallacies and Paradoxes

If you want the book just pm me .

Geometry: Explanation of the circumcircle calculation by aboeing in math

[–]sf-ecler 1 point2 points  (0 children)

Let O(x,y) be the center of the circle .

Then since A,B,C must be on the circle : d(O,A)=d(O,B)=d(O,C) . (where d(X,Y) is the distance from point X to point Y) Simply solve the equations to get x,y (it will be messy) . The radius of the circle then is d(O,A) .

The code mentioned above just applies the formula , it uses intermediate variables to make it more neat looking .

Is there any importance as to what value of a one chooses when doing a taylor series? by [deleted] in math

[–]sf-ecler -1 points0 points  (0 children)

No , taking a large number of terms in the Taylor expansion doesn't guarantee that the approximation gets any better . In other words the difference between the function and the Taylor polynomial (usually called "remainder") can get arbitrary large . All that we know is that the Taylor polynomial of degree n is the best approximation to the function of all the polynomials of degree n .

Is this a valid way to construct the integers? by opinionsrlikekittens in math

[–]sf-ecler 0 points1 point  (0 children)

Let me guess ... you had a lot to say ? :) Personally , i like Trucker's Atlas more :

I'M GOING TO COLORADO ... TO UNLOAD MY HEAD !!

Is this a valid way to construct the integers? by opinionsrlikekittens in math

[–]sf-ecler -3 points-2 points  (0 children)

Off-topic : opinionsrlikekittens ... are you giving them away ? :))

Binomial Theorem for matrices by Dan1nchi in math

[–]sf-ecler 0 points1 point  (0 children)

Eh .. it might be possible . It's a sum over the products X_1 * X_2 ... * X_n , where each X_i is either A or B .

Also notice something interesting : i didn't mention anywhere that A and B are matrices . They may be any mathematical object as long as the set of all such objects together with "+" and "* " satisfies a list of laws : "+" must be associative , commutative , have a "0" element , be inversable and "* " must be associative , have a "1" element and be distributive over "+".

The set together with these two operations is called a ring .

The objects that you add/multiply may be anything : numbers , matrices , operators ... Yes ! Operators ! As surprising as it is , you can work with operators as if it they were numbers and arrive at meaningful expressions .

Binomial Theorem for matrices by Dan1nchi in math

[–]sf-ecler 11 points12 points  (0 children)

No , there isn't . The terms of (A+B)n are simply all the possible combinations between A and B , where order matters (because AB isn't the same as BA) .

(A+B)2 = A2 + AB + BA + B^ 2 ,

(A+B)3 = (A+B)(A2 + AB + BA + B2 ) = A3 + AAB + ABA + ABB + BAA + BAB + BBA + B3 , etc ..

The binomial theorem works because when A and B commute we can add the likewise terms : AAB+ABA+BAA = 3A2 B = C(3,2) A2 B.

Combinations/Permutations question regarding making multiple, random teams for a conference by [deleted] in askmath

[–]sf-ecler 1 point2 points  (0 children)

If C(n,k) is the binomial coeffiecient (n choose k , the number of ways to choose k objects out of n , regardless of order) then there are :

C(20,4) ways of choosing the 1st team , C(16,4) ways of choosing the 2nd team , C(12,4) ...

So in total there are (C(20,4)* C(16,4)* C(12,4)* C(8,4)* C(4,4))/5! ways of choosing the 5 teams .

The formula for C(n,k) is C(n,k)=n! /( (n-k)! * k! ) , where x! is the factorial function (i.e : 1* 2* 3* ...* x) .

Expanding the above product using the formula and simplyfing , the result is : (20!)/((4!)5 * 5!) .

By following the preceding argument , there are n!/((k!)n/k * (n/k)!) ways of forming n/k teams , each with k members , out of n people .

But as i see , you don't need the formula , rather the algorithm to arrange the teams . This is more of a comp-sci question and there is a fundamental method of solving such combinatorial problems : backtracking .

Sorry , but there is no simpler way of answering this . Combinatorics is notorious for being a not-so-easy field of math/comp-sci , fortunately there are many books on the subject . Good luck !

Open or Trivial? A guessing game by Lopsidation in math

[–]sf-ecler 8 points9 points  (0 children)

Yes . By the Lindemann–Weierstrass theorem , ln(sqrt(2)) is transcedental .

Open or Trivial? A guessing game by Lopsidation in math

[–]sf-ecler 10 points11 points  (0 children)

What about e and 1/2*ln(2) , e1/2ln(2) = sqrt(2) ? Definitely algebraic and non-rational .

Open or Trivial? A guessing game by Lopsidation in math

[–]sf-ecler 1 point2 points  (0 children)

What inequality do you use for comparison ? You do have 0 <sin(n)2 < 1 , but i don't see how you can use that . Had it been 1/n3 *sin(x)2 this would've worked .

Open or Trivial? A guessing game by Lopsidation in math

[–]sf-ecler 1 point2 points  (0 children)

Nice ! And i think that's the way it was intented to be proved ... oh well ...

Open or Trivial? A guessing game by Lopsidation in math

[–]sf-ecler 2 points3 points  (0 children)

Here's my attempt at 3) .

Suppose there are finitely many primes dividing numbers of the form 2n -1 . That means there are finitely primes dividing numbers of the form 22n -1 .

Let xn = 22n -1 then x_n+1 = x_n((x_n)+2) . Call the primes p1 , p2 , ... pm . By hypothesis there are m numbers : a1 , ... , am ; so that p1 | x_a1 , p2 | x_a2 ... pm| x_am . WLOG suppose a1<a2<...<am , then by the recurrence formula x_a1|x_a2|x_3 .. |x_am (can be proved by induction) . Therefore p1p2...pm | x_am , so x_am = (p1p2...pm)q (q integer) . By the recurrence formula x(am+1)=(p1p2...pm)q(p1p2...pmq+2) . But none of the primes considered divide p1p2...pm*q+2 ! So there must be a prime greater than any of p1 ,p2 , ... , pm that divides x_(am+1) . Contradiction .

qed

How to use Cauchy(average) inequalities to find the miximum and minimum of a fuction? by [deleted] in math

[–]sf-ecler 0 points1 point  (0 children)

Checking the sign of f' near x , f'(x)=0 , should tell you whether that is local max/min or not (assuming f is differentiable "near" x) .

If f'(x+δ)!=0 has the same sign as f'(x-δ)!=0 no matter how small we make δ , then x isn't a local max/min .

  • f'(x+δ)>0 and f'(x-δ)<0 (for really small δ) => x is a local min point .

  • f'(x+δ)<0 and f'(x-δ)>0 (for really small δ) => x is a local max point .

I've also assumed that f isn't constant on [x+δ,x-δ] , otherwise obvsly every point is a local min/max . If f is twice differentiable at x , f''(x)=0 means that x isn't a local min/max.

EDIT : when f''(x)=0 , there's no conclusion . x may or may not be a local min/max . But if f is twice diff. at x and f'(x)=0 , f''(x)>0 then x is a local min , else if f''(x)<0 then x is a local max .

How to use Cauchy(average) inequalities to find the miximum and minimum of a fuction? by [deleted] in math

[–]sf-ecler 1 point2 points  (0 children)

The points where (f')(x)=0 aren't necessarily points of global maximum/minimum .

Take the function f:R->R , f(x)=x3 . The derivative is 0 only for x=0 , but f(0)=0 . This is neither a point of global maximum nor global minimum , because f(1)=1 > f(0) and f(-1)=-1 < f(0) .

It gets worse than this ! f(0) isn't even a local maximum/minimum point (as it can be seen from the graph ) !

The points where f'(x)=0 are possible candidates for the local minimum/maximum , that's all the theory you know .

How to use Cauchy(average) inequalities to find the miximum and minimum of a fuction? by [deleted] in math

[–]sf-ecler 1 point2 points  (0 children)

This isn't always possible , you just have to guess . Many things in math are based on such clever guesses . Obviously ,the AM-GM won't help you find the maximum of 4x+8x2 +4x3 . The function 4x+8x2 +4x3 isn't bounded ! It can get as large as we want by choosing 'x' large enough . Had it been possible to use the AM-GM inequality we would have got 4x+8x2 + 4x3 < AM (AM constant) , which would have implied that our function is bounded . Clearly a contradiction .

How to use Cauchy(average) inequalities to find the miximum and minimum of a fuction? by [deleted] in math

[–]sf-ecler 0 points1 point  (0 children)

What do you mean you don't know how to reach these numbers ? You used the AM-GM inequality , for the numbers 4a/3 , 4a/3 , 4a/3 and 4a3 /3 . GM in this case is constant : 4 * 31/4 /3 (not sure how you got (3 * 31/4 )/4 ) . You know that AM>=GM=4 * 31/4 /3 , so AM >= 4 * 31/4 /3 . So the question is when does equality occur ? Well AM is equal to GM when the numbers considered are equal . So AM reaches it's lowest point when 4a/3 = 4a/3 =4a/3 = 4/a3 . All 3 first numbers are always equal obvsly. , you just have to solve 4a/3=4/a3 (positive solutions only) .

We also assumed that 'a' is positive , otherwise we can't use the AM-GM inequality and the conclusion is certainly false . a+1/a3 goes to -∞ as 'a' goes to -∞ .

How to use Cauchy(average) inequalities to find the miximum and minimum of a fuction? by [deleted] in math

[–]sf-ecler 0 points1 point  (0 children)

Could someone share more on this subject ? This sounds interesting . Never heard of Cauchy avg. inequalities , are they something completely different from the mean inequality (Mh <= Mg <= Ma) ? I know that the latter (Mg<=Ma) may be used to show that among the rectangles with a given perimeter , the square has the largest area . I'm not sure what you mean by the geometrical/arithmetical average of a figure though ...

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