## 11977 Reputation

7 years, 254 days

## International Mathematics Competition fo...

Maple

This year, the International Mathematics Competition for University Students  (IMC) took place online (due to Coronavirus), https://www.imc-math.org.uk/?year=2020

One of the sponsors was Maplesoft.

Here is a Maple solution for one of the most difficult problems.

Problem 4, Day 1.

A polynomial  with real coeffcients satisfies the equation

, for all real .

Prove that  for   .

A Maple solution.

Obviously, the degree of the polynomial must be 101.

We shall find effectively p(x).

 > restart;
 > n:=100;
 (1)
 > p:= x -> add(a[k]*x^k, k=0..n+1):
 > collect(expand( p(x+1) - p(x) - x^n ), x):
 > S:=solve([coeffs(%,x)]):
 > f:=unapply(expand(eval(p(1-x)-p(x), S)), x);
 (2)
 > plot(f, 0..1); # Visual check: f(x)>0 for 0
 > f(0), f(1/4), f(1/2);
 (3)
 > sturm(f(x), x, 0, 1/2);
 (4)

So, the polynomial f has a unique zero in the interval (0, 1/2]. Since f(1/2) = 0  and f(1/4) > 0, it results that  f > 0 in the interval  (0, 1/2). Q.E.D.

## A riddle ! ...

Maple 2018

Can you guess what P() produces, without executing it?

P:=proc(N:=infinity) local q,r,t,k,n,l,h, f;
q,r,t,k,n,l,h := 1,0,1,1,3,3,0:
while h<N do
if 4*q+r-t < n*t
then f:=if(++h mod 50=0,"\n",if(h mod 10=0," ","")); printf("%d"||f,n);
q,r,t,k,n,l := 10*q,10*(r-n*t),t,k,iquo(10*(3*q+r),t)-10*n,l
else q,r,t,k,n,l := q*k,(2*q+r)*l,t*l,k+1,iquo(q*(7*k+2)+r*l,t*l),l+2
fi
od: NULL
end:


I hope you will like it (maybe after execution).

## Happy New Year!...

Maple
with(plots):
S:=cat("Happy New Year 2020!   "\$3):
N:=length(S): a:=0.77*Pi: h:=2*Pi/N:
display(seq(textplot([cos(a-k*h), sin(a-k*h),S[k+1]],
rotation=-Pi/2+a-k*h, 'font'=["times","roman",24]), k=0..N-4), axes=none);


## Putnam Mathematical Competition 2019...

Maple 2018

Maple can easily solve the B4 problem of the Putnam Mathematical Competition 2019  link

B4.  Let F be the set of functions f(x,y) that are twice continuously differentiable for x≥1, y≥1 and that satisfy the following two equations:

For each f2F, let

Determine m(f), and show that it is independent of the choice of f.

 > # Solution
 > pdsolve({ x*diff(f(x,y),x)+y*diff(f(x,y),y) = x*y*ln(x*y), x^2*diff(f(x,y),x,x)+y^2*diff(f(x,y),y,y) = x*y });
 (1)
 > f:=unapply(rhs(%[]), x,y);
 (2)
 > h := f(s+1, s+1) - f(s+1, s) - f(s, s+1) + f(s, s);
 (3)
 > minimize(h, s=1..infinity);
 (4)
 (5)
 >

## Algorithms...

Maple

There are no efficient algorithms for this.
How would you simplify by hand the expression

512*b^9 + (2303*a + 2304)*b^8 + (4616*a^2 + 9216*a + 4608)*b^7 + (5348*a^3 + 16128*a^2 + 16128*a + 5376)*b^6
+ (4088*a^4 + 16128*a^3 + 24192*a^2 + 16128*a + 4032)*b^5 + (1946*a^5 + 10080*a^4 + 20160*a^3 + 20160*a^2
+ 10080*a + 2016)*b^4 + (728*a^6 + 4032*a^5 + 10080*a^4 + 13440*a^3 + 10080*a^2 + 4032*a + 672)*b^3
+ (116*a^7 + 1008*a^6 + 3024*a^5 + 5040*a^4 + 5040*a^3 + 3024*a^2 + 1008*a + 144)*b^2
+ (26*a^8 + 144*a^7 + 504*a^6 + 1008*a^5 + 1260*a^4 + 1008*a^3 + 504*a^2 + 144*a + 18)*b + 9*a^8
+ 36*a^7 + 84*a^6 + 126*a^5 + 126*a^4 + 84*a^3 + 36*a^2 + 9*a + 1

to  (a+2*b+1)^9 - a*(a-b)^8   ?

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