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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 p with real coeffcients satisfies the equation

p(x+1)-p(x) = x^100, for all real x.

Prove that p(x) <= p(1-x) for   0 <= x and x <= 1/2.

 

A Maple solution.

Obviously, the degree of the polynomial must be 101.

We shall find effectively p(x).

 

restart;

n:=100;

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);

proc (x) options operator, arrow; (94598037819122125295227433069493721872702841533066936133385696204311395415197247711/16665)*x-37349543370098022593228114650521983084038207650677468129990678687496120882031450*x^3-1185090416633200*x^87+5974737180020*x^89-(86465082200/3)*x^91+133396340*x^93-597520*x^95+2695*x^97-(50/3)*x^99+x^100-(2/101)*x^101+(16293234618989521508515025064456465992824384487957638029599182473343901462949018943/221)*x^5-69298763242215246970576715450882718421982355083931952097853888722419955069286800*x^7+(113991896447569512043394769396957538374962221763587431560580742819193991151970540/3)*x^9-(450021969146981792096716260960657763583495746057337083106755737535521294639081800/33)*x^11+3451079104335626303615205945922095523722898887765464179344409464422173275181060*x^13-648776866983969889704838151840901241863730925272452260127881376737469460326640*x^15+(1224135636503373678241493336115166408006020118605202014423201964267584789018590/13)*x^17-(32609269812588448517851078111423700053874956628293000710950261666057691492700/3)*x^19+(17369174852688147212979009419766100341356836811271344020859968314555332168046/17)*x^21-79714896335448291043424751268405443765709493999285019374276097663327217200*x^23+(26225149723490747954239730131127580683873943002539194987613420614551124468/5)*x^25-294965074792241210541282428184641838437329968596736990461830398732050600*x^27+(186430797065926226062569133543332579493666384095775768758650822594552980/13)*x^29-608766986011732859031810279841713016991034114339196337222615083429200*x^31+22758671683254934243234770245768111655371809025564559292966948184145*x^33-755022138514287934394628273773230341731572817528392747252537299270*x^35+(380420681562789081339436627697748498619486609696130138245054547645/17)*x^37-596110444235534895977389751553577405150617862905657345084592800*x^39+(186546013247587274869312959605954587283787420112828231587660264/13)*x^41-313678397368440441190125909536848768199325715147747522784400*x^43+6254306446857003025144445909566034709396500424382183891144*x^45-114204496639521606716779723226539643746613722246036949600*x^47+1916927215404111401325904884511116319416726263341690260*x^49-29677354167404548158728688629916697559643435320275800*x^51+(93950257927474972838978328999588595121346462082404180/221)*x^53-5650787690628744633775927032927548604440367748960*x^55+69888520126633344286255800412032531913013033640*x^57-806279422358340503473340514496960223283853200*x^59+8696895011389170857678332370276446830499368*x^61-87900576836101226420991143179656778525600*x^63+(10844299000116828980379757772973769420469/13)*x^65-7447304814595165455238549781183862150*x^67+(1065245686771269279784908613651828005/17)*x^69-497741911503981694520541768814800*x^71+3738596479537236832468307626580*x^73-26593490941061853727808593704*x^75+179403449737703736809514420*x^77-1149393958953185579079600*x^79+(21007540356807993839074/3)*x^81-(121855249152521399900/3)*x^83+(3818021878637120462/17)*x^85 end proc

(2)

plot(f, 0..1); # Visual check: f(x)>0 for 0<x<1/2

 

f(0), f(1/4), f(1/2);

0, 2903528346661097497054603834764435875077553006646158945080492319146997643370625023889353447129967354174648294748510553528692457632980625125/3213876088517980551083924184682325205044405987565585670602752, 0

(3)

sturm(f(x), x, 0, 1/2);

1

(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.

 

Download imc2020-1-4.mw

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).

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);

 

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:
    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

 

For each f2F, let

 

"m(f) = min[s>=1]  (f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s))"

 

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
});

{f(x, y) = (1/2)*(x*y+2*_C1)*ln(x*y)-(1/2)*x*y-2*_C1*ln(x)+_C2}

(1)

f:=unapply(rhs(%[]), x,y);

proc (x, y) options operator, arrow; (1/2)*(y*x+2*_C1)*ln(y*x)-(1/2)*y*x-2*_C1*ln(x)+_C2 end proc

(2)

h := f(s+1, s+1) - f(s+1, s) - f(s, s+1) + f(s, s);

(1/2)*((s+1)^2+2*_C1)*ln((s+1)^2)-(1/2)*(s+1)^2-(s*(s+1)+2*_C1)*ln(s*(s+1))+s*(s+1)+(1/2)*(s^2+2*_C1)*ln(s^2)-(1/2)*s^2

(3)

minimize(h, s=1..infinity);

(4+2*_C1)*ln(2)-1/2-(2+2*_C1)*ln(2)

(4)

answer = simplify(%);

answer = 2*ln(2)-1/2

(5)

 


Download putnam2019-b4.mw

@chandrashekhar 

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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