From now on,i will post here as much inequalities involving complex numbers as possible,they seem to relate with classic geometrical inequalities,the only job to do is to find a meaning for this abstract terminology
For example,prove that $\frac{\overline{a}b+a\overline{b}}{2|a||b|}\in [-1,1]$
Let be A(a),B(b),and $\measuredangle AOB=\measuredangle XOB-\measuredangle XOA$.But $cos\measuredangle XOA=\frac{a+\overline{a}}{2|a|},cos\measuredangle XOB=\frac{b+\overline{b}}{2|b|}$ and $cos(a-b)=cosa\cdot cosb+sina\cdot sinb$ and after a few working you'll discover what means $\frac{\overline{a}b+a\overline{b}}{2|a||b|}$
Let there be math; a new place different from my mind where ideas are written down,and a possible witness for a big discovery
joi, 28 iulie 2011
luni, 25 iulie 2011
Latex formula
as you can see in the right corner that icon with latex formula,now it is easier than ever to post latex commands here :Just type latex command between dollar symbol $ $
as you can see when stoping over a formula in my last posts.
for those who are unfamiliar with latex commands here's a comprehensive dictionnary of main symbols.
as you can see when stoping over a formula in my last posts.
for those who are unfamiliar with latex commands here's a comprehensive dictionnary of main symbols.
schur's inequality
If $x,y,z \in R,t\in Z$then we have the Schur's inequality:
$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\geq 0$.It is very used when determining limits of sequences but also proves many classical inequalities:
For example$(a+b-c)(b-c)(a-c)+(a+c-b)(c-b)(a-b)+(b+c-a)(b-a)(c-a)\geq 0$,where a,b,c>0
Proof:$c+b-a=x,a+c-b=y,a+b-c=z \Rightarrow a-b=\frac{y-x}{2},b-c=\frac{z-y}{2},a-c=\frac{z-c}{2}$ which becomes $x(x-y)(x-z)+...\geq 0$ which is Schur's ineq for t=1
if $x+y+z=1,x,y,z>0$then $x^2+y^2+z^2\leq 3(x^3+y^3+z^3)$
Proof:From Schu'r inequality ,in particular t=1 we have $x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geq 0 \Leftrightarrow x^2(y+z)+y^2(x+z)+z^2(x+y)\leq x^3+y^3+z^3+3xyz \leq 2(x^3+y^3+z^3) \Leftrightarrow x^2(1-x)+... \leq 2(x^3+y^3+z^3)$ hence the conclusion
Study:$\frac{1}{(a-b)(b-c)(c-a)}(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})\leq 0$
$x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\geq 0$.It is very used when determining limits of sequences but also proves many classical inequalities:
For example$(a+b-c)(b-c)(a-c)+(a+c-b)(c-b)(a-b)+(b+c-a)(b-a)(c-a)\geq 0$,where a,b,c>0
Proof:$c+b-a=x,a+c-b=y,a+b-c=z \Rightarrow a-b=\frac{y-x}{2},b-c=\frac{z-y}{2},a-c=\frac{z-c}{2}$ which becomes $x(x-y)(x-z)+...\geq 0$ which is Schur's ineq for t=1
if $x+y+z=1,x,y,z>0$then $x^2+y^2+z^2\leq 3(x^3+y^3+z^3)$
Proof:From Schu'r inequality ,in particular t=1 we have $x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y) \geq 0 \Leftrightarrow x^2(y+z)+y^2(x+z)+z^2(x+y)\leq x^3+y^3+z^3+3xyz \leq 2(x^3+y^3+z^3) \Leftrightarrow x^2(1-x)+... \leq 2(x^3+y^3+z^3)$ hence the conclusion
Study:$\frac{1}{(a-b)(b-c)(c-a)}(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})\leq 0$
duminică, 24 iulie 2011
Distribution of decimals in some sequences
A classic problem is of showing the existing of perfect squares or powers of 2 which begin with a combination of digits,for example:
Prove that there $\exists n \in N$ so the first four digits of $2^n$be 2002.
Proof:The conclusion is equivalent to this one:$2002 \cdot 10^k \leq 2^n \leq 2003\cdot 10^k \Leftrightarrow k+lg2002\leq nlg2 \leq k+lg2003 \Leftrightarrow lg 2002 \leq nlg 2-k \leq lg 2003$
Because $lg2 \notin Q$ from Kronecker's theorem the set $\{nlg2-k|n,k \in N\}$is dense in R so it has elements in $[lg2002,lg2003)$.
Exercise:
Prove that $\exists m,n \in Z$so that the first three decimals of $m\sqrt{2}+n\sqrt{3}$ are 101
Prove that there $\exists n \in N$ so the first four digits of $2^n$be 2002.
Proof:The conclusion is equivalent to this one:$2002 \cdot 10^k \leq 2^n \leq 2003\cdot 10^k \Leftrightarrow k+lg2002\leq nlg2 \leq k+lg2003 \Leftrightarrow lg 2002 \leq nlg 2-k \leq lg 2003$
Because $lg2 \notin Q$ from Kronecker's theorem the set $\{nlg2-k|n,k \in N\}$is dense in R so it has elements in $[lg2002,lg2003)$.
Exercise:
Prove that $\exists m,n \in Z$so that the first three decimals of $m\sqrt{2}+n\sqrt{3}$ are 101
vineri, 22 iulie 2011
connection between an identity and laticial points
Prove that if (p,q)=1 then $\sum_{k=1}^{q-1}\left [ \frac{kp}{q} \right ]=\sum_{k=1}^{p-1}\left [ \frac{kq}{p} \right ]$
Considering a rectangle OABC with A(q,0),B(q,p),C(0,p) let's count the number of points with integer coordinates inside this rectangle:There are $\frac{(p-1)(q-1)}{2}$
On line OB of equation $y=\frac{px}{q}$ we don't have any points of integer coordinates(Prove it)
After that we count the number of points of integer coordinates on the line with x=k under OB,which are $\left [ \frac{kp}{q} \right ]$.If we sum them we have the left side of the identity.
If we replace p with q we have the right side of the identity.
corollary:If (p,q)=d then
Also try to prove that
Now it would it be interesting to find out a generalisation for space and points with integer coordinates in a rectangular cuboid OABCO'A'B'C' where B'(p,q,r),B(p,0,r),...
Considering a rectangle OABC with A(q,0),B(q,p),C(0,p) let's count the number of points with integer coordinates inside this rectangle:There are $\frac{(p-1)(q-1)}{2}$
On line OB of equation $y=\frac{px}{q}$ we don't have any points of integer coordinates(Prove it)
After that we count the number of points of integer coordinates on the line with x=k under OB,which are $\left [ \frac{kp}{q} \right ]$.If we sum them we have the left side of the identity.
If we replace p with q we have the right side of the identity.
corollary:If (p,q)=d then
Also try to prove that
Now it would it be interesting to find out a generalisation for space and points with integer coordinates in a rectangular cuboid OABCO'A'B'C' where B'(p,q,r),B(p,0,r),...
joi, 21 iulie 2011
Round table
At a business dinner there are 15 people.Everyone of them has some friends among each other the rest of them being foes.Prove that they cannot be sat at a round table so they would be sorrounded only by foes.What if instead of 15 there would be 14?
Sets
We've presented here:http://theodormunteanu.blogspot.com/2011/07/paths-and-combinatorics.html
some sort of proof for a few identities:
There's a much larger spectrum of identities in combinatorics which can be deduced from the initial meaning of combinations:
for example:we must calculate the number of subsets with k elements from $\{a_1,a_2,...,a_n\}$which is the number of sets with k elements from $\{a_1,a_2,...,a_{n-1}\}$+the number of sets of (k-1)numbers from$\{a_1,...,a_{n-1}\}$and the kth element being $a_n$.From the definition of combination we have $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
If we look further we will see that the same number of subsets is the no. of subsets with (k-2)elements from$\{a_1,a_2,...,a_{n-2}\}$ (the rest of 2 can be chosen from $\{a_{n-1},a_n}$)+$2\cdot$no.of sets with k-1 elements from $\{a_1,a_2,...,a_{n-2}\}$+no.of sets with k elements from$\{a_1,a_2,...,a_{n-2}}$.following the same reasoning we can conclude that
$\binom{m+n}{k}=\binom{m}{0}\cdot\binom{n}{k}+\binom{m}{1}\cdot\binom{n}{k-1}+...+\binom{m}{k}\cdot\binom{n}{0}${Try it)
Another easy application is to prove that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ which that the number of subsets of $\{a_1,...,a_n}$ is the number of sets with 0,1,2,...elements.
try also to prove that:$\binom{n}{k}=\frac{n}{k}\cdot\binom{n-1}{k-1}$
some sort of proof for a few identities:
There's a much larger spectrum of identities in combinatorics which can be deduced from the initial meaning of combinations:
for example:we must calculate the number of subsets with k elements from $\{a_1,a_2,...,a_n\}$which is the number of sets with k elements from $\{a_1,a_2,...,a_{n-1}\}$+the number of sets of (k-1)numbers from$\{a_1,...,a_{n-1}\}$and the kth element being $a_n$.From the definition of combination we have $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
If we look further we will see that the same number of subsets is the no. of subsets with (k-2)elements from$\{a_1,a_2,...,a_{n-2}\}$ (the rest of 2 can be chosen from $\{a_{n-1},a_n}$)+$2\cdot$no.of sets with k-1 elements from $\{a_1,a_2,...,a_{n-2}\}$+no.of sets with k elements from$\{a_1,a_2,...,a_{n-2}}$.following the same reasoning we can conclude that
$\binom{m+n}{k}=\binom{m}{0}\cdot\binom{n}{k}+\binom{m}{1}\cdot\binom{n}{k-1}+...+\binom{m}{k}\cdot\binom{n}{0}${Try it)
Another easy application is to prove that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ which that the number of subsets of $\{a_1,...,a_n}$ is the number of sets with 0,1,2,...elements.
try also to prove that:$\binom{n}{k}=\frac{n}{k}\cdot\binom{n-1}{k-1}$
marți, 19 iulie 2011
Sumation of arctangent
If $x,y,z\in R$ prove that $\arctan x+ \arctan y+\arctan z=\arctan {\frac{s_1-s_3}{1-s_2}}$ where
$s_1=x+y+z,s_2=xy+yz+zx,s_3=xyz$
More generally $\arctan x_1+...+\arctan x_n=$
$
$s_1=x+y+z,s_2=xy+yz+zx,s_3=xyz$
More generally $\arctan x_1+...+\arctan x_n=$
$
luni, 18 iulie 2011
Darboux property function
If $ f:R \rightarrow R$ is a function with darboux property and $\lim_{x \to \infty}f(x) \in R,lim_{x \to \ -\infty}f(x) \in R$ then f is bounded.
Inequality between sides and radius
If ABC is a triangle then the following inequality holds:$ a+b+c \leq 3\sqrt{3}R$ where a=BC,b=AC,c=AB.
Inequality with increasing function
Let be $ a\leq b\leq c \leq d$.If f is continuous on $ [a,b]$ and f' is increasing on (a,d) then prove that
$ (b-d)f(a)+(c-a)f(b)+(d-b)f(c)+(a-c)f(d) \leq 0$
$ (b-d)f(a)+(c-a)f(b)+(d-b)f(c)+(a-c)f(d) \leq 0$
Sequence of integer numbers
Find all sequences $ (a_n)_{n \geq 1}$ where $ \frac{1}{3\sqrt[3]{a_{n+1}^2}} \leq \sqrt[3]{n+1} -\sqrt[3]{n} \leq \frac{1}{\sqrt[3]{a_na_{n+1}}}$
Matrix with odd entries
Let be $ A \in M_n(Z)$with detA an odd number.Prove that A has at least $ \binom{n}{k}$ minors of order k which are odd.
A special chessboard
Let be an nxn,$ n\geq 3$ grid colored in two colours,black and white in the following way:on the secondary diagonal we have black squares and in rest white squares.we call "transformation" changing a colour on a row or column from black to white and inverse.Can we reach at a grid with 2 identical rows after a finite number of transformation?
What if we have a permutation of the black colours such that there is exactly one black square on each row and each column.
What if we have a permutation of the black colours such that there is exactly one black square on each row and each column.
Problems with buckets of water(discussion)
Let's say we have 3 unscaled buckets A,B,C of capacity 8,5 and 20 l respectively,and C is full of water.Measure 6l of water,where the only operation allowed is the spilling from one bucket to another until one is full or the other is empty.
What if we had A(8l),B(6l),C(20l) could we measure 7l of water with the same operation ?
In first case the answer is yes,the second case the answer is no:
well let's define (x,y,z) the quantity of water in (A,B,C);
Initially we had (0,0,20);
Then $ (0,0,20) \rightarrow (8,0,12) \rightarrow (3,5,12) \rightarrow (3,0,17) \rightarrow (0,3,17) \rightarrow (8,3,9) \rightarrow(6,5,9)$
More generally if we have three buckets with capacity of a , b,c respectively and another bucket large enough to transfer the water if necessary,we can measure e litres of water if and only if $latex ax+by=e$has a pair of integers as solution.
So,if $ (a,b,c)=d$ where $(a,b,c)$ means the greatest common divisor of a, b and c and $ d\nmid e$ then we cannot measure with the three buckets .So we need to introduce the fourth bucket with capacity f where $ (a,b,c,f)=1$.
There should be at least two proofs for this,and I expect them from you.
What if we had A(8l),B(6l),C(20l) could we measure 7l of water with the same operation ?
In first case the answer is yes,the second case the answer is no:
well let's define (x,y,z) the quantity of water in (A,B,C);
Initially we had (0,0,20);
Then $ (0,0,20) \rightarrow (8,0,12) \rightarrow (3,5,12) \rightarrow (3,0,17) \rightarrow (0,3,17) \rightarrow (8,3,9) \rightarrow(6,5,9)$
More generally if we have three buckets with capacity of a , b,c respectively and another bucket large enough to transfer the water if necessary,we can measure e litres of water if and only if $latex ax+by=e$has a pair of integers as solution.
So,if $ (a,b,c)=d$ where $(a,b,c)$ means the greatest common divisor of a, b and c and $ d\nmid e$ then we cannot measure with the three buckets .So we need to introduce the fourth bucket with capacity f where $ (a,b,c,f)=1$.
There should be at least two proofs for this,and I expect them from you.
Weights of coins and maybe some advanced algebra
We have 15 coins with the property that if we extract one coin,the rest of them can be separated in two heaps with the same number of coins and with the same weight.How many weighings do we need to determine the weight of each coin.
The Art of Problem Solving: Volume 1: The BASICS Solutions
The Art of Problem Solving: Volume 1: The BASICS Solutions
Paths and combinatorics
Now I'll give a new perspective of some basic formulas in combinatorics:
Firstly ,let's find the number of paths from A(0,0) to B(a,b) ,where a path means moving to the right of the Ox axis and upper on the Oy axis.
There are $ \binom{a+b}{a}$ ways to reach point B(*)
Proof: let's consider a sequence of (a+b)numbers from the set {0,1} where we have a numbers of 0 and b numbers of 1 .
So we have to find the number of pairs$ (x_1,x_2,...,x_{a+b})$where $ x_i \in {{0,1}}$ which is $ \frac{(a+b)!}{a! \cdot b!}$
A little exercise would be to prove that $ \binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1}$
Well,consider O(0,0),A(n-m,m),B(n-m-1,m),C(n-m,m-1).
The number of paths to A is the number of paths which pass through B+no. of paths which pass through C and according to (*) we have the conclusion.
Other examples:1)$ \binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$
2)$ \sum_{a\leq c,b\leq d}\binom{a+b}{a} \cdot \binom{c+d-(a+b)}{c-a}=\binom{c+d}{d}$
Firstly ,let's find the number of paths from A(0,0) to B(a,b) ,where a path means moving to the right of the Ox axis and upper on the Oy axis.
There are $ \binom{a+b}{a}$ ways to reach point B(*)
Proof: let's consider a sequence of (a+b)numbers from the set {0,1} where we have a numbers of 0 and b numbers of 1 .
So we have to find the number of pairs$ (x_1,x_2,...,x_{a+b})$where $ x_i \in {{0,1}}$ which is $ \frac{(a+b)!}{a! \cdot b!}$
A little exercise would be to prove that $ \binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1}$
Well,consider O(0,0),A(n-m,m),B(n-m-1,m),C(n-m,m-1).
The number of paths to A is the number of paths which pass through B+no. of paths which pass through C and according to (*) we have the conclusion.
Other examples:1)$ \binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$
2)$ \sum_{a\leq c,b\leq d}\binom{a+b}{a} \cdot \binom{c+d-(a+b)}{c-a}=\binom{c+d}{d}$
Determinant of matrix again
If a matrix $ A\in M_{2k}(Z),3\nmid det(A)$then there are at most $\frac{\binom{2k}{2}}{2}-1$ minors which divides by 3.
duminică, 17 iulie 2011
A refinement of Hermite-Hadamard inequality
If $af(b)-bf(a)=0$ with $f"$a continuous and concave function then $\frac{f(b)-f(a)}{2}(b-a)<\int^b_af(x)dx \leq \frac{f(a)+f(b)}{2}(a+b)$
Proof:the graph of f is above the line determined by AB having the equation $ y=\frac{f(b)-f(a)}{b-a}x$ so f(x)>$ \frac{f(b)-f(a)}{b-a}x=\frac{f(b)-f(a)}{2}(a+b)$.
For the second part of inequality the area of trapeze ABCD is higher than of the surface between lines x=a,x=b and the grapf of f hence the conclusion.
For the second part of inequality the area of trapeze ABCD is higher than of the surface between lines x=a,x=b and the grapf of f hence the conclusion.
Ranks and polynomials
If $ A,B\in M_n(C)$and rank(B)=k and $ det(A+xB)=0$ has k+1 solutions then $ \det(A+xB)=0,\forall x\in \mathbb{C}$
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