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.
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