{"id":1,"date":"2018-06-09T20:03:36","date_gmt":"2018-06-09T20:03:36","guid":{"rendered":"http:\/\/blog.metu.edu.tr\/e221325\/?p=1"},"modified":"2020-03-29T17:52:23","modified_gmt":"2020-03-29T17:52:23","slug":"notes","status":"publish","type":"post","link":"https:\/\/blog.metu.edu.tr\/e221325\/2018\/06\/09\/notes\/","title":{"rendered":"Some elementary notes from my studies."},"content":{"rendered":"

The Notes<\/h3>\n
\n\t-\n\n\n1) I've seen this, while dealing with a real analyis problem. It can be very usefull proving statements.\n\n$$\\forall x,y\\in[0,\\infty),\\quad \\sqrt{|x-y|}\\ge |\\sqrt x -\\sqrt y|$$\n\n$$\\circ--------------------------------\\circ$$\n\n2) Again, in real analysis.\n\n$$\\forall f\\neq0,\\quad \\left(f+\\dfrac1f -2\\right)=\\left(f-1 \\right)\\left(1-\\dfrac1f \\right)$$\n\n$$\\circ--------------------------------\\circ$$\n\n3)$$\\forall x_1,x_2\\in[0,\\infty),\\quad|sin^2x_1-sin^2x_2|\\le 2|x_1-x_2|$$\n\n$$\\circ--------------------------------\\circ$$\n\n4)$$ \\text{For } \\;\\; k \\in \\{0, \\cdots, m\\}\\;\\;\\text{ we have}$$\n\n$$ A_k = \\lim_{x\\to-k} \\frac{x+k}{x(x+1)\\cdots(x+m)} = \\frac{(-1)^k}{k!(m-k)!} = \\frac{(-1)^k}{m!}\\binom{m}{k}. $$$$\\Rightarrow$$$$ \\frac{1}{x(x+1)\\cdots(x+m)} = \\frac{1}{m!} \\sum_{k=0}^{m} \\binom{m}{k} \\frac{(-1)^k}{x+k} $$\n\n$$\\circ--------------------------------\\circ$$\n\n5)$$\\text{Let}\\;\\;\\;\\; S_n \\;\\;\\;\\; \\text{be permutation group and}\\;\\; A_n \\;\\;\\;\\;\\text{alternating group which is normal subgroup of}\\;\\; S_n.$$\n\n$$S_n=A_n\\cup (12)A_n\\;\\;\\;\\;\\text{or}\\;\\;\\;\\; S_n=A_n\\cup t\\circ A_n$$ is true for any transposition t.\n$$\\circ--------------------------------\\circ$$\n\n6) Additional to the 5), $$|A_n|=|(12)A_n|=|t\\circ A_n|=n!\/2$$\n\n$$\\circ--------------------------------\\circ$$\n\n7)$$ \\text{If one wants to find center of} \\;\\;\\;\\; GL_2(\\mathbb R), \\;\\;\\;\\;\\text{the trick that one can use is considering invertible matrix}\\;\\;\\;\\; \\left[ \\begin{matrix} 1 & 1 \\\\ 0 & 1 \\end{matrix}\\right] \\\\ \\text{since every element in center should commute with it we can explore the general charastics of center of}\\;\\;\\;\\; G \\;\\;\\;\\;, Z(G)\\\\ \\text{which is } Z(G)=\\left \\{\\left[ \\begin{matrix} a & 0 \\\\ 0 & a \\end{matrix}\\right] \\big | a\\in\\mathbb {R^x}\\right\\}$$\n\n$$\\circ--------------------------------\\circ$$\n\n8) A way to create isomorphism between complex numbers and certain matrices:\n\n$$ (\\star )\\left[ \\begin{matrix} 0 & -1 \\\\ 1 & 0 \\end{matrix}\\right]^2=-\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right]\\quad \\Rightarrow\\quad \\left[ \\begin{matrix} 0 & -1 \\\\ 1 & 0 \\end{matrix}\\right]=\\sqrt{-\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right]}$$\n\nUsing $(\\star)$ we have following: $$\\left[ \\begin{matrix} a & -b \\\\ b & a \\end{matrix}\\right]=a\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right]+b\\sqrt{-\\left[ \\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right]}\\approx a+b\\sqrt{-1}=a+bi$$\n$$\\circ--------------------------------\\circ$$\n\n9)A very usefull periodic identity. (i)\n\n$f:\\mathbb R\\to \\mathbb R$ be continuous and periodic with period $T>0$ then:\n\n$$(\\forall a\\in \\mathbb R),\\quad\\displaystyle\\int_a^{a+T}f(x)dx=\\displaystyle\\int^T_0f(x)dx$$\n$$\\circ--------------------------------\\circ$$\n\n10)A very usefull periodic identity. (ii)\n\n$f:\\mathbb R\\to \\mathbb R$ be continuous and periodic with period $T>0$ then:\n\n$$(\\forall a<b \\in \\mathbb R),\\quad\\lim_{n\\to \\infty}\\displaystyle\\int_a^b f(nx)dx=\\dfrac{b-a}{T}\\displaystyle\\int^T_0f(x)dx$$\n$$\\circ--------------------------------\\circ$$\n\n11) \"Every group is isomorphic to a group of permutations.\" A motivation can be given as following.\nConsidering the cayley table of a group we obtain a permutation for $g_i,g_j\\in G$ that $g_i g_j \\in G=\\{g_1,g_2,...,g_n,...\\}$\n$$\\circ--------------------------------\\circ$$\n\n12) $$\\arctan\\left(\\frac{a_1}{a_2}\\right)+\\arctan\\left(\\frac{b_1}{b_2}\\right)=\\arctan\\left(\\frac{a_1b_2+\na_2b_1}{a_2b_2-a_1b_1}\\right)$$\n$$\\circ--------------------------------\\circ$$\n\n13.i) $$\\arctan\\tanh(n+1)-\\arctan\\tanh(n-1) = \\arctan\\left(\\frac{\\tanh(n+1)-\\tanh(n-1)}{1+\\tanh(n-1)\\tanh(n+1)}\\right)=\\\\=\\arctan\\left(\\frac{\\sinh(2)}{\\cosh(2n)}\\right)$$\n$$\\circ--------------------------------\\circ$$\n\n13.ii) Like in 13.i we have $$\\arctan\\left(\\frac{\\sinh(1)}{\\cosh(2n)}\\right) = \\arctan\\tanh\\left(n+\\frac{1}{2}\\right)-\\arctan\\tanh\\left(n-\\frac{1}{2}\\right)$$\n$$\\circ--------------------------------\\circ$$\n\n14) More generally (from m.s.e.) $$\\displaystyle\\sum_j \\arctan x_j= \\arctan \\frac{\\sum_j x_j - \\sum_{j,k,\\ell} x_j x_k x_\\ell + \\sum\\text{products of five} - \\sum\\text{products of seven} + \\cdots }{1 - \\sum_{j,k} x_j x_k + \\sum\\text{products of four} - \\sum\\text{products of six} + \\cdots}$$\n$$\\circ--------------------------------\\circ$$\n\n15)\n$\\forall n\\in \\mathbb N^+$\n$$\\left\\lfloor\\sqrt{n^2+2n}\\right\\rfloor=n$$\n$$$$\n\nGenerally if $a,b\\in\\mathbb R^+$ with $a<b$ then $\\lfloor a\\rfloor \\le \\lfloor b\\rfloor$\n$$$$\nProof:\nAssume $a,b\\in\\mathbb R^+$ with $a < b $ but $ \\lfloor a \\rfloor> \\lfloor b \\rfloor $\n$$$$\n$\\lfloor a\\rfloor > \\lfloor b\\rfloor$ implies $ \\lfloor a\\rfloor \\ge \\lfloor b\\rfloor+1$ and \nnaturally from $a<b$ we got $ \\lfloor a\\rfloor \\le a < b < \\lfloor b\\rfloor +1$ and using $ \\lfloor a\\rfloor \\ge \\lfloor b\\rfloor+1$\n$$$$\nwe got : $$ \\lfloor b\\rfloor+1 \\le \\lfloor a\\rfloor \\le a < b < \\lfloor b\\rfloor +1$$\n\nbut $$\\lfloor b\\rfloor +1<\\lfloor b\\rfloor +1$$ is absurd.\n$$\\circ--------------------------------\\circ$$\n<\/div>\n<\/div><\/div><\/div><\/div><\/div>","protected":false},"excerpt":{"rendered":"

– 1) I’ve seen this, while dealing with a real analyis problem. It can be very usefull proving statements. $$\\forall x,y\\in[0,\\infty),\\quad \\sqrt{|x-y|}\\ge |\\sqrt x -\\sqrt y|$$ $$\\circ——————————–\\circ$$ 2) Again, in real analysis. $$\\forall f\\neq0,\\quad \\left(f+\\dfrac1f -2\\right)=\\left(f-1 \\right)\\left(1-\\dfrac1f \\right)$$ $$\\circ——————————–\\circ$$ 3)$$\\forall x_1,x_2\\in[0,\\infty),\\quad|sin^2x_1-sin^2x_2|\\le 2|x_1-x_2|$$ $$\\circ——————————–\\circ$$ 4)$$ \\text{For } \\;\\; k \\in \\{0, \\cdots, m\\}\\;\\;\\text{ we have}$$ $$ […]<\/a><\/p>\n","protected":false},"author":5643,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"categories":[1],"tags":[],"class_list":["post-1","post","type-post","status-publish","format-standard","hentry","category-uncategorized","category-1-id","post-seq-1","post-parity-odd","meta-position-corners","fix"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/posts\/1","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/users\/5643"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/comments?post=1"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/posts\/1\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/media?parent=1"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/categories?post=1"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e221325\/wp-json\/wp\/v2\/tags?post=1"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}