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"ପ୍ରମାଣ ଯେ π ଗୋଟିଏ ଅପରିମେୟ ସଂଖ୍ୟା" ପୃଷ୍ଠାର ସଂସ୍କରଣ‌ଗୁଡ଼ିକ ମଧ୍ୟରେ ତଫାତ

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୧୩ କ ଧାଡ଼ି:
:<math>\tan(x) = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - {}\ddots}}}}.</math>
 
ତା ପରେ ଲାମ୍ବର୍ଟ ପ୍ରମାଣ ଯେ ଯଦି ''x'' ଶୂନ୍ୟ ଛଡ଼ା ଏକ ମୂଳଦ ସଂଖ୍ୟା ତା ହେଲେ ଏହି ଅଭିବ୍ୟକ୍ତିଟି ଅମୂଳଦ. ଯେହେତୁ tan({{pi}}/4)&nbsp;=&nbsp;1, ଆମେ କହିପାରିବା ଯେ {{pi}}/4 ଆମୂଳଦ ଏବଂ ତେଣୁ {{pi}} ଆମୂଳଦ. AThis simplificationresult can also be proved using even more basic tools of Lambert'scalculus proof(integrals isinstead givenof series).<ref>{{citation | last = Zhou | first = Li | last2 = Markov | first2 = Lubomir | title = Recurrent Proofs of the Irrationality of Certain Trigonometric Values | journal = [[ProofAmerican thatMathematical πMonthly]] is| irrational#Laczkovichvolume = 117 | issue = 4 | year = 2010 | pages = 360–362|arxiv = 0911.27s_proof1933 |below]] doi=10.4169/000298910x480838}}</ref><ref name="Zhou">{{citation|last = Zhou|first = Li|title = Irrationality proofs a la Hermite|periodical = Math. Gazette|year = 2011|issue = November|arxiv = 0911.1929}}</ref>
 
== ହର୍ମାଇଟ୍ ଙ୍କ ପ୍ରମାଣ ==
ଏହି ପ୍ରମାଣ୍ଟି<ref name="Hermite">{{citation|last = Hermite|first = Charles|author-link = Charles Hermite|year = 1873|title = Extrait d'une lettre de Monsieur Ch. Hermite à Monsieur Paul Gordan | language = french |periodical = [[Crelle's Journal|Journal für die reine und angewandte Mathematik]]|volume = 76|pages = 303–311|url = http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN002155435}}</ref><ref>{{citation|last = Hermite|first = Charles|author-link = Charles Hermite|year = 1873|title = Extrait d'une lettre de Mr. Ch. Hermite à Mr. Carl Borchardt |language = french|periodical = [[Crelle's Journal|Journal für die reine und angewandte Mathematik]]|volume = 76|pages = 342–344|url = http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN00215546X}}</ref> {{pi}}ର ବୈଶିଷ୍ଠ୍ୟ ବ୍ୟବହାର କରେ ଓ ଏହା ପ୍ରମିଣ କରେ ଯେ {{pi}}<sup>2</sup> ଅମୂଳଦ. ଅମୂଳଦତାର ବହୁ ଅନ୍ୟ ପ୍ରମାଣ ଭଳି,ଯୁକ୍ତିଟି [[ପରୋକ୍ଷ ବ୍ୟତିରେକୀ ପ୍ରମାଣ (reductio ad absurdum)]] ଦ୍ୱାରା କରାଜଯାଇଛି.
 
ଧରନ୍ତୁ ନିମ୍ନଲିଖିତ [[କ୍ରମ (sequences)]] ଅଛି (''A''<sub>''n''</sub>)<sub>''n''&nbsp;≥&nbsp;0</sub> and (''U''<sub>''n''</sub>)<sub>''n''&nbsp;≥&nbsp;0</sub> ଫଂସନ୍ ର '''R''' ରୁ '''R'''କୁ, ଓ ନିମ୍ନଭାବେ ନିର୍ଧାରଣ କରାଜଯାଇଛି:
#<math>A_0(x)=\sin(x);\,</math>
#<math>(\forall n\in\mathbb{Z}_+):A_{n+1}(x)=\int_0^xyA_n(y)\,dy;</math>
#<math>U_0(x)=\frac{\sin(x)}x;</math>
#<math>(\forall n\in\mathbb{Z}_+):U_{n+1}(x)=-\frac{U_n'(x)}x.</math>
ଆନୟନ ଯୁକ୍ତିରେ ପ୍ରମାଣ କରିହେବ ଯେ:
:<math>(\forall n\in\mathbb{Z}_+):A_n(x)=\frac{x^{2n+1}}{(2n+1)!!}-\frac{x^{2n+3}}{2\times(2n+3)!!}+\frac{x^{2n+5}}{2\times4\times(2n+5)!!}\mp\cdots</math>
ଏବଂ ଯେ:
:<math>(\forall n\in\mathbb{Z}_+):U_n(x)=\frac1{(2n+1)!!}-\frac{x^2}{2\times(2n+3)!!}+\frac{x^4}{2\times4\times(2n+5)!!}\mp\cdots</math>
ଓ ତେଣୁକରି:
:<math>U_n(x)=\frac{A_n(x)}{x^{2n+1}}.</math>
ତେବେ:
:<math>\frac{A_{n+1}(x)}{x^{2n+3}}=U_{n+1}(x)=-\frac{U_n'(x)}x=-\frac1x\frac d{dx}\left(\frac{A_n(x)}{x^{2n+1}}\right),</math>
ଯାହାକି ନିମ୍ନ ସହିତ ସମାନ:
:<math>A_{n+1}(x)=(2n+1)A_n(x)-xA_n'(x)=(2n+1)A_n(x)-x^2A_{n-1}(x).\,</math>
"https://or.wikipedia.org/wiki/ପ୍ରମାଣ_ଯେ_π_ଗୋଟିଏ_ଅପରିମେୟ_ସଂଖ୍ୟା"ରୁ ଅଣାଯାଇଅଛି