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this is part of the set @the Radicals \[\] let there be a infinite Radical \[ \sqrt{17+2\sqrt{88+24\sqrt{17+2\sqrt{88+24\sqrt{\dots\dots }}}}}\] it can be written as \[\sqrt{3^2+2^2+2^2+2\sqrt{(3\times2)^2+(2\times 2)^2 +(3\times 2)^2 +2(3\times 2\times 2)\sqrt{\dots\dots }}}\] then we can say that \[ \sqrt{17+2\sqrt{88+24\sqrt{17+2\sqrt{88+24\sqrt{\dots\dots }}}}}=7\] how??? lets see\[\] let \[F(a,b,c)=a+b+c\] \[F(a,b,c)=\sqrt{a^2 +b^2+c^2 +2(ab+bc+ca)}\] \[F(a,b,c)=\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2(a^2bc+ab^2c+abc^2)}}\] \[F(a,b,c)=\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)}}\] now we can expand \(a+b+c\) like before and make it an infinite radical\[\] so we can say that\[\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{\dots\dots}}}}} =a+b+c\]

Note by Aareyan Manzoor
2 years, 7 months ago

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\[x+a=\sqrt[3]{(x+a)^3}\\x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}\] In the same way get \(x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}\) and \(x+4a=\sqrt[3]{4a(x^2+3(4a)x+(4a)^2)+x^2(x+8a)}\) and the expressions for \(x+8a,x+16a\;etc\).Replacing all of these in the original expression we get: \[\color{Red}{x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)x+(4a)^2)+x^2\sqrt[3]{\dotsm\;to\; \infty}}}}}\] \[\LARGE{\color{Green}{V}\color{Blue}{O}\color{Brown}{I}\color{Purple}{L}\color{Red}{A}\color{Black}{!!!}}\] Abdur Rehman Zahid · 2 years, 7 months ago

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@Abdur Rehman Zahid nice one Aareyan Manzoor · 2 years, 7 months ago

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@Abdur Rehman Zahid Please comment @Aareyan Manzoor Abdur Rehman Zahid · 2 years, 7 months ago

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@Abdur Rehman Zahid ok, i find these seriously fascinating, the one by you is 1 of many. it is really good as well, let me show you another \[a+b =\sqrt[3]{a^3 +b^3 +3ab(a+b)}\] \[a+b=\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{\dots\dots}}}}\] Aareyan Manzoor · 2 years, 7 months ago

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@Aareyan Manzoor How about this one: \[x=\sqrt{1+(x-1)(x+1)}\\=\sqrt{1+(x-1)\sqrt{1+x(x+2)}}\\=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}}\\=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}}\] Continuing this indefinitely we get: \[x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\dotsm}}}}}\] Abdur Rehman Zahid · 2 years, 7 months ago

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@Abdur Rehman Zahid already taken by ramnujan \[\sqrt{ax+(a+n)^2+x\sqrt{ax+(a+n)^2+(x+n)\sqrt{\dots\dots}}}=x+n+a\] Aareyan Manzoor · 2 years, 7 months ago

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@Aareyan Manzoor Yeah,realized that after posting the comment. Abdur Rehman Zahid · 2 years, 7 months ago

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@Aareyan Manzoor Nice.I was just fixated on nested radicals when I saw the one by Ramanujan.I find them seriously beautiful. Abdur Rehman Zahid · 2 years, 7 months ago

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Please note that just because something continues ad infinitum, you cannot add the \(\inf\) symbol there. That'd mean you're literally having the hyperreal number infinity in the expresssion Agnishom Chattopadhyay · 2 years, 7 months ago

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@Agnishom Chattopadhyay thanks, by the way, what is a hyperreal number,sir? Aareyan Manzoor · 2 years, 7 months ago

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@Aareyan Manzoor Hyperreal Numbers (denoted \(*\mathbb{R}\) is an extension of Real Numbers to accomodate infinite and infinitesimal numbers. It's an interesting idea, you could look it up. Agnishom Chattopadhyay · 2 years, 7 months ago

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KAGE BUNSHIN NO JUTSU Samurai Poop · 1 year, 6 months ago

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