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\[\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{\dots\dots}}}}} =a+b+c\]

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TopNewest\[x+a=\sqrt[3]{(x+a)^3}\\x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}\] 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, 9 months ago

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– Aareyan Manzoor · 2 years, 9 months ago

nice oneLog in to reply

@Aareyan Manzoor – Abdur Rehman Zahid · 2 years, 9 months ago

Please commentLog in to reply

– Aareyan Manzoor · 2 years, 9 months ago

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 \[a+b =\sqrt[3]{a^3 +b^3 +3ab(a+b)}\] \[a+b=\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{\dots\dots}}}}\]Log in to reply

– Abdur Rehman Zahid · 2 years, 9 months ago

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: \[x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\dotsm}}}}}\]Log in to reply

– Aareyan Manzoor · 2 years, 9 months ago

already taken by ramnujan \[\sqrt{ax+(a+n)^2+x\sqrt{ax+(a+n)^2+(x+n)\sqrt{\dots\dots}}}=x+n+a\]Log in to reply

– Abdur Rehman Zahid · 2 years, 9 months ago

Yeah,realized that after posting the comment.Log in to reply

– Abdur Rehman Zahid · 2 years, 9 months ago

Nice.I was just fixated on nested radicals when I saw the one by Ramanujan.I find them seriously beautiful.Log in to reply

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, 9 months agoLog in to reply

– Aareyan Manzoor · 2 years, 9 months ago

thanks, by the way, what is a hyperreal number,sir?Log in to reply

– Agnishom Chattopadhyay · 2 years, 9 months ago

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.Log in to reply

KAGE BUNSHIN NO JUTSU – Samurai Poop · 1 year, 8 months ago

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