×

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, 9 months ago

Sort by:

$x+a=\sqrt[3]{(x+a)^3}\\x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$ 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: $\color{Red}{x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)x+(4a)^2)+x^2\sqrt[3]{\dotsm\;to\; \infty}}}}}$ $\LARGE{\color{Green}{V}\color{Blue}{O}\color{Brown}{I}\color{Purple}{L}\color{Red}{A}\color{Black}{!!!}}$ · 2 years, 9 months ago

nice one · 2 years, 9 months ago

Please comment @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 $a+b =\sqrt[3]{a^3 +b^3 +3ab(a+b)}$ $a+b=\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{a^3 +b^3 +3ab\sqrt[3]{\dots\dots}}}}$ · 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: $x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\dotsm}}}}}$ · 2 years, 9 months ago

already taken by ramnujan $\sqrt{ax+(a+n)^2+x\sqrt{ax+(a+n)^2+(x+n)\sqrt{\dots\dots}}}=x+n+a$ · 2 years, 9 months ago

Yeah,realized that after posting the comment. · 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. · 2 years, 9 months ago

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 · 2 years, 9 months ago

thanks, by the way, what is a hyperreal number,sir? · 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. · 2 years, 9 months ago