Waste less time on Facebook — follow Brilliant.
×

READ ME

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, 12 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

\[x+a=\sqrt[3]{(x+a)^3}\\x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}\] 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, 11 months ago

Log in to reply

nice one

Aareyan Manzoor - 2 years, 11 months ago

Log in to reply

Please comment @Aareyan Manzoor

Abdur Rehman Zahid - 2 years, 11 months ago

Log in to reply

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, 11 months ago

Log in to reply

@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, 11 months ago

Log in to reply

@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, 11 months ago

Log in to reply

@Aareyan Manzoor Yeah,realized that after posting the comment.

Abdur Rehman Zahid - 2 years, 11 months ago

Log in to reply

@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, 11 months ago

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, 12 months ago

Log in to reply

thanks, by the way, what is a hyperreal number,sir?

Aareyan Manzoor - 2 years, 11 months ago

Log in to reply

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, 11 months ago

Log in to reply

KAGE BUNSHIN NO JUTSU

Samurai Poop - 1 year, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...