why-calculus-works
  • Introduction
  • Real Analysis
    • The World of Real Numbers
    • Limits and Derivatives
    • Integration
  • Measure Theory
  • Complex Analysis
  • Goodies
    • Calculus Computations
Powered by GitBook
On this page

Goodies

PreviousComplex AnalysisNextCalculus Computations

Last updated 6 years ago

Golden Ratio

1+1+1+…=?\sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}} = ?1+1+1+…​​​=?

How do we go about solving this problem? Set the value equal to x:

x=1+1+1+…x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}x=1+1+1+…​​​

implying

x2=1+1+1+1+…=1+xx^2 =1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}} = 1 + xx2=1+1+1+1+…​​​=1+x

Now have a quadratic equation we can solve: x2−x−1=0x^2 - x - 1 =0x2−x−1=0, implying one solution is

x=(1+5)/2x = (1 + \sqrt{5}) /2x=(1+5​)/2, which amazingly is the golden ratio!

[in progress]