Wow

Let ϕ\phi denote the golden ratio, ϕ=1+52\phi = \dfrac{1+\sqrt5}2 , prove that ∑n=1∞1ϕn=ϕ\displaystyle \sum_{n=1}^\infty \dfrac1{\phi^n} = \phi .

#Algebra

Note by Hamza A
5 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Whats ϕ\phi?? -Golden ratio ??
If that's the case then above summation is just the sum of a infinite GP with first term and common ratio 1ϕ\frac{1}{\phi}. ∴∑n=1∞1ϕn=1ϕ1−1ϕ=1ϕ−1=ϕ\therefore \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ \phi ^{ n } } }=\dfrac{\frac{1}{\phi}}{1-\frac{1}{\phi}}=\dfrac{1}{\phi-1}=\phi [Since ϕ2−ϕ−1=0⇒ϕ−1=1ϕ\phi^2-\phi-1=0\Rightarrow \phi-1=\dfrac{1}{\phi}]

Rishabh Jain - 5 years, 4 months ago

Log in to reply

Going further, if we define xkx_{k} for integers k≥0k \ge 0 as the solution to the equation ∑n=1∞1xn=xk\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{x^{n}} = x^{k} then lim⁡k→∞xk=1\displaystyle\lim_{k \to \infty} x_{k} = 1.

Brian Charlesworth - 5 years, 4 months ago

yes,it's the golden ratio

Hamza A - 5 years, 4 months ago

That is pretty cool, and ϕ\phi is the unique solution to the equation ∑n=1∞1xn=x\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{x^{n}} = x where x>1x \gt 1.

Writing the sum as S=∑n=1∞(1x)n=∑n=1∞ynS = \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{x}\right)^{n} = \sum_{n=1}^{\infty} y^{n} where y=1xy = \dfrac{1}{x} is such that 0<y<10 \lt y \lt 1, we see that we have an infinite geometric series with

S=y1−y=1x1−1x=1x−1S = \dfrac{y}{1 - y} = \dfrac{\dfrac{1}{x}}{1 - \dfrac{1}{x}} = \dfrac{1}{x - 1}.

Then to have S=xS = x we require that 1x−1=x⟹1=x2−x⟹x2−x−1=0\dfrac{1}{x - 1} = x \Longrightarrow 1 = x^{2} - x \Longrightarrow x^{2} - x - 1 = 0,

the positive root of which is 1+52=ϕ>1\dfrac{1 + \sqrt{5}}{2} = \phi \gt 1.

Brian Charlesworth - 5 years, 4 months ago
×

Problem Loading...

Note Loading...

Set Loading...