Where's Eight?

Calculus Level 2

$\left ( 1 + \frac {1}{\color{#69047E}{2}} \right ) \left ( 1 + \frac {1}{\color{#69047E}{4}} \right ) \left ( 1 + \frac {1}{\color{#69047E}{16}} \right ) \dots = \ \color{teal} ?$

Details and Assumptions :

$\left (1 + \frac 1 8 \right )$ is in fact not being multiplied.

The answer is 2.

8 solutions

Discussions for this problem are now closed

Christopher Boo
Nov 12, 2014

$\displaystyle \frac{(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{16}) ... \infty}{(1-\frac{1}{2})}$

When you got that equation, apply the rule

$(a-b)(a+b)=a^2-b^2$

The first time you will get

$(1-\frac{1}{2})(1+\frac{1}{2})=1-\frac{1}{4}$

Noticed that you can continue to apply this rule since the next term is $1+\frac{1}{4}$

$(1-\frac{1}{4})(1+\frac{1}{4})=1-\frac{1}{16}$

Do this again and again, you will find the fraction part approaches to $0$ . Hence, you will get a numerator of $1-0$ . Divide it by $1-\frac{1}{2}$ , you get $2$

This is such an elegant solution, Christopher. So in general, for $x \gt 1$ ,

$\displaystyle\prod_{k=0}^{\infty} (1 + \dfrac{1}{x^{2^k}}) = \dfrac{x}{x - 1}$ .

We can't use the same concept on the following product, though:

$f(x) = \displaystyle\prod_{k=0}^{\infty} (1 - \dfrac{1}{x^{2^{k}}})$ .

Any ideas on how to proceed with this one? Using WA we have $f(2) = 0.350184.....$ , which doesn't look that "special", so I wonder if there is a closed-form solution to this one.

Brian Charlesworth - 6 years, 7 months ago

I have no idea at all. Calling for help: @Calvin Lin

Christopher Boo - 6 years, 7 months ago

I am unaware of how to calculate that.

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – 1.5x1.25x1.125=2.109.. shouldn't the ans be great er than 2?

Sachin Arora - 6 years, 6 months ago

@Sachin Arora – Note that the third term is $1 + \frac{1}{16} = 1.0625$ instead of $1 + \frac{ 1}{8} = 1.125$ .

Calvin Lin Staff - 6 years, 6 months ago

I think you stole my solution or maybe my mind, Boo!

Rajarshi Chatterjee - 6 years, 7 months ago

You are genius!

I was taught the same concept in my coaching but still I forgot.

I expect that I don't forget that atleast now.

I upvote your solution.

Ninad Akolekar - 6 years, 7 months ago

Why if you multiply the first 3 terms the answer is more than 2? and when you multiply the other which is obvious more than 1 could make it exactly 2. But when i see your solution I admire it. I just want to know what is its explanation.

John Darryl Ocido - 6 years, 6 months ago

Vincent Dio
Nov 14, 2014

$\lim _{ x\xrightarrow { \infty } }{ \frac { \left( 1-\frac { 1 }{ { 2 }^{ { 2 }^{ 2\dots } } } \right) }{ \left( 1-\frac { 1 }{ 2 } \right) } }$

$=\frac { \left( 1-\frac { 1 }{ \infty } \right) }{ \left( 1-\frac { 1 }{ 2 } \right) }$

$=\frac { \left( 1-0 \right) }{ \left( 1-\frac { 1 }{ 2 } \right) }$

$=2$

Moderator note:

You should explain your motivation for multiplying $\frac {1 - \frac 1 2}{1 - \frac 1 2 }$ and apply the difference in squares identity: $a^2 -b^2 = (a-b)(a+b)$ . Furthermore, your use of notation is wrong. You did not mention any $x$ in your limit.

Sushant Makode
Nov 14, 2014

series simplifies into 1 + 1/2 + 1/4 + 1/8 + 1/16..... Using GP formula Sum of n terms : a*(1-r^n)/(1-r) a=1, r=1/2, n=infinity.... therefore 1 *(1-0)/(1/2) =2

Ceesay Muhammed
Nov 14, 2014

Generally, for S=(1+x)(1+x^2)(1+x^4)(1+x^8)...................., where abs(x)<1, multiplying each side by (1-x) eats the right hand side up. (1-x)S=(1-x^2)(1+x^2)(1+x^4)(1+x^8)........., =(1-x^4)(1+x^4)(1+x^8)(1+x^16)... and so on. if we go on, the right hand side is approaching 1. Therefore, (1-x)S=1, so that S=1/(1-x). For the given question, x=1/2. so S=1/(1 -1/2)=2.

U should use latex code

Mehul Chaturvedi - 6 years, 7 months ago

In my country, very few people know what latex is and how it works. Unfortunately, I'm not one of them

Ceesay Muhammed - 6 years, 6 months ago

Fox To-ong
Dec 16, 2014

by trials of computation we can get the result of 2,since the fractional part has a patern of nsquared. just mupltiply at the value ( 1 + 1/65536) the result is 1 until infinity........

Moderator note:

This solution is marked wrong. Why would you choose an arbitrary number such as $2^{16} = 65536$ when there are infinitely number of finite terms being multiplied?

Jack Barker
Dec 5, 2014

\lim _{ x\rightarrow \infty }{ (1+\frac { 1 }{ { 2 }^{ x } } } )

(1+0.999....) is approximately equal to 2

Moderator note:

This solution has been marked wrong. How does $\displaystyle \lim_{x \to \infty} \left (1 + \frac 1 {2^x} \right ) \approx 2$ ? When in fact should equals to 1.

Guilherme Rocha
Dec 4, 2014

Are the denominators increasing exponencially or was there a typo?

Moderator note:

There was no typo as the title hinted. Please refrain from posting a comment as a solution. Especially when you did not show any relevant working to the actual problem.

Richard Levine
Nov 14, 2014

If you don't happen to see the elegant algebraic solution, it is easy enough to see that this series very quickly comes to approximately 1.9999 in just 4 terms -- 3/2 * 5/4 * 17/16 * 257/256 * ... After that, you are nearly multiplying by 1 (for example, 1+1/65536 = approximately 1.000015) forever, so the product approaches 2.

Moderator note:

This solution has been marked wrong. You have only shown that the it approximates to $2$ when you multiply a small number of terms. You should in fact show that the multiplication of ALL the finite terms equals to $2$ .

Better type in latex code

Mehul Chaturvedi - 6 years, 7 months ago

Where's Eight?

The answer is 2.

8 solutions

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