Prove Series Product: 1+3^(-2^n) = 3/2(1-3^(-2^n+1))

[mishrashubham]

Homework Statement

Prove that
$$\prod_{n=0}^n 1+3^{-2^{n}}= \frac{3}{2}(1-3^{-2^{n+1}})$$

Homework Equations

The Attempt at a Solution

I am an idiot in math and I couldn't figure out where to start. I thought about using log to convert it into a summation series (I find sums easier to handle than products), but I couldn't progress. I just need a direction, a place to start from. I can hopefully take over from there.

Thanks

 

[GreenPrint]

What course is this for? Just out of curiosity, I have no idea how to answer the question, and am just wondering what course this is for. I wasn't introduced to basic series stuff tell Calculus II.

 

[mishrashubham]

What course is this for? Just out of curiosity, I have no idea how to answer the question, and am just wondering what course this is for. I wasn't introduced to basic series stuff tell Calculus II.

I don't know about the US equivalent, but I have this in Algebra, under "Sequences, Series and Progressions".

 

[NeroKid]

just use the induction it is quite easy

 

[dynamicsolo]

With infinite series or products, when in doubt, try writing out the first few terms. Set n = 1 , write the two terms in the product and multiply them out; what do you get? Now try it with n = 2 and multiply the third term times your result from n = 1. What sort of sum are you getting? Do you know an expression for the sum of a finite set of such terms?

just use the induction it is quite easy

It's true that induction would suffice to prove the relationship, which is all that is asked here. However, it would be nice to be able to establish the relationship, which in fact also turns out to be easy if you start multiplying out the terms of the product and looking at what that gives you...

 

[mishrashubham]

just use the induction it is quite easy

It really was extremely easy using induction. Thanks for the idea. I should have thought about using induction earlier.

With infinite series or products, when in doubt, try writing out the first few terms. Set n = 1 , write the two terms in the product and multiply them out; what do you get? Now try it with n = 2 and multiply the third term times your result from n = 1. What sort of sum are you getting? Do you know an expression for the sum of a finite set of such terms?

It's true that induction would suffice to prove the relationship, which is all that is asked here. However, it would be nice to be able to establish the relationship, which in fact also turns out to be easy if you start multiplying out the terms of the product and looking at what that gives you...

I always do that if I encounter series; to see if I get some pattern. But I couldn't figure it out in this question.

Thanks for the replies people.

 

[GreenPrint]

What's the name of the giant N character and which n is placed on top of and n=0 is placed on the bottom of? Is this just a different way to write sigma, and is the same way of indicating a series from n=0 to n?

 

[Galron]

What's the name of the giant N character and which n is placed on top of and n=0 is placed on the bottom of? Is this just a different way to write sigma, and is the same way of indicating a series from n=0 to n?

Pretty much yes but a series of products instead of sums.

It's the symbol for a Number Product a giant pi symbol in fact, but then that is trivial.

Product of n=0...n ...

ETA: for clarity.

oops.

 

[GreenPrint]

My bad that's for product series, sigma is for summation series... hmm interesting

 

[uart]

What's the name of the giant N character and which n is placed on top of and n=0 is placed on the bottom of? Is this just a different way to write sigma, and is the same way of indicating a series from n=0 to n?

It's a product symbol but it's not quite written correctly. It should be:

$$\prod_{k=0}^n \left( 1+3^{-2^{k}}\right) = \frac{3}{2}(1-3^{-2^{n+1}})$$

 

[Ray Vickson]

Homework Statement

Prove that
$$\prod_{n=0}^n 1+3^{-2^{n}}= \frac{3}{2}(1-3^{-2^{n+1}})$$

Homework Equations

The Attempt at a Solution

I am an idiot in math and I couldn't figure out where to start. I thought about using log to convert it into a summation series (I find sums easier to handle than products), but I couldn't progress. I just need a direction, a place to start from. I can hopefully take over from there.

Thanks

I get the answer $$1 + 3^{-2^n}$$ because your expression equals $$(\prod_{n=0}^n 1) + 3^{-2^n}.$$ However, if you meant $$\prod_{k=0}^n (1 + 3^{-2^k}),$$ then that would give a different answer.

RGV

 

[mishrashubham]

It's a product symbol but it's not quite written correctly. It should be:

$$\prod_{k=0}^n \left( 1+3^{-2^{k}}\right) = \frac{3}{2}(1-3^{-2^{n+1}})$$

My apologies.



One more thing. How would I go about solving it had I not known the right hand side?

 

[dynamicsolo]

How would I go about solving it had I not known the right hand side?

If you write the first two terms of the product, you find

$$( 1 + 3^{-2^{0}} ) ( 1 + 3^{-2^{1}} ) = ( 1 + 3^{-1} ) ( 1 + 3^{-2} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) .$$

The first three terms yield

$$( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) ( 1 + 3^{-2^{2}} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} ) ( 1 + 3^{-4} ) = ( 1 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + 3^{-5} + 3^{-6} + 3^{-7}) .$$

What sort of finite series is this? How many terms are there is each successive multiplication? Is there a formula for the sum of the general finite series?