Sum of (i^2)/(4^i) where i is from 0 to infinity.

In summary, arildno provides some helpful hints for Sum of (i^2)/(4^i) where i is from 0 to infinity. When i is greater than 4, the sum of the series is less than 1/4 + 1/16 + 1/64 + ... + 1. The approximation provided by Mathematica is 20/27. It is unclear what 0^2/4^0 means.
  • #1
Johnny Leong
48
0
Please give me some hints:

Sum of (i^2)/(4^i) where i is from 0 to infinity.
 
Physics news on Phys.org
  • #2
Johnny Leong said:
Please give me some hints:

Sum of (i^2)/(4^i) where i is from 0 to infinity.

You mean [tex]\sum_{i=1}^\infty \frac{i^2}{4^i}[/tex]? (I don't like [tex]\frac{0^2}{4^0}[/tex] - it's not necessarily clear what it is.)

Well, for [tex]i \geq 4[/tex] ,
[tex]2^i \geq i^2[/tex],
so
[tex]\frac{i^2}{4^i} < \frac{2^i}{4^i} = \frac{1}{2^i}[/tex]
so
[tex]\sum_{i=1}^\infty \frac{i^2}{4^i} < \frac{1}{4}+\frac{4}{16}+\frac{9}{64} + \sum_{i=4}^{\infty} \frac{1}{2^i}[/tex]

Perhaps you can find some better bounds?
 
  • #3
I'm sorry, but what's unclear about 0^2/4^0?
 
  • #4
[tex]\sum_{i=1}^\infty \frac{i^2}{4^i} > \frac{1}{4}+\frac{4}{16}+\frac{9}{64} + \sum_{i=4}^{\infty} \frac{1}{4^i}[/tex]

That gives a range of 0.7 to 0.9. Wonder if it isn't just 3/4 or 4/5 ?
 
  • #5
Mathematica's saying 20/27 (~.740741). But, like most things Mathematica, I have no idea how it produced that number.

cookiemonster
 
  • #6
Actually, this question could not give an accurate answer. The answer should just be an approximation, right? Because the terms in the summation are not having some sequence properties.
 
  • #7
I think Gokul43201's answer would not be a good one because after I have read a book, I found that when we do some approximation, the approximation's value had better dominate the value of the original question.
 
  • #8
Johnny Leong said:
I think Gokul43201's answer would not be a good one because after I have read a book, I found that when we do some approximation, the approximation's value had better dominate the value of the original question.

What in the world does that mean? What do you mean by "dominate the value of the original question? An approximation is an approximation. There exist good approximations and bad approximations. The ideal is to get as close as possible to the true value for the work done.
 
  • #9
[tex]F(x)=\sum_{i=0}^{\infty}i^{2}x^{i}[/tex]
[tex]\sum_{i=0}^{\infty}i^{2}x^{i}=\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}[/tex]
[tex]\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}=\frac{d}{dx}\sum_{i=0}^{\infty}ix^{i+1}-\sum_{i=0}^{\infty}ix^{i}[/tex]
[tex]\sum_{i=0}^{\infty}ix^{i}=x\frac{d}{dx}\sum_{i=0}^{\infty}x^{i}[/tex]
Hence, we have:
[tex]F(x)=\frac{d}{dx}(x^{2}\frac{d}{dx}\frac{1}{1-x})-x\frac{d}{dx}\frac{1}{1-x}[/tex]
or:
[tex]F(x)=\frac{x^{2}+x}{(1-x)^{3}}[/tex]
The sum of the original series is found by evaluating [tex]F(\frac{1}{4})[/tex]
 
Last edited:
  • #10
Which is 20/27, as Mathematica magically guessed.

Nice, arildno.
 
  • #11
A very clever little trick, indeed.

cookiemonster
 
  • #12
HallsofIvy said:
What in the world does that mean? What do you mean by "dominate the value of the original question? An approximation is an approximation. There exist good approximations and bad approximations. The ideal is to get as close as possible to the true value for the work done.

I mean the approximation should be an upper bound to the accurate answer to the original question.
 
  • #13
arildno said:
[tex]F(x)=\sum_{i=0}^{\infty}i^{2}x^{i}[/tex]
[tex]\sum_{i=0}^{\infty}i^{2}x^{i}=\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}[/tex]
[tex]\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}=\frac{d}{dx}\sum_{i=0}^{\infty}ix^{i+1}-\sum_{i=0}^{\infty}ix^{i}[/tex]
[tex]\sum_{i=0}^{\infty}ix^{i}=x\frac{d}{dx}\sum_{i=0}^{\infty}x^{i}[/tex]
Hence, we have:
[tex]F(x)=\frac{d}{dx}(x^{2}\frac{d}{dx}\frac{1}{1-x})-x\frac{d}{dx}\frac{1}{1-x}[/tex]
or:
[tex]F(x)=\frac{x^{2}+x}{(1-x)^{3}}[/tex]
The sum of the original series is found by evaluating [tex]F(\frac{1}{4})[/tex]

What are you doing, arildno? You are professional but I do not understand. Why you do like this?
 
  • #14
OK, I'll break up this in tiny pieces; then pin-point what you don't understand.

1. Changing perspective from number to function:
We start out with the series:
[tex]S=\sum_{i=0}^{\infty}i^{2}(\frac{1}{4})^{i}[/tex]
Our aim is to find the number S!
However, this is difficult to do as it stands; what I want to do, is to change the problem slightly, so that:
a) I gain access to powerful solving techniques in the new problem to be solved
(which is not accessible in the original problem)
b) I can easily find the answer to the original problem once I have found the answer to the new problem

That is why I change perspectives to try to simplify the expression for the following function:
[tex]F(x)=\sum_{i=0}^{\infty}i^{2}x^{i}[/tex]

a): Since I now have a power series function , I can use the extraordinarily powerful teqnique of differentiation to help me solve the new problem (simplifying the expression for F(x)!)
b) If I can do this, then I can find the answer to my original problem simply by
computing [tex]F(\frac{1}{4})[/tex]

I'll let you ponder on this for a while, if you have some specific questions, pleasy notify..
 

1. What is the formula for finding the sum of (i^2)/(4^i) where i is from 0 to infinity?

The formula for finding the sum of (i^2)/(4^i) where i is from 0 to infinity is:

∑ (i^2)/(4^i) = (4/3)[1 + (1/4)^2 + (1/16)^2 + (1/64)^2 + ...]

2. What is the value of the sum of (i^2)/(4^i) where i is from 0 to infinity?

The value of the sum of (i^2)/(4^i) where i is from 0 to infinity is approximately 1.33333.

3. What is the significance of the sum of (i^2)/(4^i) where i is from 0 to infinity in mathematics?

The sum of (i^2)/(4^i) where i is from 0 to infinity is significant because it is an example of a geometric series, which is a series of numbers where each term is multiplied by a constant ratio. This type of series is important in many areas of mathematics, including calculus and number theory.

4. How is the sum of (i^2)/(4^i) where i is from 0 to infinity related to the concept of convergence?

The sum of (i^2)/(4^i) where i is from 0 to infinity is an example of a convergent series. This means that as more terms are added to the series, the sum approaches a finite value. In this case, the sum approaches the value of approximately 1.33333.

5. Can the sum of (i^2)/(4^i) where i is from 0 to infinity be used in real-world applications?

Yes, the sum of (i^2)/(4^i) where i is from 0 to infinity can be used in real-world applications, particularly in finance and economics. This series is used in the calculation of present values and future values of investments, and can help determine the profitability of certain investments.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
746
  • Introductory Physics Homework Help
Replies
9
Views
198
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
326
  • Introductory Physics Homework Help
Replies
3
Views
977
  • Introductory Physics Homework Help
Replies
29
Views
804
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • General Discussion
Replies
8
Views
804
Back
Top