- #1
Johnny Leong
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Please give me some hints:
Sum of (i^2)/(4^i) where i is from 0 to infinity.
Sum of (i^2)/(4^i) where i is from 0 to infinity.
Johnny Leong said:Please give me some hints:
Sum of (i^2)/(4^i) where i is from 0 to infinity.
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.
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.
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]
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 + ...]
The value of the sum of (i^2)/(4^i) where i is from 0 to infinity is approximately 1.33333.
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.
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.
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.