- #1

mohabitar

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How were they able to derive this?

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In summary, the conversation discusses a geometric sequence and its sum, denoted by s_p. The question is whether there is a linear relationship between s_{p+1} and s_p and how to find the sum starting at 0 instead of 1. The conversation also discusses a known formula for this type of problem and how to change the index of the summation. Ultimately, the solution is to subtract 1 from the sum starting at 1.

- #1

mohabitar

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How were they able to derive this?

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- #2

Mark44

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The sequence in this sum is a geometric sequence.mohabitar said:

How were they able to derive this?

- #3

fzero

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- #4

JonF

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What would (1/2)*S_p look like?

What about S_p - (1/2)*S_p?

Now note: S_p - (1/2)*S_p = S_p(1 – ½) = (½)S_p So ask yourself what 2*(½)S_p = S_p looks like, and you should get your desired result

- #5

mohabitar

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Sorry the last post confused me. I mean is there a known formula for something like this?

- #6

mohabitar

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- #7

fzero

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[tex]\frac{1}{2}s_p = \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^p} + \frac{1}{2^{p+1}}[/tex]

[tex]s_{p+1} = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{p-1}} + \frac{1}{2^p}+\frac{1}{2^{p+1}}[/tex]

What relations between these expressions can you write down?

- #8

mohabitar

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None, I really don't see what you did from the 2nd step to the third step.

- #9

JonF

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I think subtracting them is easier to see.

S_{p} = 1/2 + 1/2^{2} + 1/2^{3} + 1/2^{4} + 1/2^{5} + … + 1/2^{p}

½S_{p} = 1/2^{2 }+ 1/2^{3} + 1/2^{4} + 1/2^{5} + … + 1/2^{p} + 1/2^{p+1}

if we take S_{p} - ½S_{p}, which terms would cancel out? Which would be left over? I color coded it to help you see it.

S

½S

if we take S

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- #10

fzero

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mohabitar said:None, I really don't see what you did from the 2nd step to the third step.

The 3rd step is just the expansion of the definition

[tex]s_{p} = \sum_{k=1}^p \frac{1}{2^k}[/tex]

applied to p+1.

- #11

mohabitar

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- #12

JonF

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- #13

fzero

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mohabitar said:

We're telling you how to find the sum without referring to another result. If you know the answer for

[tex]S_p = \sum_{k=0}^p \frac{1}{2^p},[/tex]

then explain where you got it and we'll tell you how to use it here.

- #14

mohabitar

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- #15

JonF

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Consider this example for changing indexes, actually write out the terms

∑

∑

- #16

HallsofIvy

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The only difference between [itex]\sum_{n=0}^\infty r^n[/itex] and [itex]\sum_{n=1}^\infty r^n[/itex] is that the first is missing the first term, [itex]r^0= 1[/itex]. To find [itex]\sum_{n=1}^\infty r^n[/itex], just find [itex]\sum_{n=0}^\infty r^n[/itex] and subtract 1.

For example, to find [itex]]\displaytype\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n[/itex], first find [itex]\displaytype\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n= \frac{1}{1- \frac{1}{3}}= \frac{3}{2}[/itex].

Now subtract 1: [itex]\frac{3}{2}- 1= \frac{1}{2}[/itex]

In your original problem, with a finite sum, to find [itex]\displaytype\sum_{k=1}^p\frac{1}{2}^p[/itex], start from

[tex]\displaytype\sum_{k=0}^p \frac{1}{2^p}= \frac{1- \left(\frac{1}{2}\right)^{p+1}}{1- \frac{1}{2}[/tex]

[tex]= 2(1- \frac{1}{2^{p+1}})= 2- \frac{1}{2^p}[/tex]

Now subtract 1:

[tex]\left(2- \frac{1}{2^p}\right)- 1= 1- \frac{1}{2^p}[/tex]

For example, to find [itex]]\displaytype\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n[/itex], first find [itex]\displaytype\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n= \frac{1}{1- \frac{1}{3}}= \frac{3}{2}[/itex].

Now subtract 1: [itex]\frac{3}{2}- 1= \frac{1}{2}[/itex]

In your original problem, with a finite sum, to find [itex]\displaytype\sum_{k=1}^p\frac{1}{2}^p[/itex], start from

[tex]\displaytype\sum_{k=0}^p \frac{1}{2^p}= \frac{1- \left(\frac{1}{2}\right)^{p+1}}{1- \frac{1}{2}[/tex]

[tex]= 2(1- \frac{1}{2^{p+1}})= 2- \frac{1}{2^p}[/tex]

Now subtract 1:

[tex]\left(2- \frac{1}{2^p}\right)- 1= 1- \frac{1}{2^p}[/tex]

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A closed form summation is a mathematical expression that represents the sum of a finite series of numbers. It is also known as an exact solution, as it provides an explicit formula for calculating the sum rather than relying on iterative or recursive methods.

Understanding the derivation of closed form summation allows us to efficiently and accurately calculate the sum of a series without having to manually add up each individual term. It also provides insight into the underlying mathematical principles and can be applied to various fields such as physics, engineering, and finance.

The derivation of closed form summation involves using mathematical techniques such as algebra, calculus, and geometric series to manipulate the series expression into a simpler and more manageable form. This process may also involve identifying patterns and using known formulas to simplify the expression.

No, closed form summation can only be applied to series that have a finite number of terms and follow a specific pattern or sequence. Some examples of series that can be solved using closed form summation include arithmetic series, geometric series, and power series.

Yes, there are alternative methods such as using recursive formulas or numerical approximations such as the trapezoidal rule or Simpson's rule. However, closed form summation is considered the most efficient and accurate method for calculating the sum of a series, especially for larger series with a large number of terms.

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