How Do You Solve the Floor Integral with Logarithmic and Geometric Sequences?

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In summary, the integral is a sum of a geometric series which is the sum of the harmonic sequences of 1, 1/2, 1/3, 1/4, 1/5, ....
  • #1
WhatTheYock
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Doing integrals for just plain curiosity! So, given the integral:

∫dx/floor(1-logbase2(1-x)) from 0 to 1

I have looked at the graph of the integral, and I notice what seems to be an infinite number of areas under the curve (from 0 to 1/2 the area is 1/2, from 1/2 to 3/4 the area is 1/8, etc). How can I get a general expression for the nth rectangle (perhaps a summation) and a numerical answer for the integral? Thanks for any help!
 
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  • #2
If there are indeed an infinite number of areas then the integral will be an infinite series.

Assuming that your values are right (I didn't check), then 1/2+ 1/8+ 1/64+ ... , all powers of 1/2 so that is a geometric series. There is a formula for sums of geometric series that you probably know.
 
  • #3
HallsofIvy said:
If there are indeed an infinite number of areas then the integral will be an infinite series.

Assuming that your values are right (I didn't check), then 1/2+ 1/8+ 1/64+ ... , all powers of 1/2 so that is a geometric series. There is a formula for sums of geometric series that you probably know.

I think the sum is actually 1*(1/2)+(1/2)*(1/4)+(1/3)*(1/8)+(1/4)*(1/16)+(1/5)*(1/32)+... That's a little harder. It's the integral of a geometric series.
 
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  • #4
Yes, but after some more inspection of the graph I observed that the heights of each rectangle start as 1 and form the harmonic sequence (1, 1/2, 1/3, 1/4, 1/5, ...) and the length form a geometric sequence (1/2, 1/4, 1/8, 1/16, ...). So I get the sum from 1 to infinity of (1/n)*(1/2^n) which comes out to ln(2) which I think is the answer...
 
  • #5
WhatTheYock said:
Yes, but after some more inspection of the graph I observed that the heights of each rectangle start as 1 and form the harmonic sequence (1, 1/2, 1/3, 1/4, 1/5, ...) and the length form a geometric sequence (1/2, 1/4, 1/8, 1/16, ...). So I get the sum from 1 to infinity of (1/n)*(1/2^n) which comes out to ln(2) which I think is the answer...

Why, yes, I think it is.
 

FAQ: How Do You Solve the Floor Integral with Logarithmic and Geometric Sequences?

1. What is a "Much Harder Floor Integral"?

A Much Harder Floor Integral is a mathematical concept that involves finding the area under a curve on a floor that is not perfectly flat. It is commonly used in engineering and physics to calculate the work done in moving an object across a non-uniform surface.

2. How is a Much Harder Floor Integral different from a regular integral?

A Much Harder Floor Integral is different from a regular integral in that it takes into account the varying difficulty of movement across a non-uniform surface. This means that the integral will be more complex and require additional calculations.

3. What types of surfaces can be used for a Much Harder Floor Integral?

A Much Harder Floor Integral can be used on any surface that is not perfectly flat, such as a bumpy road, a sloped floor, or a staircase. The surface does not need to be two-dimensional, as long as it is a continuous and measurable surface.

4. How is a Much Harder Floor Integral calculated?

To calculate a Much Harder Floor Integral, the surface is divided into small sections and the difficulty of movement in each section is calculated. These values are then multiplied by the distance traveled in each section and added together to find the total work done.

5. What are the practical applications of a Much Harder Floor Integral?

A Much Harder Floor Integral has many practical applications, such as calculating the energy required to move a car across a bumpy road, or the work done in climbing a mountain. It can also be used in engineering to design more efficient machines and structures that take into account the difficulty of movement across non-uniform surfaces.

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