Some problems of physics and mathematics can be solved using the properties of numerical series. The two simplest numerical sequences studied in schools are algebraic and geometric. In this article, we consider in more detail the question of how to find the sum of an infinite progression of a geometric decreasing.
Geometric progression
These words mean such a series of real numbers, the elements a i of which satisfy the expression:
a i = a i-1 * r
Here i is the number of the element in the row, r is a constant number, which is called the denominator.
This definition shows that, knowing any member of the progression and its denominator, you can restore the entire series of numbers. For example, if the 10th element is known, then dividing it by r, we get the 9th element, then, dividing it again, we get the 8th and so on. These simple considerations allow us to write an expression that is valid for the considered series of numbers:
a i = a 1 * r i-1
An example of a progression with denominator 2 can be this series:
1, 2, 4, 8, 16, 32, ...
If the denominator is -2, then a completely different series is obtained:
1, -2, 4, -8, 16, -32, ...
Geometric progression is much faster than algebraic, that is, its members grow rapidly and decrease rapidly.
Sum of i progression members
To solve practical problems, one often has to calculate the sum of several elements of the considered numerical sequence. The following formula is valid for this case:
S i = a 1 * (r i -1) / (r-1)
It can be seen that to calculate the sum of i members it is necessary to know only two numbers: a 1 and r, which is logical, since they uniquely determine the entire sequence.
Descending sequence and the sum of its members
Now consider a special case. We assume that the denominator modulus r does not exceed unity, i.e., -1 <r <1. In this case, each subsequent term will be less than the previous one. If -1 <r <0, then the sign of the elements will alternate, but their modulus will constantly decrease. In this case, one speaks of a decreasing sequence of numbers.
It is interesting to consider a decreasing geometric progression because the infinite sum of its members tends to a finite real number.
We obtain the formula for the sum of an infinite geometric progression decreasing. This is easy to do if we write out the expression for S i given in the previous paragraph. We have:
S i = a 1 * (r i -1) / (r-1)
Consider the case when i-> ∞. Since the denominator modulus is less than 1, raising it to an infinite degree will give zero. This can be checked using the example r = 0.5:
0.5 2 = 0.25; 0.5 3 = 0.125; ...., 0.5 20 = 0.0000009.
As a result, the sum of the terms of the infinite geometric progression of the decreasing takes the form:
S ∞ = a 1 / (1-r)
This formula is often used in practice, for example, to calculate the area of shapes. It is also used to solve the paradox of Zeno of Elea with the tortoise and Achilles.
Obviously, consideration of the sum of an infinite progression of a geometric increasing (r> 1) will lead to the result S ∞ = + ∞.
The task of finding the first member of the progression
We show how to apply the above formulas as an example of solving the problem. It is known that the sum of an infinite geometric progression is 11. Moreover, its 7th term is 6 times smaller than the third term. What is the first element for this number series?
First, we write out two expressions for determining the 7th and 3rd elements. We get:
a 7 = a 1 * r 6
a 3 = a 1 * r 2
Dividing the first expression by the second, and expressing the denominator, we have:
a 7 / a 3 = r 4 => r = 4 √ (a 7 / a 3 )
Since the ratio of the seventh and third terms is given in the condition of the problem, we can substitute it and find r:
r = 4 √ (a 7 / a 3 ) = 4 √ (1/6) ≈ 0.63894
We calculated r with an accuracy of five significant decimal places. Since the obtained value is less than unity, it means that the progression is decreasing, which justifies the use of the formula for its infinite sum. We write the expression for the first term in terms of the sum S ∞ :
a 1 = S ∞ * (1-r)
We substitute the known values into this formula and get the answer:
a 1 = 11 * (1-0.63894) = 3.97166.
Zeno's famous paradox with fast Achilles and slow tortoise
Zeno of Elea - a famous Greek philosopher who lived in the V century BC. e. A number of its apogee or paradoxes have reached the present, in which the problem of the infinitely large and infinitesimal in mathematics is formulated.
One of the famous paradoxes of Zeno is the competition between Achilles and the tortoise. Zeno believed that if Achilles would provide some advantage to the tortoise in the distance, then he could never catch her. For example, let Achilles run 10 times faster than an animal crawling, which for example is 100 meters ahead of him. When the warrior runs 100 meters, the turtle will crawl away by 10. Having run again 10 meters, Achilles will see that the turtle has crawled away another 1 meter. You can argue this way to infinity, the distance between the competitors will really decrease, but the turtle will always be in front.
This paradox led Zeno to the conclusion that there is no movement, and all the surrounding movement of objects is an illusion. Of course, the ancient Greek philosopher was mistaken.
The solution to the paradox lies in the fact that an infinite amount of constantly decreasing segments tends to a finite number. In the above case, for the distance that Achilles ran, we get:
100 + 10 + 1 + 0.1 + 0.01 + ...
Applying the formula for the sum of an infinite geometric progression, we obtain:
S ∞ = 100 / (1-0.1) ≈ 111.111 meters
This result shows that Achilles will catch up with the tortoise when she crawls only 11.111 meters.
The ancient Greeks did not know how to work with infinite quantities in mathematics. However, this paradox can be resolved if you pay attention not to the infinite number of gaps that Achilles must overcome, but to the finite number of steps the runner needs to achieve the goal.