Answer:
5
Explanation:
Now we will look at two famous irrational numbers, and e and calculate their approximate decimal expansions. As you may remember, pi is used in geometry and has been around for over 4000 years. It is used to find area of a circle, surface area and volume of a sphere, cone, and cylinder. Its value is used in many real world calculations such as mechanics, architecture, nature, art, and even medicine. The value e is called Euler’s number and is very important in exponential functions such as finance. Let’s approximate the value of these two numbers, e and .
Example 5:
To determine the decimal expansion of , we will use the fact that the number is the area of a unit circle together with counting the total squares inside a circle on a grid. Remember, . The unit circle is a circle with a radius of 1. Since the area of a unit circle is equal to and we will be counting squares, we can decrease our work by focusing on the area of just of the circle. Take a look at the illustrations below to help you understand our premise.
Unit Circle Area using grids
A = (1)2
A = Area of circle = total number of squares inside circle
Let’s take a look at of the unit circle using a 10 x 10 grid. It is difficult to count the number of squares inside the quarter circle because some are not whole squares.
We have inner squares inside the quarter circle, and outer squares outside of the quarter circle. Next, we mark a border inside the circle and outside the circle, as close as possible to the circle. The inside border should enclose all the whole squares inside the quarter circle, (red). The outside border contains all the whole squares within the quarter circle and parts of the squares that are outside the circle (black). Now let’s estimate the area of the unit circle to approximate .
Let r2 equal all inner squares and s2 equal all the squares within the black border and quarter circle. There are 69 inner squares and 89 inner and outer squares. Algebraically, r2 < < s2.
If we consider the area of the square with side length equal to 10 squares of the grid paper, then the area of r2 = and area of s2 = .
Therefore,
r2 < < s2
We know this is accurate because 2.76 < 3.14 < 3.44. Of course we can improve our estimate of by including more of the partial grid squares inside the quarter circle. By combining the partial squares there appears to be about 7 more inside the circle. We then add 7 to the 69. This gives us,
We can reason the same way as before to estimate s2 to 80, instead 86.
We are getting closer to approximating pi. Suppose we divide each square horizontally and vertically instead of having 100 squares, we have 400 squares.
Multiplying by 4 throughout, we have
By looking at partial squares we can estimate r2 = 310 and s2 is 321. Then, the inequality is
We can continue the process to get closer and closer estimates,
3.14159 < < 3.14160
and then continue on to get an even more precise estimate of . By now, we have a very close one already.
We conclude by making one final observation about and irrational numbers. When we take the square of an irrational number such as , we are doing so without knowing the exact value of . Since we can use a calculator to show that 3.14159 < < 3.14160, we also know
3.141592 < < 3.141602
9.8695877281 < < 9.86965056
Notice that the first 4 digits, 9.869 are the same on both sides of the inequality. Therefore, we can say that ^2=9.869 is correct to 3 decimal digits.
Example 6:
Estimate the value of the irrational number (12.03801...)2
Solution:
The estimated value is 144.91 and correct up to 2 decimal digits.
To estimate this, we truncate to one digit higher for the number on the right. Then we square.
12.038012 < (12.03801...)2 < 12.038022
144.91585161 < (12.30791)2 < 144.91825924
The first two decimal places are the same, therefore (12.03801...)2 = 144.91 is correct up to 2 decimal digits.
Example 7:
Estimate the value of the irrational number (9.204107...)2
Solution:
The estimated value is 84.715, which is correct up to 3 decimal digits.
9.2041072 < (9.204107...)2 < 9.2041082
84.715585667449 < (9.204107...)2 < 84.715604075664
(9.204107...)2 = 84.715 is correct up to 3 decimal digits.