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# Calculate approximations to log 2

Theorem 6.5 (page 240 of Apostol) states that if and if then

with

Let and and apply this theorem to approximate with sufficient precision to obtain

By the theorem we have

where

Therefore, letting and we have

Therefore,

# Prove inequalities of the logarithm with respect to some series

Consider a partition of the interval for some .

1. Find step functions that are constant on the open subintervals of and integrate to derive the inequalities:

2. Give a geometric interpretation of the inequalities in part (a).
3. Find a particular partition (i.e., choose particular values for ) to establish the following inequalities for ,

1. Proof. We define step function and by

Since is strictly decreasing on , we have

Therefore, using the definition of the integral of a step function as a sum,

2. Geometrically, these inequalities say that the area under the curve lies between the step functions that take on the values and for each .
3. Proof. To establish these inequalities we pick the partition,

Then, applying part (a) we have

The final line follows since so the sum on the left starts with and the sum on the right only runs to . These were the inequalities requested

# Prove the inequalities 1 – 1/x < log x < x - 1

1. Define the following functions:

for . Prove that for and the inequalities

hold. Proceed by examining the signs of the derivatives and . When these are equalities.

2. Draw the graphs of the functions

Interpret the inequalities in part (a) geometrically.

1. First, we compute the derivatives of the functions,

Then, considering the derivative of ,

Therefore, the function has a minimum at . Since we can directly evaluate , this means for and . Therefore, .

Next, looking at the derivative of ,

Therefore, the function has a minimum at . Since this implies for and . Thus,

Putting these two pieces together we have established the requested inequalities:

2. The graph of the two functions is:

The inequalities in part (a) imply that the graph of must lie strictly between the graphs of and shown above (and so ).

# Application of the mean-value theorem for integrals

1. Let be a function with second derivative continuous and nonzero on an interval . Furthermore, let be a constant such that

Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

2. If prove that

1. Proof. Since we have the assumption that for all we may divide by , to obtain

Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

The function is continuous since is a composition of continuous functions (we know is continuous since it is differentiable) and is continuous (again, it is differentiable since exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that is continuous. We also know that meets the conditions of the theorem since it has derivative given by

This derivative is continuous since and are continuous and is nonzero. Furthermore, this derivative does not change since on since is nonzero on (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

2. Proof. Using part (a), we take , giving us

Thus, is continuous and never changes sign. Furthermore, (where is a given constant) and since . Thus,

# Prove some inequalities using the mean value theorem

Prove the following inequalities using the mean value theorem:

1. .
2. , for .

1. Proof. Define and . Then and are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain

The final step follows since for all

2. Proof. Let , , then and . So, by the mean-value theorem we have there exists a such that,

But, since is an increasing function on the positive real axis, and we have we know

Further, since and is positive we can multiply all of the terms in the equality by without reversing inequalities to obtain,

Therefore, substituting from above we obtain the requested inequality:

# Give a geometric proof that sin x < x for 0 < x < π/2

Consider the following figure:

Compare the area of the triangle with the area of the sector to prove

Further, prove,

Proof. Assume the radius is . For we have the triangle has base length and height the area is

The area of the circular sector is

Since the area of the triangle is less than the area of the sector we have,

Then, since we have for

for

# Use dodecagons to deduce an inequality about π

By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

First, we draw some pictures of the situation for reference.

( Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. )

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is , and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then . Then we use the fact that

Now, for the circumscribed dodecagon we have the area of the right triangle with base 1 in the diagram on the left given by

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron given by

For the inscribed dodecagon we consider the right triangle with hypotenuse 1 in the diagram. The length of one of the legs is then given by and the other is given by . So the area of the triangle is

Since there are 24 such triangles in the inscribed dodecahedron, we then have,

Since the area of the unit circle is, by definition, , and it lies in between these two dodecahedrons, we have,

# Prove that (1-a)(1-b)(1-c) >= 8abc if a+b+c=1

If and , prove that

Proof. Recall the definition of the th power mean, ,

for are positive real numbers and is any integer.
By the generalization of the Arithmetic Mean-Geometric Mean inequality (proved in part b of the linked exercise) we know that .

So, for the problem at hand, we consider gives us,

Simplifying this expression, we have

# Prove the product of sums of n numbers and their reciprocals is greater than n^2

Let and let for . Prove

Proof. First,

for all . We then apply the Cauchy-Schwarz inequality to the numbers and . So, Cauchy-Schwarz gives us,

# Prove the arithmetic mean – geometric mean inequality

We define the geometric mean of real numbers by

We also recall the definition of the th power mean, :

1. Prove the arithmetic mean – geometric mean inequality, i.e., prove where is the th power mean with (also known as the arithmetic mean).
2. For integers with , prove that if are not all equal.

1. Proof. First, if , then

Hence, for the case .
Now, if are not all equal then first we write,

Then using the previous exercise, we proceed,

Therefore, if are not all equal then we have the strict inequality , as requested

2. Proof. We’ll start with the inequality on the right first. So, we want to show for any positive real numbers not all equal, where is a positive integer. First, we’ll want to observe that if are positive real numbers, not all equal, then are also positive real numbers, not all equal. So from the definition of and letting denote the th power mean of the numbers , we have

So, the observation is that . Now, using part (a), we have,

Hence, for any positive real numbers , not all equal we have which implies (see this exercise) . This gives us the inequality on the right.
Now, for a negative integer, we must show . Since we know that and so from the inequality we just proved. So,

Where we have used the same exercise again, and the fact that