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,
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,
Consider a partition of the interval for some .
Since is strictly decreasing on , we have
Therefore, using the definition of the integral of a step function as a sum,
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
for . Prove that for and the inequalities
hold. Proceed by examining the signs of the derivatives and . When these are equalities.
Interpret the inequalities in part (a) geometrically.
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:
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 ).
Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality
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:
Thus, is continuous and never changes sign. Furthermore, (where is a given constant) and since . Thus,
Prove the following inequalities using the mean value theorem:
The final step follows since for all
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:
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
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,
If and , prove that
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
Let and let for . Prove
for all . We then apply the Cauchy-Schwarz inequality to the numbers and . So, Cauchy-Schwarz gives us,
We define the geometric mean of real numbers by
We also recall the definition of the th power mean, :
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
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