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# Prove some inequalities of sin x

For all prove that

From the first exercise of this section on inequalities, we know for all . But since for all and we have the inequality on the right immediately,

For the inequality on the left let

Then, we’ll consider the first two derivatives of to show that it is positive for all .

We know from the inequality on the right that for all . Hence, for all . Therefore, is increasing on the positive real axis. Since

we then have that for all . Hence, is increasing on the positive real axis, and since

we have that for all . Thus, for all ,

# Prove inequalities of the log of 1 + 1/x

For prove that

Before the proof, a picture for this one is probably useful. In the graph below we are going to get the inequality on the right by comparing the area under the graph of from to (blue curve) and the area of the rectangle under the blue dashed line from to . We’ll show that the area of the rectangle is and the area under the curve is .

Proof. First, we note that

Furthermore,

Therefore we have

(In the picture this is the area under the curve between and .)
Since the function is strictly decreasing on the positive real axis (since it’s derivative is is negative everywhere) we know that for any fixed we have for every . Therefore, by the monotone property of the integral,

(Note the inside the integral is a constant here since we have chosen some fixed . For any fixed the point is the point on the curve at the left end of the interval. So, this is saying the integral of the curve from to is less than the integral of the rectangle of height from to .) This gives us the inequality on the right that we wanted,

Now, to prove the inequality on the left we know that if and only if . Thus, the given inequality holds for all if and only if it holds for all . Therefore,

if and only if

Now, consider

for all . Therefore, is increasing on the positive real axis. Since

we have that for all . Hence, we obtain the inequality on the left

# Prove the inequality 2x/π < sin x < x for 0 < x < π/2

Prove the inequality

Proof. Define a function . Then, since for we have

Since and is increasing on (since its derivative is positive) we have

For the inequality on the left (which is much more subtle), we want to show

So, we consider the function

We know

and

Now, if we can show that is decreasing on the whole interval then we will be done (since this would mean on the whole interval since if it were less than somewhere then it would have to increase to get back to on the right end of the interval).

To show is decreasing on the whole interval we will show that its derivative is negative. To that end, define a function

(This is the numerator in the expression we got for . Since we know the denominator of that expression is always positive, we are going to show this is always negative to conclude is always negative.) Now, and

for . Thus, is negative, and so is decreasing. Since we then have is negative on the whole interval. Therefore, on the interval. Hence, is decreasing. Hence, we indeed have

for all

# Prove that x – (1/3)x3 < arctan x if x is positive

Consider the function

Compute the derivative and use the sign of the derivative to prove the inequality

Proof. First, we compute the derivative

Therefore, for all . This implies is strictly increasing if , and in particular, is increasing for . This implies

# Prove formulas for the partial derivatives of xy

Define a function of two variables, with ,

Prove

Proof. First, we write,

Then we compute the derivatives using the chain rule and formulas for the derivatives of the exponential and logarithm,

And,

# Prove some inequalities of the exponential function

For and for prove the following inequalities,

where we assume in the second inequality. Use this to show

Proof. From the previous exercise (Section 6.17, Exercise #41) we know

for . But by assumption we have and ; hence, . Therefore this inequality must hold for :

Again, by the previous exercise we have

for . Since and this implies . Therefore,

Letting , we have

# Prove some inequalities using the function ex-1-x

1. Define a function:

Prove that for all and for all . Prove the following inequalities are valid for all ,

2. Using integration and part (a) prove

3. Using integration and part (a) prove

4. State and prove a generalization of the inequalities above.

1. First, we take the derivative of ,

Since is strictly increasing everywhere (since for all ) we have for and for . Thus, has a minimum at and so for all . Therefore, when ; hence,

Furthermore, when we have

Therefore, if then and so we have

2. Using the inequalities in part (a) we integrate over the interval to ,

For the other inequality we proceed similarly,

3. We use the same strategy as before, starting with the inequalities we established in part (b),

Similarly, for the other inequality,

4. Claim: The following general inequalities hold for all :

Proof. The proof is by induction. We have already established the inequalities for the case for all three inequalities. Now, to prove the first inequality assume

Then we have

This establishes the first inequality. For the second inequality we have proved the case already. Assume

Since is odd we may write for some nonnegative integer . We want to show that the inequality must hold for the next odd integer, . We have,

We want to show that the inequality holds for so we integrate both sides again,

Hence, if the inequality holds for odd , then it also holds for the next odd number, . Hence, it holds for all positive, odd integers.
The exact same induction argument works for all of the even integers (starting with the case we proved in part (c))

# Calculate an approximation of log 7 in terms of log 5

With reference to the previous three exercises (here, here, and here) use to calculate in terms of . Obtain the qpproximation

Letting we have

Therefore, we apply Theorem 6.5,

# Calculate an approximation of log 5 in terms of log 2

With reference to the previous two exercises (here and here) obtain the following inequalities for ,

If we take then we have

Therefore, we apply Theorem 6.5 to obtain

# Calculate an approximation of log 3 in terms of log 2

Use Theorem 6.5 (pp. 240-241 of Apostol) and the previous exercise to compute in terms of . To do this, observe that if then

Using this value of and establish the inequalities:

Taking and in Theorem 6.5 we have

Then, since , we combine with the previous exercise (Section 6.11, Exercise #1) to obtain,