# A digression on Calculus… From introductory to advanced?

Before anything, wish everyone a happy and prosperous new year full of joy and success! (This was posted on Dec 31)

Surprised, aren’t you? Well, you may not be expecting this, but I came across a bunch of beautiful problems. So I want to introduce a little bit of graphs (is that really necessary?).
Without wasting any more time, let’s continue with this hotch-potch discussion on Calculus. Yeah, I know that I am not organised, sorry about that.
I don’t know from where to start, so anyway, let me start with the fundamentals of calculus problem-solving. The problems that we(of course, ‘we’ means that I am going to follow the Indian curriculum of Calculus) usually solve consist of the following concepts: Graphs and Functions, Limits, continuity and differentiability, function-plotting.
Polynomials.
These may be defined as functions (that’s my way of looking at them) which have terms of integral powers and real or complex coefficients. For the sake of calculus of our level though, we only look at polynomials which is generally given by
given
The degree of a polynomial is the highest power of that is contained in If then the degree of is
Polynomials of even degree
If is even, then we can only have two types of graphs possible.

Assume that so that
And,
So the graph can be of two different looks:

So, a polynomial with even degree will either rise to the heaven, or fall down to hell on both sides – and . It’s also obvious that it will cut the axis even number of times.
Polynomials of odd degree
It’s easy to deduce that the graphs of polynomials with odd degree will fall to hell at one side, and rise up to heaven on the other. A nice way of recalling is that, if then the graph will rise up towards heaven at and vice versa. Like,

Roots
The roots of a polynomial are the points where the curve cuts the -axis. We can use Descartes sign rule to determine the number of positive or negative roots. Let be a certain polynomial, which has number of sign changes as while proceeding from the lowest to the highest power (ignoring zero coefficients), has a maximum of positive roots. It may have etc positive roots too. When this rule is applied to we can guess the maximum allowed negative roots.

The derivative
Let be a function. Then is the derivative at a point which is defined as If in an interval then the function is said to be differentiable over all points in Then is the slope of the tangent to the curve at A function is differentiable only if it’s continuous, but the reverse is not true.
Again, I do not intend on lecturing on removable and blah blah continuity, so I am assuming prior knowledge without loss of generality. After all, it’s tiring to mention everything, lol. So let’s get into some serious stuff.

Convexity, concavity, monotonicity.
A function is said to be monotonically, or strictly increasing, or decreasing if and only if and its reverses respectively hold for any in a certain interval.
A function f(x) defined on an interval is called convex (or convex downward/concave upward) if the graph of the function lies below the line segment joining any two points of the graph. The formal definition is, a function is convex if and only if for any two points and we have In case the equality doesn’t occur, the function is called to be strictly convex.
Concavity is just the opposite.
The necessary and sufficient condition for a certain function to be convex in an interval is that the function must lie above all of its tangents, ie Concavity: reverse the sign.
This may also be rephrased as for all in We may show that this is another necessary and sufficient condition for convexity.
Jensen’s inequality.
Let be a convex function of one real variable. Let and let satisfy Then

For a proof, look here.

Without any more of introduction, let me move on to one of the most important theorems in Calculus, ie Rolle’s theorem.
Rolle’s Theorem
If is continuous and differentiable at every point in an interval and if then there exists a certain such that
From graphical considerations, it is obvious that the function must have changed its slope from either positive to negative, or negative to positive at some point in the given interval. It can contradict this at the cost of its differentiability everywhere.

Mean Value Theorem (MVT for short)
If and and are continuous throughout an interval and if everywhere in the given interval, then there will always exist a point such that

Proof.
Consider the function
Since so applying Rolle’s theorem we see that
for some point Hence done.
Geometric interpretation
For the non-general version of Rolle’s theorem for we can see that the slope of the line joining and is and is the slope at So, obviously there exists a certain in the interval such that equals the slope of the chord joining the terminal points.

Extended MVT
Define a constant such that the equation

Is satisfied. Define a function as

Since , therefore and
for some
Since therefore and Substituting , we get

Continuing, we obtain,

Where This expression is known as the extended MVT.

With this, I am wrapping up all my discussions on the theorems that we study in high school (except for the extended MVT, everything is quite fundamental).
And, it also means that I will now move onto some nice problems. Most of the problems that I will discuss in this post will be involving construction of functions and applications of Rolle’s theorem, etc. So, let’s change the mood from theoretic to a little of applications.
Problems.
1. Given two functions and , continuous on differentiable on , and Prove that there exists a point such that
Solution
Define then and so, there exists a in such that But, note that and this leads us to our desired result.

2. Let be a function from reals to reals with at least two roots . Prove that for any real number there is such that .

Solution(Virgil Nicua)
If , then can choose . Suppose . In this case consider , Where . Observe that satisfies and , so by Rolle’s theorem we get our desired result.

3. Show that for two functions such that and given we have a certain between any two roots of such that

Solution
Consider Indeed, so there exists at least one between such that

Which leads to our desired result.

4. Solve in the following equation:
.
Solution
Though it’s obvious from graphs that are the only solutions, why not apply Rolle’s? Suppose has at least three distinct roots. By Rolle’s theorem, should have at least two distinct roots; however is strictly increasing and so this cannot happen. So, has at most two distinct roots, namely and

5. (Amparvardi) Let , be a function which is continuous and differentiable in . Prove that there exists a real number for which

Solution
Let us assume Note that and also

So from Rolle’s theorem, there exists at least one such that and we are done.

Let’s see some LMVT problems now.
6. Show that

for
Solution
Consider the function , it is continuous in and also differentiable in Applying LMVT, there must be one such that
But , therefore and hence we get our desired inequality.

7. Prove that for any we have

Solution(Virgil Nicua)
Since we can rewrite the given problem into

Apply LMVT to the function on the segments and So there exist satisfying and So,

8. Let be a polynomial with real roots. Given that and have the same sign, prove that it’s impossible for both roots of to be in the interval .
Solution(marvopnema)
Assume the two roots of are both real, and situated in , hence of the form , with . From Vieta’s relations we have and .
Then , hence and are of different signs.

9. While and find the limit

Solution(hsbatt)
By LMVT,

For some
Now, let’s sandwich the expression between and
Because both the limits of the left and the right bound are therefore the function bounded in-between must also have a limit and

10. Without integrating, show that the sum

;

converges.
Solution
Consider where and Observe that By LMVT, there must exist a certain such that Since is increasing, we get

Or,

Summing up from to we get

Or,

So, the sequence is increasing and bounded above and below. So we see that and converges.

11. Let be a monic polynomial of degree , with real coefficients and all its roots real and different from zero. Prove that for all , at least one of the coefficients is different from zero.
Solution
Lemma 1
If all the roots of are real, then all the roots of are real.
* The proof of this lemma is clear, because between two consecutive distinct roots of there is a root of (using Rolle’s theorem), and we have that if is a root of with multiplicity , then it’s a root of with multiplicity .
Lemma 2
At least one of is different from
* Suppose that the (possibly repeated) roots of are , and that .
Then from Viete’s formulas we get that:

As the are nonzero, this immediately implies that , a clear contradiction. This concludes the proof of lemma 2.
We now prove that the statement is true by strong induction on :
Base cases:
If then there is nothing to prove, and if then the result is true because (because is not a root).
Induction step: Assume that .
Case 1:
In this case we have that is a polynomial with all roots real and nonzero (using lemma 1), so we may use the induction hypothesis on (we divide by just to make the polynomial monic, a minor technicality to allow us to use the hypothesis). As the coefficients of this polynomial are nonzero multiples of the coefficients of (except ), we conclude that there are no consecutive zero coefficients in .
Case 2:
In this case we have that is a polynomial with all roots real and nonzero (using lemma 1 twice), so we may use the induction hypotheses on (again, dividing by is just a technicality), because . As the coefficients of this polynomial are nonzero multiples of the coefficients of (except ), again we conclude that there are no consecutive zero coefficients (using lemma 2 for the coefficients ).
This concludes the proof of the desired statement.

12. (a) Prove that

has only one real root .
(b) Prove that is decreasing, and .
Solution to Part (a)
Let

The key to proving that has exactly one real root is to show that has no real roots.
Note that for Hence all zeros must occur for negative
By Taylor’s Theorem (with Lagrange mean-value form of the remainder),

But for which implies that
Hence, has no zeros. Now consider As an odd degree polynomial, it must have at least one zero. However, by Rolle’s Theorem, if it has two or more zeros, then its derivative must have at least one zero. But which has no zeros. Hence, has exactly one zero.
Now, given any we know that tends to uniformly on This implies that there exists an depending on such that for has no zeros in This establishes that the sequence of roots, must tend to as
Solution to Part (b)
Let .
We have that .
Note that , since .
Hence, . Since is a polynomial of odd degree and has a unique root, we must have that

Consequently,

Since is a polynomial of odd degree and has a unique root, we must have that , i.e. is strictly decreasing.

And, I finish the post here only. 12 Problems for this post!
Also, sorry for not being able to normalise. The introduction part is too easy, and the last problem uses Taylor.

## 2 thoughts on “A digression on Calculus… From introductory to advanced?”

1. Hi, man! I’ve seen your posts around Art of Problem Solving, and, reading your signature, I’ve come to this blog.

Congratulations, it’s a pretty nice work you’ve been doing!

Also, I’ve got a brazilian maths blog, and I think that I could tell to my readers – that aren’t that much, but they do exist – about your blog, and even put on my links bar. If we could start a partnership, exchange links on blogs, I think it would be great to both of us, what do you think?

Well, the adress to my blog is http://amatematicapura.blogspot.com/. Although translating isn’t good, there’s a translate button, and I think that, even with translations it’ll be understandable.

If you want to contact me, just leave an e-mail: joaopedroblogger@hotmail.com.

See ya!

• Potla says:

Thanks a lot for your interest. 🙂