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Sunday, May 21, 2017

Rolle's Theorem and the Mean Value Theorem

Hi everyone! I hope you all had a great weekend. Rolle's Theorem and the Mean Value Theorem are two theorems of Calculus that can seem similar in a way, as they share a common hypothesis. Today, we will explore what these theorems state and examine interesting problems that could be approached with these theorems.

First, let's go over what the theorems state!
1> Rolle's Theorem:
If function f is continuous on the closed interval [a, b], differentiable in the open interval (a, b) and
f(a) = f(b), there exists a number c in (a, b) such that f '(c) = 0.

2> Mean Value Theorem (MVT):
If function f is continuous on the closed interval [a, b] and differentiable in the open interval (a, b), there exists a number c in (a, b) such that f '(c) = {f(b) - f(a)} /( b - a)

Now, let's look at some example problems.

Example 1> Prove that x^3 + x - 1 = 0 has exactly one zero.

Solution> First, we have to use the Intermediate value theorem to prove that there is a zero. For those of us who are not familiar with the Intermediate Value Theorem (IVT), the IVT states that if f is continuous on [a, b] and f(a) < w < f(b), there exists a number c between a and b such that f(c) = w. Let f(x) = x^3 + x - 1. Since f is a polynomial function, it is continuous on its domain (-∞, ∞). Thus f is continuous on [0, 1]. Notice f (0) = -1, f (1) = 1. Thus by the IVT, there exists a number c in (0, 1) such that f (c) = 0. Now, we will use proof by contradiction. Suppose that x^3 + x - 1 has two zeros at a and b. Without loss of generality, let a < b. x^3 + x - 1 is continuous on [0, 1] and since f '(x)= 3x^2 + 1 and this is never undefined, f (x) is differentiable in (0, 1). Since a and b are zeros, f(a) = f(b) = 0. Therefore, by Rolle's Theorem, there exists a number d in (0, 1) such that f '(d) = 0. But, f '(x)= 3x^2 + 1 ≥ 1 and this is a contradiction. Therefore, our hypothesis that f has two zeros were incorrect and therefore we have exactly one zero.

Example 2> Suppose that f(0) = -3 and f '(x) ≤ 5 for all values of x. How large can f (2) possibly be?

Solution> Since f is differentiable in (-∞, ∞), f is continuous in (-∞, ∞), and thus f is continuous on [0, 2]. Therefore, by the Mean Value Theorem, there exists a number c in (0, 2) such that f '(c) = {f(2) - f(0)}/2 = ( f(2)+ 3 )/2 ≤ 5. So, f(2) ≤ 2*5 - 3 = 7. Therefore, the largest f (2) can possibly be is 7.

I just wanted to introduce you all to these theorems and set of questions, because I was intimidated by these proof type questions using these theorems when I saw them. I was confused as of when to use the MVT, IVT, or Rolle's Theorem. Familiarity with the theorems are crucial, and it is important that we remember how to approach each of these problems using the theorems. I hope you all have a great rest of the day and please contact me if you do have any questions.

Mean Value Theorem; for a given arc between two endpoints,
there is at least one point at which the tangent to the arc is parallel
to the secant through its endpoints.

Sunday, May 14, 2017

Understanding Early Number Systems


Hi! I apologize for not being able to post for the past few weeks; I was a little busy with all the AP testing, but I hope you all had a wonderful weekend and...
In this post, we will be exploring the basic idea and patterns behind some of the earliest number systems!


The Egyptians and the Babylonians are known to have come up with one of the first number conventions. These conventions often used symbols that represented specific numbers, known as “Nodal numbers.” Other numbers would be formed by grouping nodal numbers together in certain way. This system is known as the “Additive System.”


The Egyptians adopted this system in their number conventions. For example, in ancient Egyptian notation, the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 19, 40 would be represented by:


From this notation, we can observe that 1 and 10 were nodal numbers; 1 would be represented by a vertical dash, while 10 would be represented by a symbol that looks like an upside-down U.


The Babylonians used 1, 10 and 60 as their nodal numbers. The Babylonians used a sexagesimal positional number system, in which the value of a particular digit depends both on the digit itself and its position within the number. It has been proposed that 60 was used as the base number, as 60 = 2^2 x 3 x 5, which makes it divisible by 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, and 30.


Other Fun Facts:
  • The Babylonians understood the concept of nothingness, but it was not seen as a number. The Babylonians understood “nothingness” as absence of a number, and thus represented nothingness with a space at first. Later on, a placeholding symbol was used to show the nonexistence of a digit in a certain place value.
Babylonian digit 0.svg
This is the placeholding symbol
used by the ancient Babylonians to
represent 0.

  • Nodal numbers are also used in Roman Numerals, in which many of us are familiar with. In the Roman number system the nodal numbers are 1, 5, 10, 50, 100, 500, 1000, represented respectively by the signs I, V, X, L, C, D, M. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 19, 40 would be represented by I, II, III, IV, V, VI, VII, VIII, IX, X, XIX, XL in Roman numerals.


I hope you all enjoyed this post, and please contact me if you have any questions!


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