Where Are Limits Needed in Calculus? Sheraz Iqbal Ithaca College May 10, 2012
Abstract Developing ideas in Where are Limits Needed in Calculus, presented by R. Michael Range; we are able to find a new method to calculating the slope of tangent lines of polynomials algebraically without the use of limits. Additionally, we are able to show further use with this process by finding ways to calculate critical points of polynomials.
Table of Contents I. Introduction II. III. IV. Derivatives Using Geometry Derivatives Using Algebra Finding Critical Points V. Basic Rule I Power Rule VI. VII. Basic Rule II Quotient Rule Basic Rule III Chain Rule VIII. Other Functions IX. Caratheodory's Definition of Differentiability X. Conclusion
I. Introduction The idea to find derivatives in elementary calculus consist of formulas derived by using limits and is motivated by appealing to small quantities. The idea is to compute the rate of change as the limiting value of the ratio of the differences as becomes infinitely small. Through research and systematic use of an algebraic approach, we will see there is a very simple and natural way of finding the derivative that avoids the complications of limits. Through previous knowledge in mathematical topics, we will be able to apply concepts to formulate a method to finding the derivative without the use of limits. We will look at concepts such as geometry, where we will apply ideas such as tangent lines and points of intersections. Next we will look at algebra, using fundamentals of polynomials such as division and commutability. Then we will look at basic rules such as the product rule, quotient rule, and chain rule, and expand from the fundamentals of polynomials. Finally we will look at some more general functions such as inverse functions, the Taylor Series, and functions with more than one variable. Throughout each topic, there will be lemmas and definitions that will help us understand what is being discussed. The idea of avoiding the use of limits in calculus has its benefits, but also can have some complications. The basic idea of led to the shortcut When taught how the idea led to this shortcut, students tend to forget how limits are used in the proof. Rather than taking the limit as, we simply evaluate at (with certain algebraic techniques) and assume that the difference quotient is continuous.
II. Derivatives Using Geometry When we look at differential calculus, we are trying to find the instantaneous velocity of a particle at a particular moment given its position as a function of time. Ideally, we are trying to find a simple method that allows us to get the value of the derivative at a point. When looking at a polynomial function, say we can find the derivative at a point, say, by the following evaluation: However, if we look at his from a graphical standpoint, we will see that there is another way to find the derivative. First we look at the graph of
Next we look at the equation of a general non-vertical line that intersects this graph at the point. Thus, we have the equation,. Since this equation is quadratic in, it can have another solution. We want to find out what has to be in order to get only one solution, which is the point of intersection. Having only one point implies that we have a double root, otherwise we would be looking at a secant line or a line that does not intersect the function at all. Even the slightest shift in the tangent line can affect the locality of intersecting points. An example of the secant line and non-intersecting line is shown as follows:
III. Derivatives Using Algebra When trying to find the points of intersection, algebraically, of a line with the graph of a function like the one previously mentioned, we have a quadratic equation which has either no (real) solutions, one single solution (which has a multiplicity of two or a double zero), or two distinct solutions. The perk of using an algebraic process is that it avoids the use of limits altogether. Let s look at the case of a polynomial of degree. Then we look at the equation of the line through a local point,, and find out where the line is tangent to the polynomial giving you the double root. Thus, we have our polynomial,, and our line which is also written as. Now, through the following algebraic process we want to find the double root when : Note: Polynomial factorization is where we will factor an from since is a root. Then we can say the following is true, where is a polynomial of degree, and it can easily be computed for any polynomial by the division algorithm. Then going back to our original algebraic process we have the following: By adding and subtracting the we can rearrange our equation in the following way:
Note: We can rewrite and through the polynomial factorization we have where is some polynomial of degree. Finally, we can finish our algebraic process as follows: Once we get to this stage in our process, when, we have a zero of multiplicity greater than one at. In other words, is the value of the slope we want in order for the line to be tangent to at. From this algebraic process, we can conclude a definition. Definition. There exists a unique line through a polynomial at the point when it is tangent to the curve, having a multiplicity of at least 2. The slope of the tangent has a value of, where is obtained from factoring. The slope of the tangent is also called the derivative of at, and can be written as. Example. Let s take a look at the function at. Then we want to find the slope of the tangent line passing through since. The equation of the line is Then following our steps as mentioned earlier, we do as follows Now, we will factor out our polynomial by dividing by. Next, we want to evaluate the following at.
By adding and subtracting the we can rearrange our equation in the following way: Now, we will factor out our polynomial k by dividing by. Then we can bring our new factored polynomial back into our original equation Finally, we can have so that we get our double root to be left in the equation. Therefore, we can conclude that q(1)=4, or the value of the derivative P (1). IV. Finding Critical Points In addition to finding values of derivatives, this algebraic method can be adapted to finding where tangent lines are horizontal. This allows us to find critical points of polynomials for optimization problems. As before we will factor the expression : When, we have a zero of multiplicity greater than one at. Now we want to find the values of for when Thus, we will look at the following
and find the double roots of. To do so, we will introduce synthetic division into our algebraic process in the following example. Example. Let s take a look at the function When we look at this function and subtract off its value at x=a, we have: Now we will perform synthetic division and and divide by using the coefficients of : + We will perform synthetic division a second time since we are looking to find the double root. 0 + 4 4 From this we can take our remainder,, and set it equal to zero. Finally we have that when and we have our critical points of our polynomial, allowing us to find such things like maximums and minimums. This allows us to further apply and find use for this algebraic method.
V. Basic Rule I Product Rule We can use algebra to go even further in the depths of Calculus to avoid the use of limits. Taking the product rule as our first approach, we will see how pure algebra will allow us to evaluate the derivative of a (rational) polynomial at a point. Looking at the product rule carefully, we start with two functions at, and So, we have and. Now, if we take the product of the two functions we have Then we want to evaluate. We want to factor our an on the right hand side as this will indicate our double root. There may be multiple ways to solving this problem, but here we simply add and subtract to get Then by simple rearrangement, we get the following: But we know that and. So, And by factoring our we get
Now we can go ahead and say since we have reached out double root. Therefore And if you remember the product rule you know this is exactly the same form as when you take the derivative of the product of two functions. Example. Let s look at the following and see how this method comes into play. Let and. We will use our technique to find the value of the derivative at. Note, we want to factor an out of both of these equations. Thus, and These will come into play later in our algebraic process. So, now let We want to evaluate at. Thus, Now we will factor out an and get Then, We figured out what and were earlier. Finally, we can evaluate at
If we check our answer with the product rule done with limits, we will find that the answers are the same. VI. Basic Rule II Quotient Rule Using a similar approach to that of the product rule, one can find the value of the derivative at a particular point algebraically. Note that both of these rules imply that derivative of any polynomial (rational) function will be a polynomial (rational) function. So again we have and So, we have and. Now let. So then we have Again we want to factor our an on the right hand side as this will indicate our double root. There may be multiple ways of solving this problem, but for now, we will add and subtract to get the following: The by simple rearrangement we get But we know that and. So, And by factoring our we get
Now we can go ahead and say since we have obtained a double root. Therefore And if you remember the quotient rule you know this is exactly the same form as when you take the derivative of the quotient of two functions. Example. Let s look at the following example to better understand how the quotient rule works. Let and. We will use our technique to find the value of the derivative at. Note, we want to factor an out of both of these equations. Thus, and These will come into play later in our algebraic process. So, now let We want to evaluate at. Thus, We will purposely not simplify the as it will save steps to simplify later on
There may be multiple ways of solving this problem, but for now, we will add and subtract to get the following: Then by simple rearrangement we get Notice how we make sure the negative sign is distributed to the 8 in the since we are adding it. Now we will factor out an out and get
Then, We figured out what and were earlier. Finally, we can evaluate at If we check our answer with the quotient rule done with limits, we will find that the answers are the same. VI. Basic Rule III Chain Rule The chain rule can be a bit tricky, but it is actually quite simple to find the value of the derivative at a particular point algebraically. So we start off by supposing and are two (rational) functions with defined and defined at and thus the composition is defined at. Then we have and Now let. Then By substitution, we have But we know, so then Since we factored out an, we can let. Thus
and And if you remember the chain rule you know this is exactly the same form as when you take the derivative of the chain of two functions. Example. Let s take a look at the following to better understand this method. Let,, and. We will evaluate at. We know that. So, so and, so Now going back to our composition function, by substitution we have But we know, so at we have Since we factored out an, we can let. Thus and so we can solve If we check our answer with the quotient rule done with limits, we will find that the answers are the same.
VII. Algebraic Functions The idea of using standard algebraic functions (like quotient), compositions, and finite inverses, with the appropriate domain, allows us to define and differentiate constant functions and. With the arguments and generalizations made with basic functions, we are able make the following statement. Factorization Lemma. If f and a is in the domain of f, then there exists q A defined on the domain of f such that Corollary. Given f A and the above factorization, then For some other k A which is defined on the domain of f, and hence at a in particular. Example. Let s have be the inverse of on. Then f is differentiable on and So now, let. We want so. Then the composition, is defined on, is in A, and is differentiable on its domain, with for Now, after looking at how this process works, we would like to see how it all applies to the fundamental theorem of calculus. Even though integration goes beyond algebra because of its use of limits, we will expand on our use of the algebraic method to apply to nonalgebraic
functions defined by integrals. We can start by stating some properties that can be done using the intuitive concept of area under the curve of a function: (i) Linearity in the integrand. Basically, what we are saying is the following: (ii) Additivity in the interval of integration. In other words, (iii) The fact that the integral of a constant over equals. We could also say (iv) The basic estimate: if on, then. Simply put, Theorem. Suppose f A is defined on the open interval I and let c be a point in I. Define the function F on I by For any a I the line given by intersects the graph of at with an appropriately defined multiplicity greater than or equal to. In other words, the function is differentiable at with. Proof. We can use the factorization lemma to prove this theorem. We know, where A. Using the properties of integrals, we can say the following:
Where. Given a bounded closed interval such that a is in its interior, there exists such that for t J and thus with and with. We can then estimate that for. Thus, if were a function of A with a zero at a of multiplicity at least 2, our estimate would hold true as the generalization of multiplicity greater than or equal to 2. Finally, we will look at the restrictions on class A functions. A function f defined on an interval I is algebraic if there exists a polynomial such that for all I. We will not focus on implicitly defined functions since their existence is beyond the scope of algebra. Rather, we will look at the algebraic curve and determine its tangent at a regular point on the curve by our algebraic process. Note, a regular point is defined if at least one of the partial derivatives of P at is nonzero. This is where our restriction comes into play, for a nonregular point may not have a tangent that exists such as at. Assuming we are working with polynomials of degree, and, we will let Then we will say that from that. Similarly, that, where we are using standard notations for partial derivatives. Now, after analogous factorization, which we will show through example in a bit, of and, we will get the following:
Where for some polynomial of degree. Since is a regular point, let us assume that and consider the line through. The points of intersection of this line with the graph are the solutions of. Then it follows that for some polynomial and. We will get to our double root once. This is also why we assumed that because. Thus, is the unique line through which has a double root when it intersects the curve. Example. Let us look at at the point (1,1). So then. Now, we want to pull out and. Thus, we get Now, through analogous factorization, we want Where for some polynomial of degree. So, we will have the following; Now, we want to substitute for. Thus, we will get
Since we have reached the double root, we would have. VII. General Functions Our method for finding the double root could also be applied to functions such as power series if we do not consider the fact that they converge. Suppose the following is a function and we will fix a point. If we want to analyze near the point, we will substitute in the series representing, expand each term in the sum by the binomial theorem, and rearrange. Thus, we have for some coefficient which depends on and. Since it follows that Where
Similar to how we got the double root through an algebraic process, the line intersects the function f at a double point when. Thus, the value of the point at the derivative is given by. Going back to our original series, we would have And if you remember how to differentiate this series as a polynomial, you would get a similar result. Example. Let s look at the function at x=1. We know this can be also expressed as: Thus, we want to factor out from this series. We will then have Next, we want to put this into the form. Then we will get Then we can say the following: Therefore, our solution is, which is what we should get through normal differentiation.
VIII. Caratheodory's Definition of Differentiability When noticing just how far one can get without using limits in Calculus, one may question, Why don t we teach our students this alternative method of using double roots to finding the derivative at a point? As you may have noticed throughout this research, there were several instances where we had to make assumptions and ignore factors for which our method of finding the double root may have not worked. It is important however to see how our method connects to the modern definition of derivatives based on limits. The factorization lemma allows us to see how a remainder,, or the difference between the function and the tangent line, can be given by the following representation: This could be rewritten as the following when we have found the double root at : so that f is differentiable at a according to our method, and where the derivative. However, the use of limits helps by including all restrictions that may be at hand through the following approximating property: The function is differentiable at if the remainder has a factorization where or equivalently, extends continuously to with. This property helps to extend our double root method by extending to functions of several variables, whether real or complex, as well as to calculus in infinite dimensions. We can then combine our remainder formula with the linear term just mentioned to get the following definition.
Definition. The function is differentiable at if there exists a factorization, where the factor is continuous at. The value is called the derivative of at. This definition helps us to translate our equation from into for This allows us to interpret as a rate of change for and that the derivative is wellapproximated. Finally, looking at our method in terms of continuous functions, we want to look at one last definition of differentiability. Definition. The function is differentiable at if where the functions are continuous at for. This definition not only hides the remainder and emphasizes the factorization, it allows one to prove all standard differentiation theorems. It is unfortunate, however, that this definition is not heavily used in teachings of calculus. I. Conclusion When I first saw this topic, I quickly jumped up to take it on for my year-long research project. I have always been a huge fan of Calculus, in particular, using limits to find the derivative. However, during my research, I had no idea what this topic was about. This made me even more interested in my research as I learned a completely new method to finding the value of the derivative at a point.
I was very fortunate to be working with my mentor. Every week he would give me new readings to read, along with practice problems to better understand the material. His office hours were very flexible in the sense that whenever I stopped by, he would be there always ready to answer my questions. I am interested in seeing if he or anyone else has done any research on this topic so that maybe I can research even more. All in all, I was very pleased with the research that I have done. I learned a whole new method to finding the derivative and was able to apply it to several applications. I was very happy to find myself connecting my research to a wide variety of math I have done in high school and previous years in college. I was able to connect in areas such as Geometry, Calculus I, Calculus II, Calculus III, Analysis, and Algebra. I mostly enjoyed doing the computations to each problem as I was getting them right when I check my answers and it showed that I understood the reading. I hope that this topic is researched someday again, and that students who read this paper can find interest to also do further research. References: 1. R. Range, Michael, Where Are Limits Needed in Calculus? Mathematical Association of America. Pgs. 404-417.May 2011