Table of Contents Introduction 1 An explanation of the Taylor Series 1 Equation to model blood flow 2 The Taylor Series and calculus 2 Derivatives of functions used in this exploration 2 Using the power rule to take derivatives 3 Using the chain rule to take derivatives 3 1st and 2nd derivatives 3 Deriving the Taylor series with only one variable 4 Example with f(x)=cos(x) 5 Partial derivatives and the Taylor series 7 Using partial derivatives to find Taylor series 7 Polynomial equation for blood flow using Taylor Series 8 Calculating partial derivatives 8 Conclusion 10 References 13 1 Introduction While carrying out my initial research for an exploration topic, I was particularly interested in models related to biology, particularly blood flow and pressure, as I aim to enter the medical field myself. I wanted to investigate Poiseuille’s Law. This law is used by doctors to calculate the blood flow rate in vessels. After digging around for more information on Poiseuille’s Law, I realised that the mathematics used to derive the equation was far beyond my level. I was taken by surprise when I saw how many variables were involved and I wished for a simpler equation to work with. It’s interesting that the application of maths to the real world has resulted in thousands of such complex equations which seek to model a certain phenomenon. But the issue is that they tend to be so infuriatingly complex that they put people off. That’s why I found the Taylor series so intriguing. I was drawn to the Taylor series because it provides a means of simplifying said equations by converting them to polynomial form. This conversion makes it far easier to input values and obtain an answer. Hence, the aim of this investigation is to provide a means to simplify a seemingly complex equation into its simple polynomial form. The equation I will create a Taylor approximation for is one used to model blood flow in the arteries. I chose this equation because, as aforementioned, I wanted to work with a model used in the medical field. Furthermore, the model uses the cosine function which makes it challenging to input values to obtain an answer quickly. Thus, generating a Taylor approximation may be useful if a quick approximation is needed regarding a patient’s blood flow. This exploration will explain aspects of calculus namely taking derivatives using the power rule, what the first and second derivatives tell us about a function, and how partial differentiation is done. Next, the one-variable and two-variable Taylor series will be explained. Finally, the function used to model blood flow will be approximated using the two-variable Taylor series. Materials used: TI-84 (Texas Instruments), Desmos, Graphing Calculator 3D (graphing app) ​ ​ ​ ​ ​ An explanation of the Taylor Series The Taylor series is a most intriguing part of calculus. It aims to simplify complex functions into polynomial functions by adding the sum of an infinite amount of polynomial terms (Chamberlain, 2016). Therefore, by finding an infinite number of derivatives and by having an infinite number of terms, we can write a polynomial function which is a close approximation of the original function centred around a specific region. To understand how derivatives can be used to do so, the first and second derivatives will be explained followed by the derivation of the Taylor series. 2 Equation to model blood flow The equation for modelling blood flow in the arteries, according to “Mathematics and blood flow” (1999) is: W (x, t) = 2.5cos(20 · π · t)cos(2 · π · x) + 2.5 In which x = space and t = time (“Mathematics and Blood Flow,” 1999). First, more about the heart and blood flow will be explained. The heart is made up of four chambers—right atrium, left atrium, right ventricle and left ventricle (British Heart Foundation, n.d.). Deoxygenated blood (which has travelled around the body) flows into the right side of the heart. This blood is then pumped into the lungs to get oxygenated after which it returns to the heart’s left side to be pumped to the rest of the body (British Heart Foundation, n.d.). The tubes that pump blood away from the heart are called arteries; the tubes that pump blood towards the heart are called veins. The equation tells us the rate at which blood flows at a certain point in the arteries at a given time (“Mathematics and Blood Flow,” 1999). The amplitude of the function shows the maximum blood flow. The frequency of the cosine function is the heart rate. The Taylor Series and calculus Derivatives of functions used in this exploration The exploration requires knowledge of the derivatives of sinx and cosx . d ● dx (sin(x)) = cos(x) d ● dx (cos(x)) =− sin(x) 3 Using the power rule to take derivatives The Taylor series depends on the power rule to take derivatives of functions. The power rule states that for any function f(x) = xn , df n−1 dx (x) = n · x . Using the chain rule to take derivatives The chain rule is used to differentiate composite functions. The chain rule states: d dx [f(g(x))] = f′(g(x))g′(x) . So the function f(x) = cos(2x) , the composite function would be g = 2x . Therefore, both the cosine function and 2x would need to be differentiated. d dx [f(g(x)) = 2 · (− sin(2x)) =− 2sin(2x) 1st and 2nd derivatives dy The first derivative of a function, f′(x) or dx , represents “the slope of the tangent line to the function at the point x” (“The first and second derivatives”). In other words, the first derivative shows us whether a function is increasing or decreasing and the rate at which it is doing so. Therefore, a positive value for the first derivative (a positive slope) indicates that at point x, f(x) is increasing. A negative value indicates the opposite. If the first derivative is equal to zero, f(x) may be constant (such as in the case f(x) = 2 ) or the point x may be a maximum or minimum point (“The first and second derivatives”). df If dx (k) < 0 , f(x) is decreasing at x = k df If dx (k) = 0 , f(x) may have reached a maximum/minimum point at x = k df If dx (k) > 0 , f(x) is increasing at x = k d2y The second derivative of a function, f′′(x) or dx2 , tells us “if the first derivative is increasing or decreasing” (“The first and second derivatives”). Therefore, the second derivative tells us whether the function is concave up or concave down at point x. d2f If dx2 (k) < 0 , f(x) is concave down at x = k d2f If dx2 (k) = 0 , we cannot make an inference about the concavity of the function. d2f If dx2 (k) > 0 , f(x) is concave up at x = k 4 Deriving the Taylor series with only one variable According to Chamberlain (2016), the following steps can be used to derive a polynomial equation using the Taylor series: 1. Write an nth term function. ​ ​ 2. Take the first, second and third derivatives. (There is no limit to how many derivatives can be taken. However, for the sake of brevity, only three shall be taken.) 3. Substitute values obtained from derivatives. We will start with an nth term function. ​ ​ 2 3 n f(x) = a0 + a1(x − k) + a2(x − k) + a3(x − k) + ... + an(x − k) In the function above, a represents a constant and k is the x-coordinate around which the approximation begins. The constant a ensures that the value of the polynomial function matches the original. a0 is simply the y-intercept. By substituting x = k , the value can be obtained easily: 2 3 n f(k) = a0 + a1(k − k) + a2(k − k) + a3(k − k) + ... + an(k − k) f(k) = a0 If we rewrite this using factorials, f(k) a0 = 0! To find the values of the other constants, derivatives must be taken using the power rule. Then we can substitute x = k to obtain the value of the constant. (Only the first four terms will be used to explain this.) df 1 2 dx (x) = a1 + 2 · a2(x − k) + 3 · a3(x − k) df 1 2 dx (k) = 1 · a1 + 2 · a2(k − k) + 3 · a3(k − k) = 1 · a1 We can rewrite this in factorial form. df dx (k) = a1 · 1! df (k) Therefore, a = dx . ​ 1 1! Similarly, we can find the other derivatives and hence, the values of the constants. d2f 1 dx2 (x) = 1 · 2 · a2 + 2 · 3 · a3(x − k) d2f 1 dx2 (k) = 1 · 2 · a2 + 2 · 3 · a3(k − k) = 1 · 2 · a2 = a2 · 2! 2 d f (k) dx2 a2 = 2! 5 d3f 0 dx3 (x) = 1 · 2 · 3 · a3(x − k) d3f 0 dx3 (k) = 1 · 2 · 3 · a3(k − k) = 1 · 2 · 3 · a3 = a3 · 3! 3 d f (k) dx3 a3 = 3! If we substitute these constants into the polynomial function: 2 3 f(x) = a0 + a1(x − k) + a2(x − k) + a3(x − k) 2 3 df d f (k) d f (k) f(k) dx (k) dx2 2 dx3 3 f(x) = 0! + 1! (x − k) + 2! (x − k) + 3! (x − k) According to Chamberlain (2016), by writing this as an infinite sum, we end with: ∞ f n(k) n ∑ n! (x − k) n=0 a0 tells us the y-intercept of the function. a1 (first derivative) ensures that the slope of a tangent line match at x = k .