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

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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.

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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)

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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

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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 . The next term ensures that the rate at which slope changes and the concavity (the second derivative) is the same as for the original function and so on and so forth for the rest of the terms.

Example with f(x) = cos(x) Suppose we have the function f(x) = cos(x) and we want to take an approximation of the function at the point x = 0 . Using the method outlined above, we can find the constants needed for the polynomial approximation g(x) .

Our nth formula would be: g(x) = a + a x + a x2 + a x3 + ... + a xn ​ ​ 0 1 2 3 n Derivatives of Substituting x = 0 Formula for constant Final constant f(x)

f(x) = cos(x) f(0) = cos(0) = 1 f(0) a = 1 = 1 0! 0 1

df df df (0) 0 (x) =− sin(x) (0) =− sin(0) = 0 dx a1 = = 0 dx dx 1! 1

d2f d2f d2f −1 1 2 (0) a = =− dx2 (x) =− cos(x) dx2 (0) =− cos(0) =− 1 dx 2 2 2 2!

d3f d3f d3f 0 3 (0) a = = 0 dx3 (x) = sin(x) dx3 (0) = sin(0) = 0 dx 3 6 3!

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d4f d4f d4f 1 4 (0) a = dx4 (x) = cos(x) dx4 (0) = cos(0) = 1 dx 4 24 4!

d5f d5f d5f 0 5 (0) a = = 0 dx5 (x) =− sin(x) dx5 (0) =− sin(0) = 0 dx 5 120 5!

d6f d6f d6f −1 1 6 (0) a = =− dx6 (x) =− cos(x) dx6 (0) =− cos(0) =− 1 dx 6 720 720 6!

d7f d7f d7f 0 7 (0) a = = 0 dx7 (x) = sin(x) dx7 (0) = sin(0) = 0 dx 7 5040 7!

d8f d8f d8f 1 8 (0) a = dx8 (x) = cos(x) dx8 (0) = cos(0) = 1 dx 8 40320 8!

Therefore, the approximated function would be: 1 2 3 1 4 5 1 6 7 1 8 1 2 1 4 1 6 1 8 g(x) = 1 + 0x − 2 x + 0x + 24 x + 0x − 720 x + 0x + 40320 x + ... = 1 − 2 x + 24 x − 720 x + 40320 x +...

There are several interesting patterns to note with the polynomial approximation of f(x) = cos(x) . When substituting x = 0 into the derivatives, a cyclic pattern can be seen where the values change between 1, 0 and − 1 . As a result, only even-numbered derivatives are part of the dnf approximated function. Also, the values of the derivatives are in the range − 1 < dxn (x) < 1 which makes sense because that is the range of the cosine and sine function.

1 2 1 4 1 6 1 8 Note that the bottom-most equation (in blue) is g(x) = 1 − 2x + 24x − 720x + 40320x .

Using Desmos, the approximated function g(x) was graphed. accurate in the domain − 3 < x < 3 . If we were to keep going, adding more terms, the function g(x) would become closer and closer to f(x). Ironically, the more terms used, the more complex the equation becomes which, to an extent, defeats the purpose of the Taylor series. It is evident that the more terms we use, the

7 closer the approximate will be to the original function. The original function f(x) is in black. The g(x) function changes from a simple linear approximation to a more complicated x8 function. The approximation is only accurate around the x = 0 region.

Partial derivatives and the Taylor series The Taylor series can also be used to approximate multivariable functions. To do so, we need to understand partial derivatives. The partial derivative is denoted using ∂ rather than d (Heil, n.d.) and is used to differentiate multivariable functions. An example of a multivariable function is 2 2 df(x) f(x) = x y + xy . We could find the derivative of this function using dx but this only takes into account the change in x and ignores the second variable, y , and how that affects the output of the function (Sanderson, 2015). Thus, partial derivatives are used to fully grasp how changes in the two variables, in this case, x and y , affect the output of the function. This is done by treating all except one variable as a constant to see how that one variable affects the output of the function (Sanderson, 2015)

According to Heil (n.d.), given a two-variable function z = f(x, y) , there are two first-order partial derivatives. ∂z 1. ∂x = zx where the function is differentiated with respect to x , keeping y as a constant. ∂z 2. ∂y = zy where the function is differentiated with respect to y , keeping x as a constant.

The subscript of z is to show which variable the function is differentiated with respect to.

We can also take the second-order partial derivatives of the function z = f(x, y) . According to Heil (n.d.), there are four second-order partial derivatives. ∂2z ∂f 1. ∂x2 = zxx where the first-order partial derivative ∂x is differentiated with respect to x , keeping y as a constant. ∂2z ∂f 2. ∂y2 = zyy where the first-order partial derivative ∂y is differentiated with respect to y , keeping x as a constant. ∂2z ∂z 3. ∂x∂y = zxy , a mixed partial-derivative where the first-order partial derivative ∂y is ∂z differentiated with respect to x or where the first-order partial derivative of ∂x is ∂ ∂z ∂ ∂z differentiated with respect to y . This is the case because ∂y ( ∂x ) = ∂x ( ∂y ) (Heil, n.d.). The ∂2z fourth partial derivative is ∂y∂x = zyx . Since this produces the same value as the other mixed partial-derivative, it’s possible to multiply one or the other by 2.

Derivatives of higher orders can be found but it becomes more and more complicated to do so. Hence, for the sake of brevity once again, second-order partial derivatives must suffice.

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Using partial derivatives to find Taylor series The formula to find the Taylor series of a two-variable equation, according to Heil (n.d.) is: 1 2 2 f(x, y) = a0 + a1(x − 1) + a2(y − 1) + 2! [a3(x − 1) + a4(x − 1)(y − 1) + a5(y − 1) ] + ...

The formula follows the same intuition as the formula for the one-variable Taylor series. The first constant a0 is the value obtained when (x0, y0) are substituted into the function f . The second constant uses partial derivatives for the linear approximation. So, the first partial derivatives must be found. The values (x0, y0 ) must be substituted into the partial derivatives to obtain the value of the constant. In matching up the first partial derivatives, the slopes of the 1 function will be the same. Also, both of the first partial derivatives are multiplied by 1! similar to the one-variable Taylor series. This is because of the cascading effect that occurs with the power rule when derivatives are taken i.e. for a term with the nth power, the derivatives will be ​ ​ multiplied by n, (n − 1), (n − 2) and so forth. The second partial derivatives must then be found 1 and (x0, y0 ) substituted into them. All of the second partial derivatives are multiplied by 2! for reasons explained above.

Polynomial equation for blood flow using Taylor Series 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.

The partial derivatives will be calculated using the method explained in the section Partial ​ derivatives and the Taylor series. There are two first-order partial derivatives for a given ​ function and three second-order partial derivatives.

Calculating partial derivatives

Partial Derivative Calculation

st 1 -order​ The function will be partially differentiated with respect to x , keeping t as a ​ constant. Using the chain rule, ∂w wx = ∂x = 2.5cos(20 · π · t) 2πsin(2 · π · x) =− 5πcos(20 · π · t)sin(2 · π · x) The function will be partially differentiated with respect to t , keeping x as a constant. Using the chain rule, ∂w wt = ∂t = 2.5 · 20π[− sin(20 · π · t)cos(2 · π · x)] =− 50πsin(20 · π · t)cos(2 · π · x)

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nd 2 -order​ The function will be partially differentiated with respect to x , keeping t as a ​ constant. Using the chain rule, ∂2w 2 wxx = ∂x2 =− 5πcos(20 · π · t)[2πcos(2 · π · x)] =− 10π cos(20 · π · t)cos(2 · π · x)

The function will be partially differentiated with respect to t , keeping x as a constant. Using the chain rule, 2 w = ∂ w =− 50π · 20πcos(20 · π · t)cos(2 · π · x) tt ∂t2 =− 1000π2cos(20 · π · t)cos(2 · π · x)

The mixed partial derivative shall be calculated by differentiating w with t ​ respect to x . This mixed partial derivative will be multiplied by 2 when used in the Taylor approximation. ∂2w wxt = ∂x∂t =− 50π · 2πsin(20 · π · t)[− sin(2 · π · x)] = 100π2sin(20 · π · t)sin(2 · π · x)

Next, the partial derivatives need to be evaluated around a certain region. I will evaluate the function at x = 1, t = 1 . This value was chosen at random, to keep the equations fairly simple to work with. The blue boxes indicate the first-order partial derivatives, the purple—second-order. Partial Substituting x = 1, t = 1 Exact Value (3 Derivatives of sig. fig.) W (x, t)

W (1,1) 2.5cos(20 · π · 1)cos(2 · π · 1) + 2.5 a0 = 5.00 = 2.5cos(20π)cos(2π) + 2.5

∂w ∂x (1,1) − 5πcos(20 · π · 1)sin(2 · π · 1) a1 = 0.00 =− 5πcos(20π)sin(2π)

∂w ∂t (1,1) − 50πsin(20 · π · 1)cos(2 · π · 1) a2 = 0.00 =− 50πsin(20π)cos(2π)

∂2w − 10π2cos(20 · π · 1)cos(2 · π · 1) a =− 98.7 ∂x2 (1,1) 3 =− 10π2cos(20π)cos(2π)

∂2w 100π2sin(20 · π · 1)sin(2 · π · 1) a = 0 ∂x∂t (1,1) 4 = 100π2sin(20π)sin(2π)

2 ∂ w − 1000π2cos(20 · π · 1)cos(2 · π · 1) a =− 9870 ∂t2 (1,1) 5 =− 1000π2cos(20π)cos(2π)

10 If the approximated function is represented by g(x, t) . 1 2 2 g(x, t) = a0 + a1(x − 1) + a2(t − 1) + 2! [a3(x − 1) + a4(x − 1)(t − 1) + a5(t − 1) ] 1 2 2 = 5.00 + 0.00(x − 1) + 0.00(t − 1) + 2! [− 98.7(x − 1) + 0(x − 1)(t − 1) − 9870(t − 1) ] = 5.00 − 49.4(x − 1)2 − 4935(t − 1)2

The values (1,1) can also be substituted into both equations to see how close the values are. W (1, 1) = 2.5cos(20 · π · 1)cos(2 · π · 1) + 2.5 = 5.00 g(1, 1) = 5.00 − 49.4(x − 1)2 − 4935(t − 1)2 = 5.00

Using the application “Graphing Calculator 3D,” this was the graph produced.

The orange graph is of g(x, t) approximated around (1,1) whereas the bluish-purple graph is of W (x, t) . g(x, t) seems to be a pretty accurate approximation of the W (x, t) as can be seen by how the curvature matches the original function. The approximated function becomes less and less accurate further away from (1,1). To make this approximation even more precise, a possible improvement is to not round the constants ( a0 , a1 , a2 etc.). Substituting (1,1) into both equations also result in the same value (of 5.00).

Conclusion Once I had settled on exploring the Taylor series, I wanted to model a one-variable equation with a real-life application. Finding a complex one-variable equation was difficult because they are generally quite simple. An approximated Taylor function would serve no purpose if the original equation was easy to substitute values into. Thus, I decided to delve deeper into multi-variable equations and learn about partial differentiation. The aim of this exploration was to create an

11 approximate function of an equation used to model blood flow in arteries using the Taylor series. After graphing the approximate function on the same graph as the original function and substituting in the values (1,1), the two functions had the same result. So it is possible to conclude that I was successful at achieving my aim.

After creating the 3D graph, I realised that perhaps my exploration didn’t have any sort of benefit or application to the real-world. Sure, the model is used to calculate blood flow but upon some reflection, I don’t think doctors would jeopardise a patient’s life by using approximations. I will admit that my rationale for approximating the blood flow function was because it had some relevance to medicine which I want to pursue in university. Further research assured me that although my exploration may not be highly applicable to real-life, the Taylor series is quite useful to solve problems. One example is Euler’s identity which states that eiπ + 1 = 0 (Khan, 2011). This identity was derived by finding the Taylor series of eix and rewriting it in terms of sin(x) and cos(x) . This identity is often dubbed as the most “beautiful equation” because it links five of the most important mathematical concepts in one formula (Khan, 2011). Another application of the Taylor series is in calculators. Calculators don’t have the answers to every single computation in their memory chips; that’s why some calculators use the Taylor series to compute certain problems (Bourne, 2018). For instance, to find the cosine of a miniscule angle, the angle is inputted into the Taylor series of the cosine (Bourne, 2018).

This exploration furthered my understanding of mathematics in two ways. At a superficial level, my understanding of calculus has improved significantly. A simple example of this is the dy notation used when taking derivatives. The dx notation and what it represented always confused me. I was unclear on what went at the top, what at the bottom, and whether or not to take x (in the denominator) to the nth power. But this exploration required me to constantly use this ​ ​ notation. I became much more familiar and comfortable with it. However, the biggest lesson I’ve learnt from this exploration is that mathematics may often appear to be a muddled mess but there is a certain simplicity to be appreciated about it by linking concepts together. With an understanding of what derivatives tell us (for example, the first derivative tells us the slope, the second derivative tells us the concavity), the Taylor series makes intuitive sense. By matching derivatives of a parent function and using them to obtain constants, which are then substituted into an nth term, it makes sense that the function obtained is a rough approximation of the parent ​ ​ function. Finding these links between concepts and approaching maths as a holistic subject is something I have learnt to do and will continue to do for the rest of my studies.

I still have several questions about the Taylor series: Is it possible to find an approximate function of equations with three or more variables? What elicited the need for the Taylor series in the first place? If calculators use the Taylor series, does that mean all problems solved using a

12 calculator are, to an extent, inaccurate, in which case can we ever come to a “correct” solution when complex functions are involved? I think those are ideas certainly worth exploring at a later time. References

Bourne, M. (2018, April 6). 3. How does a calculator work? Retrieved October 9, 2018, from

https://www.intmath.com/series-expansion/3-how-calculator-works.php

Chamberlain, A. (2016, December 10). An easy way to remember the Taylor series expansion. Retrieved September

9, 2018, from Medium website:

https://medium.com/@andrew.chamberlain/an-easy-way-to-remember-the-taylor-series-expansion-a7c3f91

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Desmos [Computer software]. (n.d.). Retrieved from https://www.desmos.com/

Doe. (2016). Blood flow diagram labeled [Image]. Retrieved from ​ ​ http://lazysupply.co/blood-flow-diagram-labeled.html

The first and second derivatives [Reading transcript]. (n.d.). Retrieved September 7, 2018, from Dartmouth College ​ website: https://math.dartmouth.edu/opencalc2/cole/lecture8.pdf

Heil, M. (n.d.). Functions of multiple [two] variables [Lecture transcript]. Retrieved September 22, 2018, from ​ ​ University of Manchester website:

https://personalpages.manchester.ac.uk/staff/matthias.heil/Lectures/2M1/Material/Chapter2.pdf

How your heart works. (n.d.). Retrieved October 8, 2018, from British Heart Foundation website:

https://www.bhf.org.uk/informationsupport/how-a-healthy-heart-works

Khan, S. (2011, May 17). Euler's formula & Euler's identity [Video file]. Retrieved from Khan Academy database. ​ ​ Mathematics and blood flow. (1999, April 19). Retrieved October 6, 2018, from

http://math.arizona.edu/~maw1999/blood/index.html

Nourian, S. (2000). Graphing calculator 3D [Computer software on CD-ROM].

Sanderson, G. (2015). Introduction to partial derivatives. Retrieved January 1, 2019, from

https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivative-and

-gradient-articles/a/introduction-to-partial-derivatives

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My own notes (please ignore this)

Sources→ USE NOODLETOOLS TO CITE!!!: ​ https://math.dartmouth.edu/opencalc2/cole/lecture8.pdf (September 7, 2018) ​ https://medium.com/@andrew.chamberlain/an-easy-way-to-remember-the-taylor-series-expansion-a7c3f9 101063 (September 9, 2018) ​ https://www.whitman.edu/mathematics/calculus_online/section03.01.html (September 14, 2018) ​ https://personalpages.manchester.ac.uk/staff/matthias.heil/Lectures/2M1/Material/Chapter2.pdf (September 22, 2018) → keep using this, it’s really helpful :-) —> ask for help here using this ​ notation http://desmos.com (September 29, 2018) ​ http://math.arizona.edu/~maw1999/blood/index.html (October 6, 2018) ​ http://tutorial.math.lamar.edu/Classes/CalcI/DiffTrigFcns.aspx (October 8, 2018) —> this is the ​ (Dawkins, 2018) person. https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html (October 8, 2018) http://lazysupply.co/blood-flow-diagram-labeled.html (October 8, 2018) —> citation is (Doe, 2016) AND ​ this was used for the heart diagram :-) https://www.bhf.org.uk/informationsupport/how-a-healthy-heart-works (October 8, 2018) ​ https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-14/v/euler-s-formula-and-euler-s -identity (October 9, 2018) ​ https://www.intmath.com/series-expansion/3-how-calculator-works.php (October 9, 2018) ​

Stuff to use: http://www2.gcc.edu/dept/math/faculty/bancrofted/teaching/handouts/partial_derivatives.pdf —> for ​ more partial derivatives! FUN STUFF :-)

For EVEN more information on partial derivatives: http://www.columbia.edu/itc/sipa/math/calc_rules_multivar.html