Embed Size (px)
Citation preview
CHAPTER 5
INTEGRATION IN THEORY
5.1 AREA APPROXIMATION
5.1.1 SUMMATION NOTATION
Fibonacci Sequence First, an example of a famous “sequence” of numbers. This is commonly attributed to themathematician Fibonacci of Pisa, although it is believed that he did not invent it. The sequence begins with 0 and 1,then the next number is found by adding the previous two. The first several terms: 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987, . . ..
if you let A(n) be the n’th element in the sequence, then A(n) = A(n−1)+ A(n−2) [i.e., the sum of the previoustwo elements. This is an example of a “recursively” defined sequence. It requires that you know A(0) = 0 andA(1) = 1.
This sequence is very interesting, and involves the golden-ratio, which unfortunately goes beyond the scope ofthis course, except to say that the sequence has a “closed-form” definition in addition to the recursive definition.The golden-ratio:
ϕ= 1+p5
2≈ 1.6180339887
Believe it or not, it can be shown that A(n) = ϕn − (1−ϕ)n
p5
, test it on your calculators!
5.1.2 SIGMA Σ
When you must add many terms, it is often convenient/necessary to use summation notation. We use the Greekletter sigma (Σ) to indicate that we are taking a summation. In the previous example, we could abbreviate addingany number of terms in the Fibonacci sequence, lets say the first 16 terms:
0+1+1+2+3+5+8+13+21+34+55+89+144+233+377+610+987 =A(0)+ A(1)+ A(2)+ A(3)+ A(4)+ A(5)+ A(6)+ A(7)+ A(8)+ A(9)+ A(10)+ A(11)+ A(12)+ A(13)+ A(14)+ A(15)
=15∑
n=0A(n)
Since A(0) = 0, why not just call it∑15
n=1 A(n). We will study summations in intricate detail in this course.
Figure: Fibonacci Spiral
5.1.3 SUMMATION FORMULAS
The fact thatn∑
i=1C =C ·n is clear, pretty much from the definition of multiplication, for example 2+2+2+2+2 = 2·5.
Andrew Dynneson [email protected] 11
To see thatn∑
i=1i = n(n +1)
2, visualize this as dot-counting:
You want to count the number of red-dots. So, you make a congruent triangle of green dots inverted, and placedon top of the red-triangle. Then, the result is a rectangle, the base of which has n+1 dots, and the height has n dots.Then, there are a total of n(n +1) dots in the rectangle, exactly half of which are red.
Source:
www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof
Next, to see thatn∑
i=1i 2 = n(n +1)(2n +1)
6, see the diagram below:
One of the pyramids pictured in the top-left has the desired volume. Thus, when you put the three together,slicing half of the top piece, and reflecting it, it forms a perfect rectangular prism. Then, the following formula isachieved:
3 ·n∑
i=1i 2 = n(n +1)(n +1/2)
Simplification yields the desired formula. [4]
The Evidence thatn∑
i=1i 3 =
(n∑
i=1i
)2
can be seen in the following diagram[4]:
Andrew Dynneson [email protected] 12
Then substitute the previous formulan∑
i=1i = n(n +1)
2, to achieve the desired result:
n∑i=1
i 3 =(
n∑i=1
i
)2
=(
n(n +1)
2
)2
= n2(n +1)2
4.
5.1.4 RIEMANN SUMMATIONS
Andrew Dynneson [email protected] 13
Figure 5.2.1: Mario digital image at 800x magnificication, with zoom-smoothing off
5.2 THE FUNDAMENTAL THEOREM OF CALCULUS
Now that you have completed these activities, you are ready to derive one of the most powerful results in Calculus!What you have shown essentially is that the derivative of the area under a curve function is equal to the functionof the curve itself. In other words, if you know the anti-derivative, you can find the area. Today I will discuss howthese theories fit together to form the Fundamental Theorem of Calculus.
The first claim is the Riemann Sum, that:
area( f (x), x = a, x = b) ≈n∑
i=1f (xi )∆x
Recall that ∆x = b −a
n, and xi = a + i ·∆x.
Then, by the Pixel Theory;
area( f (x), x = a, x = b) = limn→∞
n∑i=1
f (xi )∆x =:∫ b
af (x)d x
Here theΣ changes to an∫
, to indicate a refinement from discrete to continuous intervals. For the same reason,the individual points xi changes to continuous points x on the interval [a,b]. Similarly, the notation for∆x changesto a d x, to indicate that in the limit ∆x → 0. Therefore the d x encompasses the idea of a direction (or variable) ofintegration, as well as a limit as those rectangles approach lines. This also establishes an intimate relationshipbetween Series and Integration, and these two topics will be discussed in detail throughout Calculus II.1
Next, we let the end-point b be a variable, and at this stage we undergo a change in variable to that of t , andsimply integrate up to the end-point x. Thus the area can be represented as a function of x, which encompassesthe theory of Riemann:
F (x) := area( f (t ), t = a, t = x) =∫ x
af (t )d t ≈
n∑i=1
f (ti )∆t
1This relationship is so entwined so that a “Series” is sometimes referred to as an “Integration” in the literature.
Andrew Dynneson [email protected] 14
The magic happens when I set ∆x = ∆t . Then, I claim that F (x +∆x) is simply one more rectangle’s worth ofarea further-out in the approximation of that area. Then, F (x +∆x)−F (x) actually approximates the area of thatsingle rectange!
The area of that rectangle is f (x) ·∆x, the height times the base. Hense: F (x +∆x)−F (x) ≈ f (x) ·∆x. Next:
F (x +∆x)−F (x)
∆x≈ f (x)
Now this is very close to the definition of derivative. Indeed, taking the limit∆x → 0, this simultaneously calcu-lates the derivative of the area function, and since∆x =∆t , this also causes the Riemann approximation to becomeexact, changing the wavy-equal to an equality:
F ′(x) = lim∆x→0
F (x +∆x)−F (x)
∆x= f (x)
The first part of the Fundamental Theorem is stated precisely: Let f be a continuous function on the interval[a,b], and F (x) = ∫ x
a f (t )d t . We know that F (x) is continuous on [a,b] and differentiable on (a,b).
F ′(x) = d
d x
∫ x
af (t )d t = f (x)
What this means is that the area of a curve is changing with a rate that is exactly equal to the value of the curveitself. This result has an almost mystical quality about it. Another way of stating the result is that the area functionis an anti-derivative of the curve function.
The second part of The Fundamental Theorem states that if f is a continuous function on the interval [a,b],and F is an anti-derivative of f on [a,b], then we can calculate the area exactly:∫ b
af (x)d x = F (b)−F (a)
Andrew Dynneson [email protected] 15
The proof of this is based on what we have already shown, namely that F ′(x) = dd x
∫ xa f (t )d t . Let F (x) be any
anti-derivative of f (x). Then we know that there is a constant C such that:
F (x)+C =∫ x
af (t )d t
Then, substitute x = a into this function to get:
F (a)+C =∫ a
af (t )d t = 0
Since there is no area within a vertical line at x = a [an axiom of geometry]. But this implies that C = −F (a),which yields:
F (x)−F (a) =∫ x
af (t )d t
Now, simply substitute x = b into this function to get the desired result:∫ b
af (t )d t = F (b)−F (a)
In other words, the area from x = a to x = b can be found by finding an anti-derivative of the curve function,then evaluating at a and b, and subtracting these two values.
Andrew Dynneson [email protected] 16
CHAPTER 6
THE EXPONENTIAL
The constant e becomes approached as the number of compoundings n become large. Here, the rate is 100%, andthe time interval t = 1. Then, we let n grown larger and see that the compounding equation begins to approach theideal of “continuous compounding” by way of computation:
In Calculus, we will be able to get-at this number more exactly. Next, once we see that (1+1/n)n ≈ e, we canadapt this approximation to see that the compounding equation approximates continuous compounding for nlarge enough.
Let r be the desired rate. Replacing n with n/r is okay, because if our rate is less than 100%, then n/r becomeslarger since r < 1, and our calculation converges even faster. On the other hand, if r > 1, then it is true that n/rbecomes smaller, but I claim that we only need to take n out further in that case to get our calculation to convergeto the desired level of accuracy. Thus the claim is that:(
1+ 1
n/r
)n/r
≈ e
The next step is to divide 1/(n/r ) = r /n. Also, taking the r ’th power of both sides yields:(1+ r
n
)n =((
1+ 1n/r
)n/r)r ≈
er
Next, take the t ’th power of both sides to introduce time-increments into the equation:(1+ r
n
)nt ≈ er t . Fi-nally, multiply both sides of the equation by the initial value P , and we have successfully derived the continuous-compounding equation:
P(1+ r
n
)nt≈ Per t
6.1 DEMOIVRE: POLYGONS IN THE COMPLEX-PLANE
One of the many many reasons that DeMoivre’s theorem is so useful is that it gives us a really cool way to programpolygons.
Consider the example of an heptagon. Dividing the unit-circle into seven equal arcs reveals each to be θ = 2π/7
Andrew Dynneson [email protected] 17
