In mathematics a series is defined as the sum of a sequence of numbers. It can be express by using the notation ∑ a_n, where ∑ represents “sum” and “a_n” is the n-th term of the sum, which is a generalized way of expressing the terms of the sum.

∑ a_n= a_1+a_2+a_3+a_4+….+a_n

Since  ∑ a_n is a sum, we can group terms, change the order or the terms, without varying the result of ∑ a_n. (i.e.  ∑ a_n= a_4+a_3+(a_1+a_2)+….+a_n)

When writing ∑ a_n we also express the limits or the sum, in other words, we express when do we start adding (let’s say a_0)  and when do we finish (let’s say a_5). When we do this, ∑ a_n has an specific value. However, a sum can also have only one limit, this occurs when a sum starts a a point “n” but never ends. In this case, where the sum is infinite, two things can occur: The total sum will keep increasing to infinity (±∞)  and we say the series diverges, or the total sum can get close to a certain number k and we say the series converges to k.

The idea of a infinite series converging to a value k can be difficult to make sense, since it may seem that adding infinite terms would imply that the sum will always increase. This is true, however if the terms you add become smaller each time, then you don’t “progress” as much, in fact, let’s say you have 1 and you add +0.1+0.01+0.001+0.0001+…. then you never reach let’s say 2. You reach a number K (between 1 and 2).

A function f(x) is described as analytic around a point a, when f(x)  can be expressed as a power series ∑Cn (x-a)^n around a with a positive radius of convergence R.

This means that at any point c if |c-a|≤R, then the series  ∑Cn (c-a)^n converges.

Ex. f(x)=e^x  can be expressed as ∑((x-0)^n )/n! center of convergence 0.

By using ratio test on ∑(x^n)/n!, the radius of convergence is found to be R=∞

Then f(x)=e^x is an analytic function around 0.

What are the implications/uses of this function property?

If f(x) can be expressed as a power series around a point, then arithmetic operations can be made term by term in the series, as well as differentiation and integration. One use of  these operations  can be found when the power series of a given function cannot be described, but  the power series of the derivative or integral exists.

Ex. Evaluating ∫e^x² dx can be “easily” done by knowing

e^x=∑ (x^n) /n! which is analytic around 0 (from the previous example). Because it is analytical we can operate on it.

So: e^x²=∑ (x^2n) /n! =1+x²+(x^4)/2!+(x^6)/3!+…+x^(2n)/n!

And we can take the integral of every term in 1+x²+(x^4)/2!+(x^6)/3!+…+x^(2n)/n! ->  ∫e^x² dx

It is said that you learn more from your mistakes than from your successes. This thought can be applied in both personal and academic sides. As for today, I am going to focus on my past midterm.

Question 1 (f) was for me a challenging question. The reason for this is that this question requires not only knowing how to apply integration techniques, but also , as one of my instructor would say, it requires being a little bit smart. Overall, this exercice requires to join and wisely use almost all techniques studied in class.

A reasonable first aproach would be to use substitution method with what it’s inside the square root. However, that wouldn’t have been an effective approach. Looking carefully t²+1 could be substitute by using trigonometry equivalences. t= tan(θ). With this substituion we will be left with ∫cos(θ)/sin²(θ) dθ. Now, another integration technique is necessary: substitution. This method can be very useful when knowing how to make a good substitution.

Tips for the final:

• Practice is the key. Intuition when evaluating integrals comes from practice. Not all integrals with similar forms can be solved similarly but after practice there are patterns that you start to notice.
• Trigonometric substituions. Practice when and how to use trigonometric substitutions, how to change the limits of the integral, how far should I go with my substituions, and when to stop and use a different method.

Evaluating integrals might seem challenging sometimes, especially when we encounter very complex looking expression or sometimes even when we encounter simple looking integrals such as ∫sec(x) dx, which turns out to involve some cleverness when evaluating it. Being “good” at integration might be “innate” for some people, but often a lot of practice is the key. When practicing you might notice that you get an “instinct” on how to solve certain types of integral even at first sight and how to apply certain methods, like when and how to apply  integration by susbtitution method.  Here are some of the tips I got from my hours of practice in how to make a good substitution:

• When encountering a binomio in the form 3: (ax+b)^n. Solving an expresion like this will take a lot of time and effort. In this case, a good way to proceed is:

∫(3x-34)^32 dx

u= 3x-34 ; let the binomio = u, then du=6 dx

∫(3x-34)^32 dx = ∫(u)^32 du/6 = (1/6) ∫u^32 du = (1/6) [u^(32+1) /(32+1)]

• When encountering a fraction in the form: x^(n-1)/x^n +k  or x/x+k (k is a constant). How to proceed:

∫ x^4 / (x^5-3x²) dx=∫ x^4 / [ x² (x³-3) ] dx = ∫ x² / (x³-3) dx

u=x³-3 ; let u= the denominator, the du=3x²dx → dx=du/3x²

∫ x² / (x³-3) dx = ∫ x² / u ( du / 3x² ) = ∫ 1/u ( du/3 ) = 1/3  ∫ 1/u du = 1/3 log |u|

• When encountering log(x). This expression can appear in different ways and you might not choose to evaluate the integral always the same, but a good way to start is:

∫14x² / [x³ log²(x)] dx = ∫14/xlog²(x) dx = 14 ∫ 1 / xlog²(x) dx

u=log(x) ; choose u=log(x), then du=1/x dx → dx=x du

14 ∫ 1 / xlog²(x) dx = 14 ∫ (x du)/(xu²)  = 14∫ 1/u² du = 14 u‾¹/-1 = 14 (-1/u)

Claim: A function with removable discontinuities is integrable.

• What does having a removable discontinuity mean?

A function f(t) has a removable discontinuity at a if lim t→a and f(a) are defined but not equal.

• What does integrable mean?

A function f(t)  is integrable on an interval [a,b] if the lim n→∞ of the sum of the rectangles (height= f(t*) × width=(a-b)/n) is the same no matter the choice of t*. 

t* is a point taken from one interval.

Given the fact that all continuous functions are integrable a function like f(t)=3  is integrable and for any rectangle the height=3.

Lets  calculate the integral of f(t)=3 from [0,5]. We divide the interval into equal length subintervals and multiply them by a sample point t* of height=3. The area under the curve=6. We can also get to this result by analyzing the figure that x=0, x=5, y=3, and the t-axis make. A rectangle and we know the area of a rectangle (A=width×height)

Now lets calculate the area under the curve for the same interval [0,5] for a similar function with a removable discontinuity

f(t)=    3 when t≠3   and   =5 when t=3

Again we divide the interval into equal length subintervals and multiply them by a sample point t*. Now not all t* will give f(t*)=3. We can pick t* to give f(t*)=5 and the area of that rectangle will be different from the rest of the rectangles. However, consider the difference that this different rectangle will have on the total sum, is it significant? In the case of our rectangle, does the rectangle stop being a rectangle after taking a single point out? No, it doesn’t. The discontinuity is negligible. And what if the function/rectangle has two, three, four, or ten points of discontinuity? It is still a rectangle. But what if the function has an infinity number of discontinuity points? Then, we can make all of our t* to give a different value and the area will not be the same.

Provided that the number of discontinuity points is finite, f(t) is integrable on the interval [a,b]

Finding the limit of one function can be considered easy depending on the complexity of the function. But what happens when we want to combine two or more functions? Then some of the methods we used before can make finding the limit more complicated. In order to work with these types of limits in a more simplified way, we need to use different properties of limits. One of these properties is the property of taking the limity of the addition or substraction of two functions. We say that if lim f(x)=L as x→a   and   lim g(x)=M as x→a. Then  lim [f(x)-g(x)] =L-M as x→a

Claim:  lim [f(x)-g(x)] =L-M   as   x→a

Proof:

|f(x)-L|< ε/2 for  δ_1 and  |g(x)-M|<ε/2 for δ_2

We want to prove that lim [f(x)-g(x)] =L-M   as  x→a    i.e.   |f(x)-L|-|g(x)-M|<ε   for the smallest  δ between δ_1 and δ_2

|f(x)-L-[g(x)-M]|=|[f(x)-L]+[-g(x)+M]|

By triangle inequality |[f(x)-L]+[-g(x)+M]| ≤ |f(x)-L|+|-g(x)+M|

Note that |-g(x)+M|=|g(x)-M|

|f(x)-L- ( g(x)-M )|≤ |f(x)-L|+|g(x)-M|

We know that |f(x)-L|< ε/2  and  |g(x)-M|<ε/2

|f(x)-L- ( g(x)-M )|< ε/2 +ε/2

|f(x)-L- ( g(x)-M )|< ε  

Your roommate is a baking lover. She is baking a cake for your birthday. You are a curious person and decide to take a look at the cake while it is in the oven. When you take a look at your cake you realize three things. First, your roommate chose a rectangular prism cake tin. The base of the tin is a rectangle of size 20x30cm. The height of the tin is 10cm. However, the tin is only half full. Second, you see that due to the baking soda, the upper surface of the cake is going up at a rate of 0.5cm/min. Third, your roommate chose to prepare a chocolate cake. You want to know the rate at which the volume of the cake is changing when the tin is full.

The volume of a rectangular prism is given by:

V=Area of the base x height

Since the volume is changing only due to the change in height. the rate of change of the volume, by implicit differentiation, is given by:

dV/dt = (20cm) x (30cm) x dh/dt

dV/dt = 600cm x 0.5cm²/min

dV/dt = 300 cm³/min

The Intermediate Value Theorem states that if a function f(x) is continuous on [l, r]; then for any number L within f(l) and f(r), there exists a number a in [l, r] such that f(a)=L

This statement might seem complicated when you first read it, but the basis of it is simple to explain.

Imagine you want to travel from Vancouver BC, to Seattle, USA by car. The only way you can do it is by crossing the border. So, even if you travel all the way to Toronto or Montreal, if you want to go to Seattle you must cross the border. So, at some point during your trip you have to be at the border. This is the basis of the intermediate value theorem (IVT).

The function in our example is the trajectory of the car, which is continuous from Vancouver (l) to Seattle (r). Between your position in Vancouver ( f(l) ) and Seattle ( f(r) ), you must cross the border (L).

This argument relies explicity on the concept of continuity because the first assumption of the  IVT is that the  the function has to be continuous between [l, r].

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