Limit

Limit sounds abstract but is actually a very concrete concept.  Limit describes a phenomenon where the variable is arbitrarily close to a value but never actually reaches it.  For example, if the limit of y is 6 when x is approaching to 3, then y is getting closer and closer to 6 when x is getting closer and closer to 3. But y can never reaches 6 in this process. But can y be 6 at the point of 3, or even, can x be 3? We don’t care. We don’t even talk about it in the context of limit. There is even some cases where y gets closer and closer to 6 without ever touching it on the whole real numbers interval.

But how can it be possible? How can y never reaches 6?

Well it is possible. Imagine as x moves to the right, y get from 10 to 9 ,say, and get from 9 to 8.5 as x moves even further to right for the same distance in the previous move. And then for every same distance x move to infinity (right), y decreases from 8.5 to 8.3, then from 8.3 to 8.25, then from 8.25 to 8.2425…..

See y can keep decreasing (get closer and closer to 6) but since the speed of the decrease is also decreasing, it is possible that it never reaches 6.

A function being analytic at a means the Taylor Series for the function at a converges to the function itself in an open interval centered at a. Taylor Series fails to describe what happen at the point a and the inside of the function. In question 2, the function is not analytic at a=0 because the Taylor Series at a=0 (the Maclaurin Series for it) is not convergent to it.

For example, function y=x^2 is analytic at any points on its domain. Its infinitely differentiable and its Taylor Series at every point is convergent to x^2.

Another example for non-analytic function is y=|x|. It is not analytic at x=0. Because its not eifferentiable at that point , we cannot even write a taylor series for it.

This concept is important since calculus aims to describe the inside of a function and it is both necessary and interesting to know when it does not work.

All the analytic functions must be infinitely differentiable but not all the infinitely differentiable functions are analytic.

In my opinion, the most difficult question on the midterm is Question 1. Although Question 1 is considered as the most basic fundamental computational skills that we need to master before talking about other more advanced calculus problem that demand more thinking or different kinds of thinking, the question took me 60 minutes in a 90 minutes’ exam. I thought that as long as I “knew how to do” the computation, I would be able to do them quickly in an exam but it turned out that i was wrong. My computational speed did not fit the requirement. For other questions, the ways of thinking them were already covered in assignments or other materials, which made the thinking itself not hard under an exam condition, and underneath all that thinking , is still, computational skills, that actually calculate the answers!

Tips:

1. Mathematics learning does need practice! Do not think it is sufficient to just “know how to do it”. The fact that the level of education changes that I am not in high school anymore and I am now doing things that are cooler than just computation does not change a core factor of mathematics–computation.
2. I should have managed my time more reasonably in the midterm and spent more time for the other questions.

1. Remember to take the derivative of what you want to substitute in the integral. Because when we use the method of substitution, we actually change the variable, the whole process changes (the variable and the limits). We need to transfer it completely to get the correct value.
2. If you see a polynomial  under a root sign that you cannot  take its square root directly , think about substituting the variable with some form of trigonometric  functions. There are some characteristics of trigonometric functions that are easy to  be  converted into a form that can be taken square root of,  are favorable for elimination of the variables elsewhere and are easy to differentiate.
3. If it doesn’t work for one way of substitution, don’t quit, try another. For example, if t doesn’t work, try t square; if sin(x) doesn’t work,  try cos(x),etc. It is hard for beginners to foresee if a method could work, because unexpected things happen all the time. Sometimes a variable can be eliminated; and sometimes we think it can be but it actually can’t.

For interval [a,b], we can partition it into n subintervals. And as we take the length of the interval as width, the y value of sample points as height, the area under the curve can be seen as formed by rectangles. As we know, the area of the rectangle can be calculated by taking the product of width and height, and adding up each area together. According to the definition of integral, as we arbitrarily  narrow down the width of each rectangle, ie, the width  of the rectangles arbitrarily approaches to 0 , and the height remains the same, the area for each rectangle will approach to 0.

Though as  the width approaches to 0, the number of the discontinuous points remains the same, which means the number of the rectangles does not change, and the height of the rectangles does not change, only the width goes to 0. Since there are only finite number of rectangles, while the width decreases infinitely, “infinitely” is somewhat powerful than “finite”, thus the total areas of those rectangles contains the “bad points” approaches to 0 as well.

We have a function f(x)=log(1/e^x)-ex^3

Evaluate df(x)/dx

f(x)=log1-log(e^x)-ex^3=-x-ex^3

df(x)/dx=-1-3ex^2

Assume the UBC fountain pool is a cylinder with the radius of 3m. Water is added into the pool at a rate of 0.5m^3/min. At what rate (m/min)does the height of the water raise?

Let the height be h.

The volume V=pi3^2h=9pih

Differentiating both sides,

dV/dt=9pi dh/dt

We know that dV/dt=0.5

dh/dt=0.5/9pi=1/18pi

Thus the height changes at a rate of 1/18pi m/min

(a) Label the right edge of the circle point M, and the left edge point N.If the temperature at M is the same as that at N, then we have antipodal points where the temperature are the same.

But what if the temperature at M is lower than that at N?

Imagine a stick longer than the diameter of the circle, and its midpoint is fixed at the center of the circle. So the two contact points of the stick and the circle are antipodal points.Label the two points A and B. Now imagine the stick rotates from the x-axis anticlockwise, with A going  from M to N, and B going from N to M.It stop after rotating 180 degrees.So at the initial time, temperature at A is lower than temperature at B;temperature at B is higher than temperature at A.After the rotation, A has higher temperature than B.During the whole process the temperature can only change continuously, and the stick cannot be bent. Thus there must exist a certain instant, at which the temperature at A and that at B are the same.

(b) My argument rely implicitly on the continuity of the change of temperature. In real life, temperature changes continuously, and I use a real life way to interpret the proof, in which the stick is assumed to rotate smoothly (also related to continuity) .

1. The change of temperature as the altitude increases can be considered as a function. Generally, for every 100 meters’ increment at the altitude, the temperature decrease by 0.6℃. If we assume the temperature at the sea level is 25℃,the temperature(T,℃) vs altitude(x,m) function can be written as: y= -0.006x+25

Since it is a linear function, it does not have a horizontal asymptote. Assume it has a horizontal asymptote y=a

so the horizontal asymptote does not exit.

2. There is a pancake, its area is 1.If we cut it into halves.Then after the first time we cut it , the area of each of the pieces is 1/2.After the third time,the area of each of the pieces is 1/4.

…..

If we write down the area after every time we cut it ,it will be

1/2,1/4,1/8,……,

It does not converge.

3.If we sum up all the terms above,we get 1/2+1/4+1/8+…

By the geometric test,we know that the series converges to 1.

This is to claim that I have done my homework.