Author: jsilvia7

I will try to write a proof that proves the limit as x approaches 2 of the function f(x)=(-4+x^2)/(2-3x+x^2) equals 0. I thought this limit seemed wrong. The limit should be 4, not 0. 

What I do not understand as of right now is that it seems that the limit of (-4+x^2)/(2-3x+x^2) as x approaches 2 would be 4 because that is where the hole is in the function. 

What is easy about this is that I really like the example given to us beforehand. What will be hard is trying to figure out exactly what the formulas are for this problem to work out correctly and give me the answer of lim x→2 ((-4+x^2)/(2-3x+x^2))=0. The example we were given was that the limit as x approaches -2 of the function (x+2)/(x^2+3x+2) equals -1. The reason or this even though there is a hole at x=-2 is because there is a constant, C such that f(x)-(-1)<C|x-(-2)|. I am assuming that the -1 is for the limit and the -2 is for the x that the limit is approaching. Additionally, there is a seemingly random fraction, ½, that is set as greater than or equal to the absolute value of x+2. I tried plugging in different values of x to try to obtain ½, and I believe I got x=-3 for the example problem, but this does not help me because it seems this integer comes from nowhere. 

I have done some researching and some speculating and I have come to the conclusion that we use ½ to set a boundary for our x to be in. We do not want to use one in this case or else the limit could be approaching -3 or -1. So to restrict our x value, we set the change in x to be ½ and see what x value the limit approaches. 

 

A much simpler way of proving that the limit as x approaches 2 of the function ((-4+x^2)/(2-3x+x^2)) is to take the derivative of the numerator and the denominator and then plug in 2 for x. Taking the derivative of the numerator gives us 2x and taking the derivative of the denominator gives us 2x-3. Then we get 2(2)/(2(2)-3) which equals 4/1 which equals 4. 

The limit as x approaches 8 of the function (x+4)/(x^2 – 10x +10) is equal to -2. We can prove this simply by plugging in 8 for x in the function. (8+4)/(8^2 – 10(8) +10) = 12/(64-80+10) = 

12/-6 = -2. 

What was hard about trying to write this proof is that I have never thought about finding a limit in this way. We were taught to find where the hole is in a function and then we find the limit that way. At least the second function was easier to solve for the limit because I just needed to plug in the x value.

I am convinced that B(R^N) is a sub-algebra of R^N. B(R^N) is closed under the operations of addition, scalar multiplication, and multiplication. This makes sense because B(R^N) represents the set of all bounded real-valued sequences. It seems that you can treat these sets of sequences as vectors and therefore addition, scalar multiplication and multiplication would all be possible. 

1. <s, s> > 0 for all s not equal to the empty set in the set of all bounded real-valued sequences. The inner product of two bounded sequences s, t: N → R is equal to <s, t> = the summation of s(n)t(n)/2n while n is greater than or equal to 1. This would mean that <s, s> = the summation of s(n)s(n)/2n = s2(n)/2n while n is greater than or equal to 1. The multiplication between two bounded sets is possible, therefore s(n)s(n) = s2(n)
2. Similar to how the dot product between two vectors is the same whether it is a dot b ot b dot a, <s, t>=<t, s> for all s,t in the set of B(R^N). <s, t> := the summation of s(n)*t(n)/2^n for n is greater than or equal to 1. <t, s>:= the summation of t(n)*s(n)/2^n for n is greater than or equal to 1. Because of the commutative property of multiplication, s(n)*t(n) = t(n)*s(n), therefore <s, t> = <t, s> for all s, t in the set of B(R^N).
3. This statement, “<s_1+s_2, t> is greater than or equal to <s_1, t> + <s_2, t> for all s_1, s_2, t in the set of B(R^N)” is true because <s_1+s_2, t> := the summation of (s_1(n)+s_2(n))*t(n)/2^n for n is greater than or equal to 1 and <s_1, t> + <s_2, t> := the summation of s_1(n)*t(n)/2^n + s_2(n)*t(n)/2^n which equals (s_1(n)*t(n) + s_2(n)*t(n))/2^n which equals (s_1(n)+s_2(n))*t(n)/2^n because of the distributive property.
4. <λs, t> = λ<s, t> for all s, t in the set of B(R^N) and all λ in the set of real numbers because <λs, t>:= the summation of λ*s(n)*t(n)/2^n for n is greater than or equal to 1. λ<s, t> := λ(s(n)*t(n)/2^n) hich equals λ*s(n)*t(n)/2^n.

What I notice about this sequence is that it converges to 4. It is infinitely increasing and is bounded from 3 ≤ y(n) < 4. It is not a Cauchy sequence because it does not oscillate even though with each iteration, the values get closer and closer together. 

Below is a graph of the iterations of y(n). what can be observed by this graph is that y(n) gets closer and closer to 4 with each iteration. The line is very smooth and there are no kinks or waves. The sequence increases quite quickly as it approaches 4 with the first 20 or so iterations, then the increase of y(n) gets slower.

Below is an array of the iterations of y(n).

Below is the code that I wrote for finding the values of y(n)

For all x in the set of positive rational numbers, L(x) is a lower cut if L(x) ;= {a/b is in the set of positive rational numbers such that a/b < x}. 

• L is a non-empty subset of Q+
• L is bounded above
• L has no maximum element
• If a/b is in the set L and 0 < c/d < a/b then c/d is in the set L

A set with these four properties is a lower cut of Q+. We know that L is not an empty set because from another proof that I wrote out before, if a/b < c/d are rational numbers then the set of rational numbers x/y such that a/b < x/y < c/d is  not finite. Therefore there exists at least one element and in fact there is an infinite set of elements in L because of this proof. L is  bounded above by x, whatever positive rational number that is decided upon. L has no maximum element because there is an infinite set of elements in the set of L that is between 0 and x. As for the last requirement for a set to be a lower cut of the positive rational numbers, it seems quite obvious that if a/b is in the set L and 0 < c/d < a/b then c/d is in the set L. 

The function x |→ L(x):Q+ → R+ is one-one. For every element in the set of L, there is only one positive real number that matches the element in L. 

If a/b < c/d are rational numbers then the set of rational numbers x/y such that a/b < x/y < c/d is  not finite. The reason why is because if we happen to find a rational number in between a/b and c/d, let’s say, c/d – a/b = bc/bd – ad/bd = (bc – ad)/bd. This is a rational number between a/b and c/d. Now we have a/b < (bc – ad)/bd < c/d. We can repeat this same procedure to find a rational number between a/b and (bc – ad)/bd. (bc – ad)/bd – a/b = (bc – ad)/bd – ad/bd = (bc – 2ad)/bd. We can keep repeating this procedure for an infinite amount of times and get a different rational number each time that is between a/b and c/d. 

Let S be a finite set, {a, b, c, d, … , n}, a, b, c, d, … , n are in the set of real numbers, S can have any number of elements. S cannot be Dedekind infinite. Dedekind infinite would imply that there is a bijective function from S to T, where T is an infinite set of numbers and T is a proper subset of S. T, which should be a proper subset of S if S is Dedekind infinite, has infinitely more elements than S, therefore S is not Dedekind infinite. 

Let N be the set of natural numbers and S the set of squares of natural numbers. I will prove that |N| = |S|. The set N is infinite. For each element in the set of N, there is a matching square number which is in the set of S. Think of the functions y = x and y = x^2. The set of natural numbers and its squares are infinite and each number is N maps a number is S, therefore there is a bijective function Φ: N → S and |N| = |S|.

Let N x N denote the set of all ordered pairs {{m, n}: m, n in the set of N}. |N x N| = |N| because for every ordered pair, there will be a matching element in the set of N.

The sequence does seem to have a limit as a continuous function. In the first example, it approaches the function, y = sin(x). This is because the first table forms a list of functions that are the Taylor series functions of sin(x). 

The sequence “converges” in the sense that as there are more terms of the Taylor series for sin(x), it appears more and more like sin(x). 

We could analyze the convergence by observing |sin(x) – fn(x)| as n goes to infinity. This is a list of functions of the Taylor series for sin(x), so as n approaches infinity, |sin(x) – fn(x)| will approach 0.

sqrt(5) is not rational. I will prove this by contradiction. Let’s say that sqrt(5) is rational, that is sqrt(5) = m/n for some m, n in the set of positive integers and m and n are coprime (so that we know it is a unique solution). If we square both sides of the equation, we get 5 = (m/n)^2 = m^2/n^2. Then we can multiply both sides of the equation by n^2 and we get 5n^2 = m^2. The equation is saying that 5 divides m^2 and therefore divides m. This means that some positive integer p exists such that 5p = m. So now 5n^2 = m^2 = (5p)^2 = 25p^2 and then we can divide 5n^2 = 25p^2 by 5 and get n^2 = 5p^2. Then divide both sides by p^2 and obtain 5 = n^2/p^2. Now take the square root of both sides and we get sqrt(5) = n/p. This leads to a contradiction because from the original equation, sqrt(5) = m/n and now we have sqrt(5) = n/p. In the original equation, m is greater than n and in the second equation, n is greater than p. In other words, this is saying p < n < m. n and p are another pair of integers other than m and n that have a quotient of sqrt(5) which contradicts our assumption that m and n are unique. 

The square root of 17 is not rational. I will prove this by contradiction. Let’s say that sqrt(17) is rational, that is sqrt(17) = m/n for some m, n in the set of positive integers and m and n are coprime (so that we know it is a unique solution). If we square both sides of the equation, we get 17 = (m/n)^2 = m^2/n^2. Then we can multiply both sides of the equation by n^2 and we get 17n^2 = m^2. The equation is saying that 17 divides m^2 and therefore divides m. This means that some positive integer p exists such that 17p = m. So now 17n^2 = m^2 = (17p)^2 = 289p^2 and then we can divide 17n^2 = 289p^2 by 17 and get n^2 = 17p^2. Then divide both sides by p^2 and obtain 17 = n^2/p^2. Now take the square root of both sides and we get sqrt(17) = n/p. This leads to a contradiction because from the original equation, sqrt(17) = m/n and now we have sqrt(17) = n/p. In the original equation, m is greater than n and in the second equation, n is greater than p. In other words, this is saying p < n < m. n and p are another pair of integers other than m and n that have a quotient of sqrt(17) which contradicts our assumption that m and n are unique. 

The square root of 9 is rational and the types of proofs above will not show that the square root of 9 is irrational. This is because if we assume that sqrt(9) = m/n, m, n are in the set of positive integers and are coprime, we can see that 9 = (m/n)^2 = m^2/n^2. If we let m = 3 and n = 1, 3^2/1^2 = 9/1 = 9. 

The cubed root of 2 is not a rational number and I will prove this by contradiction. Let’s assume that there exists an m, n such that crt(2) = m/n and m and n are coprime (this means that m and n are the smallest positive integers that exist such that m/n = crt(2). If we cube both sides of the equation, we get 2 = (m/n)^3 = m^3/n^3. We can multiply each side of the equation by n^3 to get 2n^3 = m^3. This means that 2 goes into m^3 and goes into m and therefore there is a positive integer, p such that 2p = m. Now we can substitute 2p in for m in 2n^3 = m^3 and we get 2n^3 = (2p)^3 = 8p^3. Now let’s divide both sides of the equation by 2 and we get n^3 = 4p^3 = 2*2p^3. This means that 2 also divides n^3 and therefore 2 divides n. This goes against the assumption that  m and n share no common factors other than 1 since 2 divides both m and n.

Log base 10 of 5 is not rational. I will prove this by contradiction. Let’s assume that log105 is rational, that is log105 = m/n, m and n are positive integers and are coprime. This means that 10^(m/n) = 5. We can raise both sides of the equation to the power of n and we get 5^n = 10^(m/n)^n= 10^m. There are no positive integers, m and n such that 5^n = 10^m. 5 to the n would result in a positive integer that ends with a 5 and 10 to the m would result in a positive integer that ends with a 0. 

I cannot see the diagrams on mathematica and I also tried looking up these diagrams and still could not find them. 

The first mathematica code took 122.4573370 seconds and the second one took 7.0831204 seconds. There is a big difference probably because since the second one stores the computed results in memory, it takes less time for those values to be used again. Since the first one does not save the computed results in memory, it takes a lot more time for those values to be computed again to be used. 

I have inserted both the array and the plot of x1., … , x25 below. The plot shows that the sequence converges to 2. The sequence is infinitely decreasing, but is never less than 2. 

The reason why m+n is defined for all m, n in the set of natural numbers is because the set of natural numbers is infinite. Any length m added to another length, n, will always exist. 

The binary operation +:NxN→N is commutative. We can prove this geometrically because if we first have a length m and add it to another length n, and in another situation we first have a length n and add it to another length m, the 2 lengths, m+n and n+m will be the same. I show this below. 

The binary operation +:NxN→N is associative. I prove this by using a similar method that I will also show below. 

The cancellation property for all m, n, p in the set of Natural numbers means that m + p = n + p means that m = n. I will prove this by using the fact that a + 0 = a for all a in the set of natural numbers. I will also use the associative property to prove this.

The Grothendieck group of (N, +) exists because

The reason why m x n exists for all m, n in the set of natural numbers is because the set of natural numbers is infinite and m groups of n or n groups of m will exist for all values of m and n. 

The binary operation x:NxN→N is commutative. The reason that m x n = n x m is because conceptually, m groups of n and n groups of m give the same value. 

The binary operation x:NxN→N is associative. The reason that (m x  n) x  p = m x  (n x  p)

The reason that m x 1 = m = 1 x m is because if we think of m x 1 and 1 x m as an array of m objects, whether the array is going vertical with m objects and only one column of the objects or the array is going horizontal with m objects and only one row, the array will only have m objects. 

The reason why the cancellation property of multiplication exists, that is if m x p = n x p, then   m = n for all m, n, p in the set of natural numbers, is because if we can divide both sides of the equation by p. In other words, we can multiply both sides of the equation by 1/p which is the reciprocal of p and this is the same thing as dividing by p. m x p x 1/p = m x 1 because p x 1/p = p/p and any number divided by itself is 1. Now m x 1 = m. This is the left hand side of the equation. n x p x 1/p = n x 1 because p x 1/p = p/p and any number divided by itself is 1. Now n x 1 = n. This is the right hand side of the equation. Now m = n. 

a/b is logically equivalent to a/b because if we multiply both sides of the equality by 1/b, we will get a/b x 1/b is logically equivalent to a/b x 1/b. This simplifies to a x 1 = a x 1 because b x 1/b = b/b = 1 since any number divided by itself is 1. Then a x 1 = a x 1 simplifies to a = a because any number multiplied by 1 equals itself. Also from what is said before in the previous lines of this mathematica notebook, a/b is logically equivalent c/d means a x d = b x c. In this case, because a/b is logically equivalent to a/b, this means that a x b = a x b which is true. 

If a/b is logically equivalent to c/d, then this means that a x d = b x c. If c/d is logically equivalent to a/b, then this means that b x c = a x d. Because multiplication is commutative, that means that a x d = b x c and b x c = a x d. This means that if a/b is logically equivalent to c/d then if c/d is logically equivalent to a/b.

If a/b is logically equivalent to c/d and c/d is logically equivalent to e/f, then a/b is logically equivalent to e/f. This is true because a x d = b x c and c x f = e x d. Then c = d x e / f if we divide both sides of the second equation by f. Then if we substitute d x e /f in for c in a x d = b x c, we get a x d  = b x d x e / f. I have written out the rest of the problem by hand so that it is more clear what is happening and how I did the problem.

This recursive sequence seems to converge to 2. On the 53rd iteration of an, an finally gets close enough to 2 so that there is a roundoff error where an = 2.0 instead of a real number that is something like 1.9999999… The sequence is bounded by 1.5 ≤ an < 2 and is infinitely increasing. This sequence approaches 2 quite slowly as it takes 53 iterations for the roundoff error to be small enough for python to round an to 2. The initial condition makes it so that the sequence increases infinitely and converges to 2. For all other values of a1, if a1 is equal to 2, an= 2 for all an, n = 1, 2, 3, … , to infinity. If a1 is greater than 2, the sequence will decrease infinitely and converge to 2. If a1 is less than 2, the sequence will increase infinitely and converge to 2.

an is never going to be greater or equal to 2 when a(1) = 1 because no matter how close an gets to 2, it will always be 1.99999… and then 1.99999…/2 is less than 1. Then a number really close to 1, but still less than 1, plus 1 is only still really close to 2.

This recursive sequence reminds me of fixed point iterations I learned in numerical analysis this year. When a(1) = 2, we get 2 in return for all an, n greater than or equal to infinity.

Below is some code I wrote in order to find out if the recursive sequence converges. 

I tested to see the number of for loops so that I could see how slowly or quickly an converges. Compared to some other sequences, I believe this one converges pretty quickly. This reminds me of the function e^x where the function will converge if x is less than or equal to 1 and diverge if x is greater than 1.

Below is an array of the iterations of an, n = 1, 2, 3, … , 61. 

We can see what an is with each iteration that an gets closer and closer to 2 until python’s roundoff error results in an output of 2.0 for an. Then from what we know from a fixed point iteration, for this sequence, an input of 2 will output a value of 2 for an, n is equal to 1 to infinity. The same is for the python program, since the error makes a(54) = 2.0,  an will equal 2 for n = 55, 56, 57, …, to infinity. We know that this is not actually the case though because an is infinitely increasing and is only approaching 2.

Below is a graph of a(n+1) =a(n)/2 +1, a(1) =1. 

It is clear that this sequence converges to 2 from this plot.

There was not really anything hard about this exercise. Coding up some lines for this recursive sequence was quite easy and even trying to figure out how many iterations it took for an to converge close enough to 2 for python to have a roundoff error was easy.

This recursive sequence appears to converge to 3 both when a1 = 1 or when a1 = 15. When a1 = 15, the sequence is bounded above by 7.8 and bounded below by 3. The series converges quite quickly as the sequence only needs 7 iterations of an for python’s round off error to round an to 3.0 when a1 = 15.When a1 = 1, the sequence is bounded above by 5 and below by 3. The sequence is decreasing infinitely, but is never less than 3. When a1 = 1, the series converges quickly as well, but the round off error occurs on the 6th iteration rather than the 7th, meaning that the series converges quicker than when a1 = 15. It seems that the fact that a1 = 1 rather than a1 = 115 has very little significance to the series because they both converge to 3. 

Below is an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =15.

Below is an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =1.

Below is a program that performs 16 iterations of an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =15. To perform iterations of an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =1, I simply changed an1 = 15 to an1 = 1 in line 14. 

Below is a graph of the iterations of an when a1 = 1. The series converges to 3.

Below is a graph of the iterations of an when a1 = 15.