# Quicklopedia

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• Aug 8th 2006, 08:17 PM
ThePerfectHacker
Quote:

Originally Posted by Quick
anyway, you briefly meantioned that determinants are related to the discriminant in the quadratic formula. How?

They are related in a sense that they describe the charachteristic of an equation.
---
There are 3 types systems involving linear equations: inconsistent, consistent and dependent and consistent and independent.

Inconsistent: If you are given two equations that are contradictory there are no solutions. For example,
$\left\{ \begin{array}{c}x+y=1\\x+y=2$
How can two numbers add to 1 and 2? So there are no solutions. But I want you to look at is determinant of its coefficient.
$\left| \begin{array}{cc}1&1\\1&1 \end{array} \right| = 0$. It is zero.

Consistent and Dependant. This is a system that has infinitely many solutions. For example,
$\left\{ \begin{array}{c}x+y=1\\2x+2y=2 \end{array}$
Because the second equation can be divided by two to give,
$x+y=1$ which is the same as the first. Thus for any value of $x$ you can find a $y$. Thus, there are infinitely many solutions.
Now, find the determinat of the coefficients.
$\left| \begin{array}{cc}1&1\\2&2 \end{array} \right|=0$. Also, zero.

Warning When you find the determinant of the system you need the variables to be lined up. For example,
$\left\{ \begin{array}{c}2x+y+z=1\\2y+x+z=2\\2z+x+y=3$. You need to line up the x's the y's and the z's and then take the determinant. Thus, it is,
$\left| \begin{array}{ccc}2&1&1\\1&2&1\\1&1&2 \end{array} \right|$

Consistent and Independant. This is a system with a unique solution. For example,
$\left\{ \begin{array}{c}x+2y=1\\2x+y=1 \end{array}$.
Its determinant of the coefficients is,
$\left| \begin{array}{cc}1&2\\2&1 \end{array} \right|=3\not = 0$. This leads to the following theorem.

Theorem A system of linear equations is consistent and independant (unique solution) only when, its determinant is N0N-ZER0.

Of course, when the determinant is zero it can have no solutions (inconsistent) or infinitely many solutions (consistent and dependant). However, how can you determine which one it is when determinant is zero? Later on I can show you.

Definition A system of linear equations,
$\left\{ \begin{array}{c}{ a_{11}x_1+a_{12}x_2+...+a_{1n}x_n=0\\a_{21}x_1+a_{ 22}x_2+...+a_{2n}x_n=0\\.......................... ........=0\\a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n=0$ is called homogenous.

Theorem If a homogenous system of linear equations has its determinant zero then it has infinitely many solutions.

Proof When the determinant of the system
$\left| \begin{array}{cccc}a_{11}&a_{12}&...&a_{1n}\\a_{21 }&a_{22}&...&a_{2n}\\...&...&...&...\\a_{n1}&a_{n2 }&...&a_{nn}\end{array} \right| = 0$
Then it can have either infintely many solution or no solutions (it cannot have a unique solution). But this equation does have solution, namely the trivial,
$x_1=x_2=...=x_n=0$
Thus, it CANNOT have no solutions, thus it must have infinitely many solutions. (This type of equation does appear in applied math often and it is the case when the determinat is zero because otherwise it is not interesting).

Hope you like this lecture, that should entertain you for some hours. Next, time I will show you how to solve equations with determinants..
• Aug 13th 2006, 05:18 AM
Quick
Limits
Would you please tell me what limits are and how you find them?
• Aug 13th 2006, 11:23 AM
ThePerfectHacker
This discussion would be mathematically limited (get the pun) for the sake of simplicity.
---
Assume, you have a function, and you wish to find,
$\lim_{x\to c}f(x)$. What it means is what values does $f(x)$ approach as $x$ approaches $c$? Note, the function does not need to be defined at $x=c$. I might be mistaken but I believe Pierre de Fermat (1601-1665) was the first person to use limits at a point. [Though the method of exhaustion was a favorite of Archimedes he did not compute it for points].

There are 3 ways to compute.
1)Graphical. This is a useful way to observe how the function behaves as it approaches a certain point.
2)Numerical. This is often the easiet method to use by creating a table of values.
3)Analytical. This method is the one which employs the use of theorems to deduce limits. Note, this is the only method acceptable for proofs. Remember, mathematicians do not accept graph (and do not use them) and they do not rely on numerical evidence.

The first method you probably already understand. Simply draw your graph and see how it behaves. Sometimes it might be difficult to draw a graph so it is easier to create a table of values. Before I procede to tabular method I will first define left and right limits.

There are two different types of limits.
$\lim_{x\to c^+} \mbox{ and }\lim_{x\to c^-}$. They respectively are called, limit from the right and the limit from the left. What it means is basic. You approach your point from the right, now from the right means all point larger than that point. And from the left means all point lesser than that point.
For example, compute the following limit,
$\lim_{x\to 0^+}\frac{|x|}{x}$?
Create a table of values getting closer from the right (but never reaching) and see how they affect the function.
$\left\{ \begin{array}{cc}x& f(x)\\
1&1\\
.1&1\\
.01&1\\
.001&1$

it is resonable to say the limit from the right is 1.
Now from the left, meaning,
$\lim_{x\to 0^-}\frac{|x|}{x}$?
Create a table of values getting closer from the right (but never reaching) and see how they affect the function.
$\left\{ \begin{array}{cc}x&f(x)\\
-1&-1\\
-.1&-1\\
-.01&-1\\
-.001&-1$

it is resonable to say the limit from the left is -1.
Note, an important fact.
$\lim_{x\to 0^+}\frac{|x|}{x}\not = \lim_{x\to 0^-}\frac{|x|}{x}$.
That means, that the limit,
$\lim_{x\to 0}\frac{|x|}{x}$ does not exist. Because from one side it is one value and from another side it is a different value. There is even a theorem in calculus (analysis).

Theorem. A limit of a function exists at a point if and only if the left and right limits coincide.

Now for an excerise create a table of values (from both sides) and say whether this limit exists (and what is its value)?
$\lim_{x\to 0}\frac{x}{x}$?

-=WARNING=-I do not want you to find since its limit is L then $\frac{0}{0}=L$ because you cannot substitute.

• Aug 14th 2006, 08:38 AM
ThePerfectHacker
I am very angry with you Quick. I posed a limit finding excerise for you to do and you still did not do that. I need to see how you did it to correct you is necessary.
• Aug 14th 2006, 10:42 AM
Quick
Quote:

Originally Posted by ThePerfectHacker
I am very angry with you Quick. I posed a limit finding excerise for you to do and you still did not do that. I need to see how you did it to correct you is necessary.

I'm sorry, here is my answer:

the table for $\lim_{x\to 0^-}$ is $\left\{ \begin{array}{cc}x&f(x)\\-1&1\\-.1&1\\-.01&1\\-.001&1$
therefore we say $\lim_{x\to 0^-}=1$

the table for $\lim_{x\to 0^+}$ is $\left\{ \begin{array}{cc}x&f(x)\\1&1\\.1&1\\.01&1\\.001&1$
therefore we say $\lim_{x\to 0^+}=1$

then we realize: $\lim_{x\to 0^-}=\lim_{x\to 0^+}$

therefore there is a limit, which we'll call $L$
• Aug 14th 2006, 10:46 AM
ThePerfectHacker
Quote:

Originally Posted by Quick

then we realize: $\lim_{x\to 0^-}=\lim_{x\to 0^+}$

therefore there is a limit, which we'll call $L$

Good. But since,
$\lim_{x\to 0^-}\frac{x}{x}=\lim_{x\to 0^+}\frac{x}{x}=1$
By the theorem I posted, the limit, (without left or right)
$\lim_{x\to 0}\frac{x}{x}=1$
• Aug 14th 2006, 11:07 AM
ThePerfectHacker
Discussion on limits will now countinue (get the pun).
---
Whenever you calculate limits either they exist or they do not. When they do not exist there is a special case which should be noted.

Defintion. When the limit of a function at a point increases without bound, we say the limit is $+\infty$. Warning: It does not mean that the limit exits. It does not. It rather means it is a special case when the limit keeps increasing. Furthermore, this is not a number, it is a notation used to demonstrate this instance.

Definition. When the limit of a function at a point decreases without bound, we say that the limit is $-\infty$. Warning: It does not mean that he limit exists. It does not. It rather means it is a special case when the limit keeps descreasing. Furthermore, this is not a number, it is a notation used to demonstrate this instance.

Here is a good example which shows both instances in one.
Consider the function, $f(x)=\frac{1}{x}$. I recommend to use that graphing program you have to graph it and see how it looks.

We will analyze the limits of $x=0$. As before, we will use a table from both side. First from the right.

$\left\{ \begin{array}{cc}x&f(x)\\
1&1\\
.1&10\\
.01&100\\
.001&1000
$
.
You can see that the function just keeps getting bigger and bigger. Thus, we write,
$\lim_{x\to 0^+}\frac{1}{x}=+\infty$

Now from the left,
$\left\{ \begin{array}{cc}x&f(x)\\
-1&-1\\
-.1&-10\\
-.01&-100\\
-.001&-1000$

You can see that the function just keeps getting smaller and smaller. Thus, we write,
$\lim_{x\to 0^-}\frac{1}{x}=-\infty$

The graph will show the curve goes to the heavens from the right and goes to the underworld from the left.

As, an excercise do the limits of $\frac{1}{x^2}$ at $x=0$
• Aug 14th 2006, 11:25 AM
Quick
Before going on, I would like to thank you for taking the time to do this hacker

Quote:

Originally Posted by ThePerfectHacker
As, an excercise do the limits of $\frac{1}{x^2}$ at $x=0$

from the left we get: $\left\{ \begin{array}{cc}x&f(x)\\-1&1\\-.1&100\\-.01&10000\\-.001&1000000$

therefore: $\lim_{x\to 0^-}\frac{1}{x^2}=+\infty$

from the right we get: $\left\{ \begin{array}{cc}x&f(x)\\1&1\\.1&100\\.01&10000\\. 001&1000000$

therefore: $\lim_{x\to 0^+}\frac{1}{x^2}=+\infty$

You make it sound like, even though both limits are +infty, that no limit exists...
• Aug 14th 2006, 12:18 PM
ThePerfectHacker
You did everything correct!. Except on the tables you forgot to mention minus signs from the left!
---
Quote:

Originally Posted by Quick
You make it sound like, even though both limits are +infty, that no limit exists...

That is correct, a limit which is infinite does not exist.
As, an additional point I would like to mention that sometimes you might want to write, (without left or right)
$\lim_{x\to 0}\frac{1}{x^2}=+\infty$
To indicate that both limits from the left and from the right are $+\infty$. The only advantage of this is that is saves writing space.
• Aug 14th 2006, 02:05 PM
ThePerfectHacker
Before I get to theorems about limits there is still one more limit to learn. It is similar to the one previously but the positions are reversed. I am referring to,
$\lim_{x\to +\infty}f(x) \mbox{ and }\lim_{x\to -\infty}f(x)$. As you can guess it involves finding the limit as the number gets larger and larger. While the other one involves a number getting smaller and smaller.
For example,
$\lim_{x\to -\infty}\frac{x+1}{x+2}$
Create a table of values, unlike the past there is no such thing as a limit from the left or right just simply make it smaller and smaller.
$\left\{ \begin{array}{cc}x&f(x)\\
-10&1.125\\
-100&1.0102\\
-1000&1.001\\
-10000&1.0001$

It is resonable to say the limit is one.
Thus, we write,
$\lim_{x\to -\infty}\frac{x+1}{x+2}=1$

Another example, find $\lim_{x\to +\infty}x$
We find,
$\left\{ \begin{array}{cc} x&f(x)\\
1&1\\
10&10\\
100&100\\
1000&1000$

The limit does not exist but it increases without bound, thus,
$\lim_{x\to+\infty}x=+\infty$

When you calculate a limit at $\pm \infty$ it can exist as in the first example or it cannot exists but increase/descrease without bound as in the second example. But can it be neither of these? Yes, observe (hope you know what radians are),
$\lim_{x\to\infty}\cos x$. I will create two tables of values,
$\left\{ \begin{array}{cc}x&f(x)\\\pi&-1\\3\pi&-1\\5\pi&-1\\7\pi&-1$ and $\left\{ \begin{array}{cc}x&f(x)\\2\pi&1\\4\pi&1\\6\pi&1\\8 \pi&1$
Thus, take the function along odd multiplies of $\pi$ you get -1 (no matter how large) and take the function along even multiplies of $\pi$ you get 1 (no matther how large). Thus, the values oscillate between 1 and -1. Thus, the limit does not exist nor is $\pm \infty$.
This concludes the discussion on limits.
I will attach some example in the end of this post.
---
This completes my discussion on limits. I will show some theorems involving limits. But before that, do the following examples (just post your solutions without the table).

Say whether the limits exists or they do not. If they do not say whether they are infinite limits or not. If they do say whether the value of the limits

1) $\lim_{x\to 0}\frac{x^2-x}{x-1}$

2) $\lim_{x\to +\infty}\frac{x}{2^x}$

3) $\lim_{x\to 1}\sqrt{1-\frac{1}{x}}$

4) $\lim_{x\to \pi/2}\tan x$

5a) $\lim_{x\to 0}\frac{\sin x}{x}$ where $x$ in radians.

5b) $\lim_{x\to 0}\frac{\sin x}{x}$ where $x$ in degrees.

What conclusions can you make about 5a and 5b?
• Aug 17th 2006, 05:25 PM
Quick
1) $\lim_{x\to 0}\frac{x^2-x}{x-1}=0$

2) $\overbrace{\lim_{x\to +\infty}\frac{x}{2^x}=0}^{\text{not entirely sure}}$

3) $\underbrace{\lim_{x\to 1}\sqrt{1-\frac{1}{x}}\neq\text{anything}}_{\text{I think I did something wrong}}$

4) $\lim_{x\to \pi/2}\tan x\neq\text{anything}$

5a) $\lim_{x\to 0}\frac{\sin x}{x}=1$ where $x$ in radians.

5b) $\lim_{x\to 0}\frac{\sin x}{x}=0.01745$ where $x$ in degrees.

Quote:

What conclusions can you make about 5a and 5b?
not much... the reason they're different is becuase x degrees will be different than x radians, yet becuase they are divided by x two different answers will occur
• Aug 17th 2006, 05:32 PM
ThePerfectHacker
Quick did you form tables from the left and right?
You did not exactly do the best on this test.
Try again. Read what I wrote before.
Always form a table from the left and right.
Then compute the results.
• Aug 17th 2006, 05:34 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
Quick did you form tables from the left and right?
You did not exactly do the best on this test.
Try again. Read what I wrote before.
Always form a table from the left and right.
Then compute the results.

Unfortunate eh? well, I'm a bit annoyed with what I have to do with excel to find such limits as x to 1, but I will edit my revisions in soon.
-----------------------------------------
The only thing I think I did wrong was set some limits at plus or minus infinity... (which I have changed in my post)
• Aug 17th 2006, 06:34 PM
ThePerfectHacker
Quote:

Originally Posted by Quick
well, I'm a bit annoyed with what I have to do with excel to find such limits as x to 1,

Do not use excel just use you basic calculator to do this limits.
• Aug 25th 2006, 10:29 AM
Quick
I feel like the answer is staring me in the face but I can't find what I did wrong!!!

And leaving me to figure it out isn't going to make me find the mistakes...
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