Math Principles and Misconceptions

[Sandmaster]

phi>e>pi in awesomeness
pi>e>phi in number

XP

[General Veers]

What other irrational constants are there?

Pi (ratio of circle's circumference to its diameter)
e (a.k.a. Natural Number)
Phi (a.k.a. Golden Ratio or Golden Number)
?

[Sandmaster]

Square root of 2?

1/PI (is required to obtain a square with the same area of a circle)?

I isn't irrational, it's just imaginary

[General Veers]

Hence the reason I didn't include it...

And yes! 2^1/2 and 3^1/2 are very commonly used in trigonometry for the "special" 30-60-90 and 45-45-90 triangles.

[Sandmaster]

IDK how the hell do you get something to the 1/2 power?

[General Veers]

x^1/n is "the n^th root of x"

2^1/2 is the square root of two, for example...

8^1/3 is the cube root of eight, which is two.

1024^1/10 is the tenth root of one thousand twenty four, which is two.

[Sandmaster]

Oh, that makes sense now.

and negative squares for example would be 10^-1=10/10/10 or something

[General Veers]

Negative roots create reciprocals.

10^-1=1/10=0.1

10^-2=1/10²=1/100=0.01

10^-3=1/10³=1/1000=0.001

(2/3)^-1=3/2

(2^-3/3^-2)=(3²/2³)=9/8

It helps to know how to use fractional powers when you don't have access to checkmarks that can be used as root symbols, and it helps to have negative fractional powers when...well...technically, a number with a negative power is not considered simplified and therefore ought to be simplified.

Negative powers are nice for getting rid of a division sign, however. For example, instead of 3/(x²-45x+60), you could have 3(x²-45x+60)^-1, which to me is more aesthetically pleasing.

The same for units. Instead of velocity as m/s, I could have ms^-1.

[Sandmaster]

whoops! that's what I meant!

[General Veers]

Shall I teach the introduction to differential calculus? To do so, the audience must know how to work with limits well enough, although that should be easy...

Differential Calculus

You probably know how to find the slope of a linear function: take two coordinate pairs (x₁, y₁) and (x₁, y₁), subtract the second y coordinate from the first y coordinate, subtract the second x coordinate from the first x coordinate, and find the quotient of the difference of the y coordinates (change in y) over the difference of the x coordinates (change in x), i.e.

(y₂-y₁)/(x₂-x₁)=m

What if the function is NOT linear, for example a quadratic function of the template y=a(bx-c)²+d (whose "parent" function is y=x²)?

If the function is not linear, then it will not have a constant slope. However, you can find the slope of a function (a.k.a. "a curve") at a specific point or find the general pattern for a slope for any given value of x of the function f(x). The slope of f(x) at a point (x, f(x)) would be a line that crosses the function at that point and is locally tangent to the graph (depending on the graph, the imaginary line might "accidentally" cross the graph at other places: this will be inevitable).

How do you know what the slope of the imaginary "tangent" is? First, take an approximation. How do you do that? Pick two points for a line to go through and try to make that line parallel to the imaginary tangent. For a function y=f(x), you would choose a point (x, f(x)) and a point that is h units horizontally away from it. That point's x coordinate is easy: x+h. For the y coordinate, you must remember to add h directly to x, not to f(x): you should get the coordinate f(x+h).

Obviously, you take the slope of that line using the aforementioned "rise over run" method. In doing so, you should get

(f(x+h)-f(x))/((x+h)-x)

That should simplify to

(f(x+h)-f(x))/h

That approximation might get you the slope, but usually you won't pick the perfect pair of points and therefore get the exact slope of the imaginary tangent that runs through the points. How do you find the exact slope of a curve at a specific point?

Take another approximation, but this time pick points that are closer together horizontally, i.e. have a smaller of h. The approximated slope should come closer to the actual slope of the imaginary tangent. Pick two other points even closer together (i.e. with an even smaller value of x). The approximated slope should come closer to the actual slope of the imaginary tangent. Notice that as your two points come closer together and begin "sandwiching" the point whose slope you want to find, the closer the approximated line comes to matching the actual imaginary tangent. If you were to decrease the distance between the two points to zero (and therefore have one point, the one you want to find), you would have the exact slope of the imaginary tangent.

Just use the slope formula we mentioned above, right? Not exactly. If both points are the same (i.e. when h=0), you would get the indeterminate form 0/0, which does not tell you what the slope is. However, you saw that choosing two points that are infinitely close together would create an approximation that is infinitely close to matching the exact value without getting a wierd 0/0 result. This is where limits come in.

A way to reword that penultimate (second-to-last) statement is as follows: The limit of the function of x+h and the function of x all divided by h as h approaches 0 is the slope of the imaginary tangent. Worded like that, you can create the expression below.

lim_{(h ► 0)}(f(x+h)-f(x))/h

This expression, which will provide the exact value of the slope of a curve at a given point (x, f(x)) is the definition of the derivative.

Let's take an example curve, say, f(x)=x². What is the slope of the curve at x=1?

lim_{(h ► 0)}(f(x+h)-f(x))/h
lim_{(h ► 0)}(f(1+h)-f(1))/h
lim_{(h ► 0)}((1+h)²-1²)/h
lim_{(h ► 0)}(1²+2(1)h+h²-1²)/h
lim_{(h ► 0)}(2(1)h+h²)/h
lim_{(h ► 0)}h(2(1)+h)/h
lim_{(h ► 0)}2(1)+h
2(1)+0
2(1)
2

The slope of the curve at x=1 is 2.

What is the slope of the curve at any point in general?

lim_{(h ► 0)}(f(x+h)-f(x))/h
lim_{(h ► 0)}((x+h)²-x²)/h
lim_{(h ► 0)}(x²+2xh+h²-x²)/h
lim_{(h ► 0)}(2xh+h²)/h
lim_{(h ► 0)}h(2x+h)/h
lim_{(h ► 0)}2x+h
2x+0
2x

At some arbitrary point (x, x²), the slope is twice the value of x. The derivative of x² is 2x.

Let's try an arbitrary point on the function f(x)=x⁵...

lim_{(h ► 0)}(f(x+h)-f(x))/h
lim_{(h ► 0)}((x+h)⁵-x⁵)/h
lim_{(h ► 0)}(x⁵+5x⁴h+10x³h²+10x²h³+5xh⁴+h⁵-x⁵)/h
lim_{(h ► 0)}(5x⁴h+10x³h²+10x²h³+5xh⁴+h⁵)/h
lim_{(h ► 0)}h(5x⁴+10x³h+10x²h²+5xh³+h⁴)/h
lim_{(h ► 0)}5x⁴+10x³h+10x²h²+5xh³+h⁴
5x⁴+10x³(0)+10x²(0)+5x(0)+(0)
5x⁴+0+0+0+0
5x⁴

Notice something? Whenever you take the derivative of xⁿ, you get an answer of nx^n-1. This is no coincidence: indeed that is what you get when you take the derivative of some polynomial function.

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There is so much more information to share about calculus, but this is all I will present for now...

Any questions?

[vaconcovat]

... Gobsmacked.

[General Veers]

One of two unnecessary posts removed. Thank you.

Feb 23, 2009 0:10:31 GMT -5 General Veers said:

DO NOT SPAM OR POST USELESS COMMENTS

I am trying to maintain a clean thread after having seen so many become spam breeders. Either modify your posts to ask a question, or delete the posts, please.

Thank you for your patience and understanding, both of which are highly appreciated. Your cooperation would be even MORE appreciated.

Again, thank you.

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Now that I have that taken care of, are there any questions about anything relating to any of the principles of mathematics?

[Sandmaster]

I don't know much about limits. I figured out the workings of the universe with algebra.

[vaconcovat]

Hey, my math teacher still thinks that 0 to the power of 0 is 1. in the logic of it, its like getting an empty basket of apples, and then saying that you will put these apples to the power of 0, then one apple magically appears! OMG I AM SOO ANNOYED.

[General Veers]

Alright, then.

Limits

A limit is useful for describing how a function would behave as it got infinitely close to a particular x, even if no f(x) exists at x.

Let's say that you have a function whose domain excludes a value, such as f(x)=(x²-1)/(x-1). At x=1, there is no value f(x) since you would get an indeterminate form; however, if you were to look at its graph, it would seem that it ought to be 2.

f(x) itself is indeterminate, but its limit can be found...

Take a look at the graph. What is the value of f(x) at x=0.9? What about at x=0.99? What about at x=0.999? f(x) should get closer to being 2. The value you are approaching is considered the limit of f(x) as x approaches 1 from the left.

lim_{(x ► 1-)}f(x)=2

Look at the graph again. What is the value of f(x) as x=1.1? What about at x=1.01? What about at x=1.001? f(x) should get closer to being 2. The value you are approaching is considered the limit of f(x) as x approaches 1 from the right.

lim_{(x ► 1+)}f(x)=2

Does the limit of f(x) as x approaches 1 from the right equal the limit of f(x) as x approaches 1 from the left? If so, then the limit of f(x) as x approaches 1 exists. What is this limit, if it exists? Your answer should be 2. You would write that out as follows:

lim_{(x ► 1)}f(x)=2

There is a way to analytically find the limit of f(x) as x approaches 1.

lim_{(x ► 1)}f(x)=lim_{(x ► 1)}(x²-1)/(x-1)
lim_{(x ► 1)}f(x)=lim_{(x ► 1)}((x-1)(x+1))/(x-1)
lim_{(x ► 1)}f(x)=lim_{(x ► 1)}((x-1)(x+1))/(x-1)
lim_{(x ► 1)}f(x)=lim_{(x ► 1)}x+1
lim_{(x ► 1)}f(x)=(1)+1
lim_{(x ► 1)}f(x)=2

When you find limits, you can factor out or take derivatives in order to change an indeterminate form into something determinate, like 0 or an undefined value. If, as you find limits, you get a plain infinity or an undefined value, then the limit does not exist. If the right-hand limit does not equal the left hand limit, then the limit itself does not exist.

[vaconcovat]

Thanks. Ills how that to mai maths teacher. hopefully it will work.

[General Veers]

Mar 13, 2009 1:07:12 GMT -5 @vaconcovat said:

Thanks. Ills how that to mai maths teacher. hopefully it will work.

Actually, the limits thing was for Sandmaster: what you want is in the first post. I'll retrieve it for you...

Feb 23, 2009 0:10:31 GMT -5 General Veers said:

Zero divided by zero:

x/y=z
0/0=z
0=0*z
0=0

The above expression is true no matter what value of z you have because z has no effect on the validity of the expression. The expression 0/0 is considered indeterminate. The "value" of 0/0 depends upon the mathematical context in which you find it in, in which case a knowledge of limits will greatly help to determine what value 0/0 approaches.

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Zero to the zeroth power:

Setup	Explanation
bm+n=bm*bn	Sum property of exponents
bn-n=bm/bn	Difference property of exponents
n+0=n	Zero addition property of equality
0=n-n	Subtract n from each side
00=0n-n	Substitution (Substitute 0 for n-n)
00=0n/0n	Substitution (Quotient property of exponents)
00=0/0	0 raised to non-zero number is zero
00 is indeterminate	0/0 is indeterminate

0⁰ is just another way of expressing 0/0, which is indeterminate.

[vaconcovat]

Yeah, i know, i realised that just after i posted. LOL. And for some reason i cant modifiy my post or delete it. strange.

[The Dark Master]

Do any of the algebra symbols mean any particular number?
So, i think we established the fact that 0 divided (or times...or add..or subtract) by 0= ZILCH!

[General Veers]

Usually, when I use a variable, it represents any REAL (i.e. non-imaginary and non-complex) number, unless I specify a certain parameter such as "when x is a non-zero number."

0/0 is indeterminate because it can be any answer. It does not equal nothing, it equals everything, albeit not all at once...

x/y=z is your basic division operation.

Multiply both sides by the denominator (which is y).

x=yz

Substitute x=0 and y=0 (since that would give x/y=0/0)

0=0z

Anything times zero is zero.

0=0

The expression above is always true: no matter what value of z you have, zero will always equal itself. z could be e^i*pi/30, and 0 will equal zero.