But what’s even more important than having a ball in math class is to retain that information for future use.  That’s the real fun, isn’t it?  So without further ado, I present to you, The Blog.

So we started off with a couple of warm up problems.  The SmartBoard told us to make 81^5 equal to 3^20 without actually calculating the actual values that these produce.  It also told us that by “manipulating” either one or both sides of the equation, we could make these sides equal.  The first thing that I recognized was that 81=3^4 and that I could use this to help further down the line.  If i could reduce the original equation down to this, it would be solved.

What was required to do next was to simply substitute the 81 and 3^4 into the original equation.  This gives us (81)^5 on the left, and (3^4)^5 on the right to keep it equivalent to the original 3^20.  If you take it one step further you can take the fifth root of both sides and get it to the equivalent that we knew previous to this problem.  This leaves us with 81=3^4, which we all know to be true.  Knowing this skill can now help us in future situations to simplify exponents, and even replace bases with other bases.

Speaking of replacing bases, that was our next subject.  We used the same exact situation to learn this.  The board read “81^x=3^?”.  In other words, if you want to write one exponential value that has 81 for a base, how can you write an equivalent exponential value with 3 for a base.  This was actually quite simple to figure out, as all we had to do was look at the power of 81 in the first problem (5) and easily relate it to the power of 3 (20).  Obviously, you can just multiply 81′s power by 4 to get to 3′s power, which leaves us with a solution: 81^x=3^4x.  The reverse of that is what we solved for next in the same way which got us 3^x=81^(x/4).

To further reinforce this concept, we did the same thing with 7 and 5.  This was also quite simple if you understood the topic, as all we really need to do is plug in 1 for x in the equation 7^x=5^? to get us 7=5^x.  From that point on, it’s really just a whole lot of guessing and testing on our calculators.  In class we found that x was approximately 1.21, but to be a bit more precise, it was closer to about 1.2090619551.  Again, we did the opposite with this, 5=7^x.  To acquire x, all we really need to do is find the reciprocal of 1.21 like we did in the first one with 4x to get 1/4x.  We found our final solution to be 5^x=7^.83x.

From these challenge problems came the basis of our lecture for the day, which really told us what we were to take away from today’s class: ANY exponential value can be written under ANY other base by changing the power of the original exponential value.

We also reviewed how we can find the slope on an exponential function give any point on the graph.  To show this,  we got out our handy dandy TI-81′s and graphed a simple parabola.  After zooming in a countless number of times on the vertex of it, we found that at the very bottom of it, we have a straight line that has a slope of zero.  Another way of thinking about this is that on the left side, the parabola’s slope is negative, but slowly approaching 0 until it finally does at the vertex and then becomes positive.

Well that’s all folks, hope you enjoyed and learned lots from it!

Oh yeah, and I promised Dundas a sweet dominoes video, but I decide to mix things up with something a little different.

OK Go -This Too Shall Pass – Rube Goldberg

As a reminder, a Power Function always follows the following format:  f(x)=a?x^b          

The property that was given to us states that:  If f is a power function, multiplying x by a constant results in multiplying f(x) by a constant.  Next, we came up with equations that helped to illustrate this property.

if  f(x)=a?x^b      x2=c?x1       then…

f(x2 )=c^b?f(x1)

 In the equations above, c is a constant,  x1 represents the first step in a function, and x2 represents the second step in a function. (The 1 and 2 should be subscripts!!!) 

Here is an example we did in class: 

We first had to find the constant (c) in which all the x values were multiplied by.  It turned out to be 2, so we knew that c=2.  Next, we found that the outputs of the function were also multiplied by a constant value, which in this example was 1.5.  Because both the inputs and outputs are multiplied by a constant to find the next value, we knew that it followed the Multiply-Multiply Property!  Our next goal was to find out what b was.  To do this, we plugged values into the equation above (  f(x2 )=c^b?f(x1)  ).  x2=18 because it is the second output of the function and x1=12 because it is the first input of the function.  We could then solve for b.  Once you get to the point where b is the exponent, you have to use the guess and check method or the natural log button on your calculator. 

The next property we illustrated during class was the Add-Add Property.  This property is very similar to the Multipy-Multipy property, however instead of multiplying the inputs and outputs of a function by a constant value, you add a constant value to each.  This property applies to Linear Functions, which follows the format f(x)=ax+b.  Here is the example we did in class to show this property:

Because a constant value is added to both the inputs and outputs of this function to get the next value, it follows the Add-Add Property and is a Linear Function.

That’s about it for today’s class!  I hope this helped to reinforce what we learned today

First off, we have to understand the basics. A Power Function is defined as any function in the form y=Ax^b. In power functions, A?0 and B?0.Some examples would be X² and X³, or even X^¼ and X^¾ (assuming A=1) With that knowledge, you can begin to learn special properties that can be utilized with these functions. Let’s start with the basic behaviors of their symmetry.

Alright. By now you should be quite familiar with symmetry, but in case you need a refresher, here goes. Symmetry occurs when (in this case) one half of a graph is the “mirror image” of the other half. The line the two halves are mirrored off of is called the Axis of Symmetry. As an example, we can graph the power functions X¹,X²,X³ shown, respectively, below.

X^-1 or 1/(x^1)

Power Functions

Change in A-Values

To start off you first need to take a good look at your packet and learn a few simple definitions from your packet.  The first definition you should learn is what the heck a power function is.  A #Power Function is defined as any function in the form y=Ax^B.  In this equation A does not  equal zero and B is a non-zero interger. 

The next set of definitions that are necessary for looking at #Power Functions either on a graph or your calculator, have to deal with symmetry.  I would hope that everyone knows what symmetry is, but in case your drawing a blank it is when one half of a graph is the mirror image of the other half.  The line that acts as a mirror to that function is called the axis of symmetry.  Finally a graph that is symmetric with respect to the origin is one that for every x and y value on the graph the negative of the x and y values are also on the graph.  With this information you can start to identify some characteristics of different #Power Functions.  I hope that was just a review for you guys, but if it wasn’t then your welcome. 

Another way you can identify some characteristics about these functions is by looking at the end behavior of it.  In other words examining what happens to these functions as x becomes larger or smaller.  By graphing these functions you can get a better visual on what they look like to help you detirmine what happens to the functions as x gets bigger or smaller.  End behaviors are written like so: x (right arrow) infinitey sign (looks like a side ways 8).  This would mean that as x gets bigger the function also gets bigger or appraches infinity.  (Sorry if that was confusing, but I didn’t know how to put in an arrow or infinity sign.)  =(  

 In addition to the end behavior of #Power Functions you can also use the ladder of powers to help you see how increasing or decreasing the power of a function will affect the shape of the function.  Basically as the power increases the skinier your graph will be and as the power decreases the wider the graph will be.  If you want to see it for yourself try plugging some power functions into your calculator for example x^2 compared to x^4 and see what is different in the two graphs. 

The final piece to the puzzle of #Power Functions for now is rational Power Functions.  These are simply #Power Functions where the power x is taken to is a fraction(x^m/n) , which is equivilant to taking the nth (whatever n is) root of a #Power Function(the nth root of (x^m)).  In these expressions x>0 and m and n are intergers.  (Sorry if that was confusing but I couldn’t figure out how to make the square root sign.)  =(

Now you should be able to do all these things with #Power Functions.  I hope this helped anyone who was struggling with this stuff.  If your still hungry for more well you will just have to wait until next math class.  If you really need more help here’s a great video to check out! Power Functions    And if you just want to look at some more #dominoes here you go! Dominoes!      

#Dominoes  #Takingcareofbusiness  #PowerFunctions

Prove n is greater than or equal to 2, a(x1+x2+x3+…+xn)=ax1+ax2+ax3+…+axn 

n=2,   a(x1+ax2) =ax1=ax2 (distributive property)  TRUE!

Now we have to continue the problem by assuming n=k,

a(x1+x2+x3+..+xk)+ax1+ax2+ax3+…axk

The next step is to assume k+1,

a(x1+x2+x3+…+xk+xk+1)=ax1+ax2+ax3+…+axk+axk+1

The left part of the equation can also be writtten as a((x1+x2+x3+…+xk)+xk+1)

By rewriting the left side of the equation like this we can subsititute in when we assumed n=k.

ax1+ax2+ax3+…+axk+axk+1 = ax1+ax2+ax3+…+axk+axk+1 q.e.d.

I apologize for not being abke to figure out how to insert subscripts.  All of the numbers and k’s are supposed to be in subscript form.

After the warm-up proof by induction problem we moved to the main focus of the class which was to prove identities.  The previous day in class we went over all of the diffferent properties which included pythagorean, quotient, and reciprocal.

The proof we did today in class was sinx/1+cosx —>1-cosx/sinx    

To start the problem we multiplied sinx/1+cosx by 1-cosx/1-cosx.  We did this to obtain 1-cosx in the numerator just like the other one.

The result of the multiplication was sinx(1-cosx)/1-cos(squared)x

The 1-cos(squared)x translates to sin(squared)x

It now looks like sinx(1-cosx)/sin(squared)x. Since the demoninator is squared, written out it looks like sinx(1-cosx)/sinx times sinx.  SInce there is both a sinx in the numerator and in the denominator we can elminate one of them in the deonminator.

Therefore, the final is 1-cosx/sinx=1-cosx/sinx q.e.d.

If you need more help a great website for information is http://www.purplemath.com/modules/proving.htm. 

1. sin²x + cos²x = 1
You can also write property #1 in terms of sin²x or cos²x like so:
1a) sin²x = 1 – cos²x
1b) cos²x = 1 – sin²x

After we learned the first property, we then skipped to properties #4 and 5.  These two properties are called quotient properties.
4. tan(x) = sin(x)/cos(x)
5. cot(x) = cos(x)/(sin(x)

The last three properties are called the reciprocal properties because it is 1 over a number.
6. csc(x) = 1/sin(x), which can also be written as  sin(x) = 1/csc (x)
7. sec(x) = 1/cos(x), which can also be written as cos(x) = 1/sec(x)
8. cot(x) = 1/tan(x), which can also be written as tan(x) = 1/cot(x)

As you can see, we used all 6 of the trig functions within those properties. These properties are the ones that are very important and you should begin to memorize. We then went back to properties #2 and #3, which you do not need to memorize because they are the first equation, just divided by either cosine or sine.

2. (sin²x + cos²x)/(sin²x) = 1/sin²x
 We then simplified this by
sin²x/sin²x + cos²x/sin²x = 1/sin²x
the first term sin²x/sin²x is simply, and becomes 1
the second term seems more complex, but if you look at it, it is the same as cot²x
the third term, on the right side of the parentheses is the same as csc²x
so the simplified equation of this is:
2) 1 + cot²x = scs²x
This can also be written as:
2a) 1 = csc²x – cot²x
2b) cot²x = csc²x – 1

The third property is the same as #2, except you divide by cos²x.
3. (sin²x + cos²x)/(cos²x) = 1/cos²x
Again, we could simplify this equation by writing it out as:

sin²x/cos²x + cos²x/cos²x = 1/cos²x
The first term, again, looks complex, but it is actually just tan²x
The second term, is simple, and simplifies to 1
The third term is the same as sec²x
So the simplified equation for property #3 is,
tan²x + 1 = sec²x
this can also be written as:
3a) tan²x = sec²x -1
3b) 1 = sec²x – tan²x

All the these properties can be tied in with proof by induction, so we then did a problem at the end of the class.
The problem we were given was:
cotA + tan A ? scsA · secA
the first step we took was getting rid of cotangent and tangent and turning into something we are more familiar with:
cotA  ? cosA/sinA   and    tanA ? sinA/cosA    so the new equation is:
cosA/sinA + sinA/cosA
next, to be able to add the two terms, we got a common denominator of sinAcosA. so the new equation is:
cos² A/sinAcosA + sin²A/sinAcosA
to get the first term you had to multiple cosA/sinA by cosA/cosA
to get the second term you had to multiple sinA/cosA bu sinA/sinA
the next step is to add the two terms together to get:
cos²A + sin²A/sinAcosA
The numerator of the equation is something we know! It is property #1, so therefore we know it equals 1! This helps us out because then we can simplify that equation to get:
1/sinAcosA
This is also the same as:
1/sinA · 1/cosA
And!
1/sinA = cscA
And!
1/cosA = secA
Therefore we know cotA + tanA can turn into cscA · secA
Tombstone!

2+2²+3²+4²+…. 2^K = 2^(K+1) +2

2+2²+3²+4²… 2^K+2^(K+1) = 2^(K+2) – 2

2^(K+1) + 2^(K+1) = 2^(K+2)

2(2^K+1) = 2^(K+2) Q.E.D. #TOMBSTONE

#HAVINFUNYET ?

NEXT WE LEARNED THE #PYTHAGOREAN, #QUOTIENT, AND #RECIPROCAL PROPERTIES…

#PYTHAGOREAN…

1) SIN²X+COS²X=1   1A) SIN²X=1-COS²X       1B) COS²X=1-SIN²X

2)1+COT²X=CSC²X   2A) COT²X=CSC²X-1      2B) 1=CSC²X-COT²X

3)TAN²X+1=SEC²X    3A) TAN²X=SEC²X-1      3B) 1=SEC²X-TAN²X

#QUOTIENT …

4) TAN(X)=(SIN(X)) / (COS(X))

5) COT(X)= (COS(X)) / (SIN(X))

#RECIPROCAL …

6) SIN(X)=1 / (CSC(X))       6A) CSC(X)=1 /SIN(X)

7) COS(X)=1 / (SEC(X))       7A) SEC(X)=1 / COS(X)

TAN(X)=1 / COT(X)         8A) COT(X)=1 / TAN(X)

THANNNNNNNN…..

WE DID A PRACTICE PROBLEM! HERE IT IS AMIGOS!

COT(A) + TAN(A) =                  CSC(A) + SEC(A)

( COS(A)/SIN(A) ) + ( SIN(A)/COS(A) ) = ^^

( COS²A/(SIN(A)COS(A)) ) + ( SIN²A/(SIN(A)COS(A)) ) = ^^

1/( SIN (A)COS(A) )= ^^

1/SIN(A) x 1/COS(A) = CSC(A)SEC(A)

#RIGHTRIGHTRIGHT #YOUSTILLDOIT

1. Find what the equation equals.

2+4+6+8+…+2n=n²+n

2. Assume Pk (do this by changing all the n’s to k’s)

2+4+6+8+…+2k=k²+k

3. Prove Pk+1 is true (do this by adding this to the left side and changing all the k’s on the right to k+1)

2+4+6+8+…+2k+2(k+1)=(k+1)²+(k+1)

4 Use the asumption from step 2 to substitute for the “2+4+6+8+…+2k”

k²+k+2(k+1)=(k+1)²+(k+1)

5. Simplify using #algebra

k²+k+2(k+1)=(k+1)²+(k+1) equals

k²+3k+2=k²+3k+2 q.e.d.  Therefore, we know it is TRUE!

The second one we did was 1²+2²+3²+…+n²=.  You solve this one the same way we solved the first.

1. Find out what the equation equals (In this case it was given to us since it’s hard to find by just guess and check)

1²+2²+3²+…+n²=n(n+1)((2n+1)/6)

2. Assume Pk

1²+2²+3²+…+k²=k(k+1)((2k+1)/6)

3. Show that Pk+1 is true

1²+2²+3²+…+k²+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6)

4. Insert the assumption “n(n+1)((2n+1)/6)” from step 2 in for “1²+2²+3²+…+k²”

n(n+1)((2n+1)/6)+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6)

5. Simplify using #algebra

n(n+1)((2n+1)/6)+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6) equals

(k+1)((2k²+7k+6)/6)=(k+1)((2k²+7k+6)/6)

If you still need more help, you can check out the #KhanAcademy video about Proof by Induction

 Proving 1^2+2^2+3^2+…+2n=n(n+1)(2n+1/6)

First: Prove That the First Domino Falls

n=1

1^2=1(1+1)(2*1+1/6)

simplify

4=4 The first domino falls!

Second: Assume another domino falls

 1^2+2^2+3^2+…+2k=k(k+1)(2k+1/6)

 1^2+2^2+3^2+…+2(k+1)=(k+1)((k+1)+1)(2(k+1)+1/6)

Sence  1^2+2^2+3^2+…+2n=k(k+1)(2k+1/6) We can subsitute in the problem above

 (k)(k+1)(2k+1/6)+2(k+1)=(k+1)((k+1)+1)(2(k+1)+1/6)

(k)(k+1)(2K+1/6)+2(K+1)=(K+1)(k+2)(2k+3/6)

(k+1)(2k^2+k/6)+2k+2=(k+1)(2k^2+7k+5/6)

(k+1)(2k^2+7k+5/6)=(k+1)(2k^2+7k+5/6)QED Or Tombstone

What your favorite QED or Tombstone?

 Jets and Sharks you say? In future it will be the QED’s and Tombstone’s

http://www.youtube.com/embed/3YT9riTfzNM