Precal Blog

Yes, today was the first day of a long fourth quarter; exciting, right? We’re almost there guys, about 9 weeks left….anyways, precalculus. Today we started off the quarter talking about sequences, in exploration 14-2a. This worksheet had us use previous knowledge of sequences to state future terms by discovering a pattern in a sequence of numbers. We learned that an arithmetic sequence progresses by adding a constant to the preceding term to get the next term. For example: {5,10,15}, the next term would be 5 because all you’re doing is adding five to the previous term. We also learned that a geometric sequence progresses by multiplying the preceding term by a constant to get the next term. For example: {5,10,20} the next term would be 40 because you’re multiplying the previous term by 2.
The Recursion Formula
Next, we were given a sequence of numbers:
{5,7,9,11,13,15,17,21,…,133,…}
From this sequence, we were given the common difference [d], which is the constant added to an arithmetic sequence to identify the next term, which is obviously 2. (The common difference is also considered the slope.) We then figured out that T subscript n would be the equivalent of y, and T subscript n-1 would be the previous term.
We were then able to write our Recursion Formula:
d=t(n) -t(n-1)
or if you want to write that to have it equal t(n):
t(n)=t(n-1)+d
To generalize this equation we just use one of the basic explicit formula:
y = y(1) + m (x-x(1))
and we then switch around a few terms to get it into ’sequence’ form:
t(n) = t(1) + d (n-1)

YAY! not too bad, but I have a feeling deep down that there’s going to be some complicated twists coming up. Tomorrow’s scribe is GRANT! Have fun!

Yeah. So coming back to school on a Monday. After spring break…which I spent in Spain. On a rainy day of all things. Was EXTREMELY disappointing if I do say so myself. Right, well math. Today calss was pretty decent. We spent the first half of the class working on exploration 14-2a, which dealt with patterns in sequences. After class was half over, we came back together as a whole to discuss stuff…all the expansion stuff no one ever likes. Okay so first we talked about The Recursion Formula. Beginning the discussion on this formula we were given a whole new sequence of numbers:

{5,7,9,11,13,15,17,19,21,…133,…}

With this, we were told the common difference (the constant added to a term of an arithmetic sequence to get the next term) of these numbers was what we knew to be as the slope. To find the Recursion Formula we decided d would equal e common difference. Also T subscript n would be the term we are dealing with, and T subscript n-1 would be the previous term. With all these variables we were able to come up with the Recursion Formula:

Or, solving for T subscript n:

In words, the Recursion formula says To find a term, take the revious term and add the comon difference. This formula works for smaller n values. Say we want to find what the 100th term would be. We obviously are not going to go through all 99 terms just to find the 100th. We need a more general equation. This general equation we are solving for is known as the explicit equation.

To find the explicit equation, we started out with the geeral, transformed equation of a line:

Putting this in terms of T and n, we get the final Explicit Equation, which we can use to find any term just knowing its term number:

Yeah well that was class. Pretty simple. So I suppose tomorrow’s scribe will be… KaitlynP. Have fun [=

Hey its my turn to blog apparently, so here are the yellow sheets due at the end of the quarter:

A-2

7-2

11-5

11-4

11-3

11-2

And today we took a test. So ya.

So, I know it is the weekend, and its really not my turn to blog…but I had a few ideas of how we could imrove the blog and make it even more useful then it already is.

I was looking through past comments, and it seems that before every quarter journal students are asking what yellow sheets are we responsible for. I think that we should make a list, just like the scribe list tab, but for yellow sheets. This way students can always keep track, and they exactly where they can find the list.

Also I have found that using the blog is a great way to study and review before a test. I think it would be to the advantage to all that we put up a list for the learning targets as well. Then students could post coments to that list, like give example problems that would help the students better practice the learning targets. It could kinda of be like a spot to gather the best problems from all the claims and categorize them in a way that makes studying each learning target simpler.

Anyway these are just a couple of ideas to improve our class blog. If anyone has any other ideas of how we could make it better…please comment!

Here are the slides on the problems we did today.

| View | Upload your own

I saw the most trouble with change of base. Let me give you the way to think about it one more time:

Say we are converting from base-2 to base-10. Look (click) at the diagram below and it should be clear that the unit distance (the distance between 0 and 1) on the base-10 scale is about 3 times as much as the unit distance on the base-2 scale.

So, if we knew exactly how many base-2 units it took to make a base-10 unit then we would have a conversion factor. Look closely again at the diagram. You should see that the base-10 scale is exactly `(log_10(10))/(log_10(2))` times as big. This, then is our conversion factor.

If the bases were equal our conversion would be `log_2(x)=log_10(x)`.

They are not equal so we apply the conversion factor and get: `log_2(x)= (log_10(10))/(log_10(2))*log_10(x)`.

This simplifies to give us the typical change of base formula: `log_2(x)=(log_10(x))/(log_10(2))`

This argument can be applied to the change of any base to another.

Here is a second area of trouble: Everyone is comfortable with `10^x=10000`. I think everyone is also ok with `log_10(10000)=x`. The first equation asks “What is the exponent I raise 10 to in order to get 10000″? The second equation asks the same question right? So when you see `log_3(x+1)=4` the question is “What is the exponent I raise 3 to in order to get `x+1`”? Write the question as an exponential equation: `3^a=(x+1)`. You know the answer because the logarithmic equation has a number after the equal sign. The logarithm(base 3 of `x+1`)is the exponent(4). Therefore, `3^4=x+1`.

A few of you asked for a printout of the last slide we looked at today, which I will do for you, but I also thought putting up some summary slides of what we have done may help you study and make sense of all the logarithm stuff we have been discussing.

There really isn’t an easy way out – it will take some time and meditation to understand this difficult concept.

| View | Upload your own

If you want to view in full screen, hold down Shift and click the View link. This will open a new window and then click the Full icon to the right of the arrows.

Well, here we go, my second blog. Today, we got claims as promised. Steph started us out by presenting number one. She was responsible for C, F, and L. For C, she explained that because 1/1000 is the same as 1/10^4, it was the same as 10^-4. Since we already knew that logarithms are exponents, the answer was -4. For F and L, the answers were basically given to you in the exponents (F was 16 and L was 35). Next, we had Hans shock us with his mathematical genius during number 2. For number two, we were given numbers and asked to calculate what power for ten gets us there. Hans explained how this was the opposite of Steph’s problem. So, for part A, Hans explained that -2 would be the exponent of ten to get you to five. Tony went next and demonstrated the skill he has acquired by solving problem 3 for us. Tony started by drawing the “Powers of Ten” scale and then how he used that for his estimates. His method got him very close to the calculated answers. Mr. B asked if anyone had done it a different way and Natalie raised her hand. Natalie had rounded 1956 to 2000 and then “multiplied” 2 and 1000 (which we know is three because 10^3 gets 1000). To estimate .117, Natalie used .2. We debated a bit about how to better solve this with a closer decimal, but decided that this was the closest we could get to. The next person to presented did an amazing job and used such eloquent mathematics to solve number 5. I showed how to do D, E, and F of number five. This dealt with the three forms we learned on Friday, with addition/multiplication, division/subtraction, and exponents. For part D, you need the log of 32, which can be gotten by taking the log of 2^5 because that equals 32. That is the same as 5log2. For part E, you can take the log(25/2) to get log12.5. When you divide you are really subtracting. I used the log of 25, which you get in part C of the problem, but you can also do it by taking 2log5. Then you would have 2log5-log2. The last part of my problem wanted to show 1000 using only log2 and log5. I knew that it would be 3, but I would have to prove that. So, I took 3log5 (because five to the third is 125) and 3log2 (which is 8). 125 x8= 1000, so I knew that I could add the logs. That gives us three. Hooray. No one claimed problem 7, so Mr. B took us through part B. Tony started us off by saying that 3^4=81. After being stuck for awhile, Nick said that we should apply logs to both sides, rewriting the problem as 4log3=log81. After that breakthrough, we were stuck again. Kari then spouted something off about multiplying some numbers that didn’t make 80, but her ideas lead Justin to come up with the correct multiplication of log10+log8, to get 80. Our problem said 80 and 81 were about equal, so we had solved the problem.

For the last 20 minutes of class we were given a new packet to read and work on. This was all about using a different base, specifically base two. We made a comparison between the Power of 10 line and the Power of 2 line. Kari pointed out that the separtations were equidistant, which makes it easier to find points that are not exactly on the line. We went through a couple of the problems together, but at the end of the lesson, one key point remained. All of the things that are true for base 10 are true for all other bases. Tomorrow’s scribe, once again, will be Phil.

Due to the fact that Mr B. was at a meeting during 4th hour, our class did a sketchpad activity. I noticed that few…no, change that…..very few people actually worked on the activity. But no matter, I’m sure that for some, this might have actually made the idea of Logarithms even more confusing (like me…). I got to the second page and then everything started to go down hill. The main idea, of the problems that I did, was that it is not possible to have a negative or zero y-coordinate. The y-coordinate can never reach 0 because the number decreases from .1 to .01 to .001, etc. In a way, 0 acts like an asymptote. Anyways, good night everybody…

and Ethan, you can be the scribe for tomorrow.

Today we went to the computer lab to work on the Logarithmic scale, semilog graphs, exponential eguations and more which I did not get to.

THE LOGARITHMIC SCALE

1. It is not possible to have a negative or zero y-coordinate on a logarithmic scale because [tag]logarithms[/tag] are exponents and no exponent can get 0 or negative numbers. Anyhing to the 0 power is one and anytihng to the negative power is less than 1 but greater than 0.
2. The relationshipship between the y coordinates of the point A and its dialation is the y-coordinate is squared 10^2 * 10^2 = 10^4.
3. The points on the logarithmic scale are related by multiplying the power numbers.

In conclusion from this section we learned that adding the lengths is the same as multiplying the power numbers. This was a review of what Mr. B told us yesterday.

SEMILOG GRAPHS

Semilog graphs are different. The x axis shows the exponent numbers and the y axis displays the powers(10^1,10^2,…). The intervals on the y axis are uneven because of ratios. This makes a parameter of x=1 and y=2v(both increasing at the same rate) a concave downlow sloped graph.

EXPONENTIAL EQUATIONS

Exponential equations on semigraghs from a straight linear shape. This is because of the y axises powers and the intervals in the graph.

This is as far as I get and I don’t fully understand. And I am not positive it is all correct, but I am pretty sure it is. Hope it helped you a little.

The next scribe will be Shelby.

Well okay Friday was the day of the blood drive, so we were missing some of the class. Regardless, we discussed more in-depth the concept of logarithms and how they relate using the exponential and power function’s linear scale. I am still trying to understand them and how to use them properly, because I find them to be very difficult, so I will not try to explain how to do them on here because I don’t know how myself. But I will tell you that claims are problems 1-9 in the packet (he will decided specifically which ones out of those 9 are actuals for Mon.) so you should definitely look your stuff over before class on Mon. If anyone is good with logarithms help me out because I’m trying to understand them. Next scribe is Jon.