Today in #bienprecal, we did some more solving of proof by induction equations. The first one we did was 2+4+6+8+…+2n=. The steps we used to solve this are as follows:

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

We dove right into class today. We review how to do the proof by induction. After we tried a problem we tried something a little more difficult.

 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

Well, ladies and gents, we started today with dominos again. We set up our patterns, spelling out “calc” and/or setting up straight lines. Some patterns fell over right away (success!); others, unfortunately, did not. However, it was these so-called “failures” that provided the greatest learning examples. After dominos, we were given our board problem:

1+3+5+7+ … +2n-1 = n^2

Where:

n : the number of terms on the left hand side that we use

2n-1 : how we generate the odd numbers

… : the infinate amount of odd numbers

Our statement to prove: If we add up any anount of odd numbers, that amount (n) is equal to that amount squared (n^2)

How are we supposed to prove this is true? We can’t add up odd numbers to infinity! This is where the dominos analogy comes in. Remember: Dominos are to proofs as Soda Machines are to functions.

First, let’s think about how the domino setups failed:

1. There is a domino somewhere that falls, but it doesn’t push over the one in front of it

2. The first domino never falls/gets pushed

Okay, now that we know how we can “fail” at dominos, let’s look at the information again in light of Proofs by Induction. We will often be refering back to the main number sequence above:

1+3+5+7+ … +2n-1 = n^2

1. Make sure the first domino falls

• Check and make sure that the first term on the left hand side is equal to the right hand side
• E.g. n=1; 1=1^2; True! First domino falls (first term is correct/true)

2. Assume another domino falls…

• We call this generic domino P (subscript)k; P stands for the big number sequence, k stands for the generic domino
• We don’t use n because n is what we’re trying to prove; rather, the sequence works for any n value
• However, we can plug in k (our generic number/domino) for n (our any value)

• 1+3+5+7+ … +2k-1 =k^2

3. Now, check and see if the next domino falls

• If our first generic domino is k, our second one is k+1
• 1+3+5+7+ … +2k-1+2(k+1)-1 =(k+1)^2

4. Solve algebraically

Okay, what? We’ve got two crazy equations: 1+3+5+7+ … +2k-1+2(k+1)-1 =(k+1)^2 and 1+3+5+7+ … +2k-1 =k^2 ! How are we supposed to solve anything? We’ve got that crazy “…” representing infinity! How do you algebraically solve infinity?

Wait a second, both equations have things in common! Look:

1+3+5+7+ … +2k-1        +2(k+1)-1 =(k+1)^2

1+3+5+7+ … +2k-1         =k^2

That really long bit with infinity is really equal to k^2. So, replacing that long numerical jargon, we get a condenced equation:

k^2+2(k+1)-1 =(k+1)^2

This can easily be simplified to the true statement:

k^2+2k+1 =k^2+2k+1 Q.E.D.

Don’t forget, the Q.E.D. is one way to end a true proof; the other way is to draw the tombstone.

Proof by Induction: Conquered!

For proof by induction, you  start with an equation where the sum of odd numbers  (2n-1) = n^2   in  1+3+5+….

Then there  are things to proof by induction that must be true in order to prove that this equation works for everything. The first number must work & that every other number must work.

To prove this there are 4 steps: 

1. the first “domino” has to fall, which means that the first step that you are trying to prove has to be true.                      example:  the first number is 1 so you start with that 2(1) – 1 =1^2  which is 1=1 so this step is true

Now that this step is proved to be true you may move onto step number two

2.  For this we assume that any other “domino” falls(symbol for this is Pk ) so 1+3+5+…..+2k-1=k^2

Next the question we try to prove is that this knows down the next “domino”(symbol is Pk+1):

3.  1+3+5+…+2k-1+2(k+1)-1=(k+1)^2    (K+1) is the next term in the sequence.

When writing this out you realize that you can subsitute 1+3+5+….+2k-1 for k^2 since that is what it is equal to in step 2. This now simplifies the equation to k^2+2(k+1)-1=(k+1)^2 The next step is basic algebra to see if these equation actually  each other. *It must stay in a general term because if you solve for one number then you only prove that specific number not every number possible*

4. k^2+2(k+1)-1=(k+1)^2 …. k^2+2k+2-1=(k+1)^2….k^2+2k+1=(k+1)^2…k^2+2k+1=k^2+2k+1 q.e.d.                                   after basic algebra you find out that these two equations are actually equal to each other proving this certain statement true. *When finished you should put q.e.d. or a rectangle tombstone to show that this is finished and proven true.

This link has a great video that shows another example: Little proof by induction

With the dawn of a new semester, the class was introduced to the blog today. We were told the basic rules, such as posting an important title and filling the post with information learned in class.
Besides the website we also started a project involving dominoes. First we watched a few really cool videos of domino chains and even a chain of pound coins from the UK. These videos remind me of a music video in which the band set everything up perfectly to make a huge chain of events all shot in one clip:

In our project, each group of four was assigned to create a chain of dominoes using the fifty or so dominoes given to us. The system could be formed in any shape and can even involve different elevations. We were given pretty much complete freedom to use our creativity and make a test run for our domino chain. Tomorrow we will go through the real runs, so hopefully everyone has their project ready to go!

Today was our first day of the new semester.  In pre-calculus class we watched a few videos of epic domino sequences. Below is another cool one:

We then split up into groups to create our own mini versions of these domino sequences.  My group is going to create a sinusoid shape out of the 58 dominos we were given.  We considered using the textbooks to create stairs for the dominos to climb and descend, but decided that it would be too difficult to do with so few dominos and in such a short period of time.  I am excited to present our domino sequence and interested to see what the other groups came up with.  It would be really cool if we all our ideas together to create a mega domino set up like they show in the videos we watched.  I’m interested to see how this domino activity ties into pre-calculus.  I’m thinking maybe because each concept leads to another which leads to another, and so on.  I guess we will have to see where this takes us.

Equation to look of graphs:

• Even exponent: Both ends point the same direction
• Odd exponent: Each end points a different direction
• Positive coefficient: as x??, p(x) ??
• Negative coefficient: as x??, p(x) ?-?
• Turns = degree – 1
• Branches = zeros =      degree
• A zero of multiplicity bigger than 2 flattens the graph out
• Greater the exponent the more it flattens out
• Multiplicity of root=1 will go through x-axis
• Multiplicity of root = 2 will touch x-axis
• Find the y intercept for a more accurate graph

To show the different details we learned I’ll use an example.

f(x) = -(x + 2)(2x – 1)³(x – 3)²(x +6)

degree = 7 (greatest exponent or for multiplying factors- how many factors there are)

multiplicity: -(x + 2)(2x – 1)(2x – 1)(2x – 1)(x – 3)(x – 3)(x + 6)

zeros: -2, ½, ½, ½, 3, 3, -6 (what makes each factor = 0)

y-intercept: 108 (found by putting 0 in for each x and then solving)

Local behavior- How a function behaves when the graph is zoomed in enough to see the x and y intercepts and the discontinuities/vertical asymptote

Global behavior- How the function behaves when the graph is zoomed way out, and horizontal asymptote

How do we know if we have a good model?

1.      Interpolation: Put points in your model and check how well they match the true point

2.      Extrapolate: Go outside the data set and check how well it fits the model

3.      See how close the “r” value is to 1 (the closer to 1, the better)

4.      Look at it, does the model fit a majority of the points

5.      Look at endpoint behavior

6.      Look at the residual plot and if the points are random or in a pattern (random is good, meaning that the model accounts for all sources of error)

How to do the math using your calculator:

For reach point above, you can input it into your calculator under the stat plot lists as follows—

L₃ = deviations² = (average (L₂) – L₂)²

L₄ = residuals² = (y₁(L₁) – L₂)²

r² = SL₃ – SL₄ (S represents the sum of that list)

SL₃

Now I’ll do the bluegill problem off of an assessment we did.

(Enter this data into your calculator to produce a graph.  Then using the graph and calculations you can analyze the model.  My analysis of the model is below)

As seen on the graph, the logistics function is the best fitting model for the data. Although ln could also be a good model, logistics is better for a number of reasons.  First off, logistics has a r value of .9997, while ln has an r value of .9966; also, the other regressions r values were worse than both of these (power-.9848 and linear-.9775).  Another reason to rule out power and linear is that the model misses a majority of the points while the other two hit a majority of the points; the data is more curved then the power and linear functions.  So after that step I decided to further look into only logistics and ln. When looking at the graph of the residuals both regressions are random and have no pattern, showing that both models account for all sources of error.  For interpolation, I randomly chose the point (68,95). Logistics gave me the point (68, 95.5), and ln gave me (68, 98.1); so logistics is better for interpolation.  Then I look at extrapolation and can see that logistics converges to 177.9, while ln continues to grow.  Converging makes since because bluegills don’t grow forever, so although converging is still growing a tiny amount, that is better then ln which continues to grow more rapidly for the whole graph.  Also, ln starts off with a negative total length for any initial length less then 21.83, and at about 21.84 initial length, there is no growth total growth.  Where as if you look at the logistics graph, there is no negative lengths, and it starts off at 0 initial length with a total growth after one year of 24.86; which makes since fish start off as an egg and grow.  After making these evaluations, I can say logistics the best model for this set of data.