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

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!

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

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!

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.

• 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