Today we continued working on finding sums by using Sigma Notation.
First we did our usual POD, which was a review for the final exam:
Find in Radians all 1st revolution angles where the cotθ=-√3. No Calculators or Snowman sheet.
In this problem, the first thing we should do is identify what the cotangent of an angle means. Cotangent is the reciprocal of tangent. So if, tanθ=y/x, then cotθ=x/y.
In this case, we are given that cotθ=-√3, which equals -√3/1; where x equals positive or negative √3 and y equals positive or negative 1.
If we graph this we can see that this angle will be in either the second or the fourth quadrant because it is a negative angle.
We first found the angle of the second quadrant; we know what the x and y values are, so we just draw how the triangle would look like:
These measures tell us that it is a 30-60-90 triangle, and we can find which angle is which by comparing it to the length of the leg, the 30 degrees are across the short leg of the triangle.
To find the first angle we need, we substract from 180º the 30º of the small triangle.
180-30=150º
Since we are asked to find the measure in radians, we convert 150º to radians by multiplying it by πradians/180º.
150 x πradians/180º = 5π/6
Now that we have our first angle, we still have look for the angle in the fourth quadrant. We begin by making a drawing.
We know that a full revolution measures 360º, and that the angle, or the small part, we are missing to complete that full revolution measures only 30º. So to find the fourth quadrant angle we just substract 30 from 360º.
360º - 30º= 330º
We convert that value to radians...
330º x πradians/180º = 11π/6
Next, we continued our lesson on Sigma Notation.
We were first asked to find the sum of 3 + 6 + 9 ... + 81 by using sigma notation.
We realized that this was an arithmetic series because we added 3 to each term to get the next one; also that it was a finite series, because its last term is 81.
However, we also realized that each term was equal to the number of the term times 3, as it is shown below:
Next... we were told that the explit formula in sigma notation was:
27
Σ 3i
i = 1
To get there we would have had to solve for the summand by using the definition of an arithmetic sequence, as follows:
tn = t1 + (n-1)d
tn = 3 + (n-1)3 plug ti which is 3, and distance between terms which is 3
tn = 3 + 3n - 3 Apply distributive property
tn = 3n 3s cancel out
Then we have to find the limits of the summation. We know that it starts in term 1, so we have to find the number of the last term in the series, knowing that tn = 81. We now use the definition for an arithmetic sequence to solve for n.
tn = 3 + (n-1) 3 = 81
3n = 81
n = 27. The limits of the summation is 27.
We now remmember the first property of sums, which is:
n n
Σ ca = c Σ a
i = 1 i = 1
and apply it to our explicit formula:
27 27
Σ 3i = 3 Σ i
i=1 i = 1
With this, the series becomes (1 + 2 + 3 + .... + 27) times 3.
So if we apply Gauss's formula: n/2 (n+1) , multiplied by 3, we will end up by having
S27 = 3 (27/2 (27 + 1)) = 3 (13.5 (28)) = 1,134
The next problem we solved was:
Find the sum of the first 30 multiples of 7.
In order for us to set up the explicit formula in sigma notation, we have to identify the diferent parts of the sigma... just by looking at the question, we identify the limits of the summation, which are 1 and 30. We also notice that it is as arithmetic series, and that each term is the number of the term multiplied by 7.
The series would then be: 7(1) + 7(2) + 7(3) + .... + 7(30)
giving us the summand which 7n
We write the explicit formula as:
30 30
Σ 7n = 7 Σ n
n=1 n=1
and the series as 7 (1 + 2 + 3 .... + 30)
And we solve by using Gauss's formula times 7...
S30 = 7 ((30/2)(30+1))
S30 = 7 (15 x 31)
S30 = 3,255
Next, we learned about the second property of sums which actually applies to both addition and subtraction...
Addition = The sum of a sum is the sum of the sums.
n n n
Σ a + b = Σ a + Σ b
i=1 i=1 i=1
Subtraction = The sum of a difference is the difference of the sums.
n n n
Σ a - b = Σ a - Σ b
i=1 i=1 i=1
Now that we knew this property, we could solve for a little more complex sums....
The first example we tried was:
Find 12
Σ 5k - k^2
i=1
By using the property we just learned, we separated the diference to the diference of two sums...
12 12
Σ 5k - Σ k^2
i=1 i=1
To the first sum we can apply the first property we learned:
12 12
5 Σ k - Σ k^2
i=1 i=1
We still do not have the skills to solve the second sum of the equation that involves a variable raised to a power, but we can solve for the first sum:
12
5 Σ k
i=1
This equals to 5(1 + 2 + 3 + ... + 12)
We use Gauss's Formula times 5
S12 = 5 ((12/2)(12+1))
S12 = 5 (6 x 13)
S12 = 390
Knowing this, we can plug it to the formula and get our answer...
12
390 + Σ k^2
i=1
Next, having improved our skills, we could go back to the series 3 + 2.4 + 1.92 + 1.536 + ... and solve it...
We had already solved this problem last week by using the definition for a geometric series, and got an answer of approx. 5.87.
We also knew that explicit formula in sigma notation, which is:
18
Σ 3 (.8^K-1)
K=5
First we can apply our first property of sums to get
18
3 Σ (.8^K-1)
K=5
Now, since we only want the sum of the terms 5 to 18, and not of terms 1 to 4, we can find for the total sum and subtract the value we do not want....
18 4
3 {Σ (.8^K-1) - Σ (.8^K-1)}
k=1 k=1
We now substitute each sum with the definition of Geometric Series for each, using 1 as t1 and .8 as r. The formula for a Geometric series is ti ((1-r^n)/(1-r)).
3 {1 x ((1-.8^18)/(1-.8)) - 1 x ((1-.8^4)/(1-.8))}
We solve for these values by using our calculators and it gave us:
3 (4.9099 - 2.952) = 5.87 (approx)
Which is the same answer we got last time....
The last thing we did was begin solving the following problem...
20 20
Σ 2n = 2 Σ n
n=10 n=1
We used then the second property of sums and the last problem as an example to write the equation as:
20 9
2 {Σ n - Σ n}
n=1 n=1
And that is as far as we went in the class, we will continue solving that problem tomorrow wednesday in class...
There is no new homework on the board, but the one from Monday is still up: read pgs. 506-507 and solve #1-15 odd in pg. 508.
Next scribe Caro Ibañez and Fio.
(Gaby wanted a picture alone w/ Mr. A. and in the few seconds between the camera's button was pushed and the flash went on, 5 other people where already in the picture, although mari's cloud world was a little bit higher than the rest so she took longer to pose. Say "CHEESE" Mr. A!)
