Today I have a question inolving sub-sequences and recursively defined squences. I really like these questions, since when you solve it everything just fits in perfectly. Watch it here:
A sequence an is defined as:
a1 = 11
an + 1 = -0.5an + 4.5
bn is defined for natural n's as bn = an - 3
a) Prove that bn is a geometric sequence
b) Find the sum of all even positioned terms of bn.
Let's start with a, of course.
A sequence is a geometric sequence if the ratio of every two consecutive terms is constant. So, for our needs, we can show that bn + 1/bn is a constant, which proves that bn is a geometric sequence.
By definition:
bn = an - 3
bn + 1 = an + 1 - 3
Using the definition for an and an + 1:
bn = an - 3
bn + 1 = -0.5an + 4.5 - 3 = -0.5an + 1.5
Divide both terms:
(-0.5an + 1.5)/(an - 3) = -0.5(an - 3)/(an - 3) = -0.5
The quotient of two consecutive terms is a constant, so bn is a geometric sequence.
b) Since the ratio of consecutive terms in bn is -0.5, the ratio of consecutive even-positioned terms is that ratio squared, meaning (-0.5)2 = 0.25. Now that we have that quotient, we can find the sum of those even-positioned terms.
The formula for the sum of an infinite converging geometric series is a1/(1 - q). We have q, and now we need a1. In this case, the first term (which is given the symbol a1) is b2, since it's the first even-positioned term. Let's find it:
b2 = a2 + 4.5 - 3
Using the recursive definition:
a1 + 1 = -0.5a1 + 4.5
a2 = -0.5 * 11 + 4.5 - 3 = -1
So:
b2 = -1 - 3 = -4
Plug it in the formula:
-4/(1 - 0.25) = -4/0.75 = -4/(3/4) = -16/3
The sum of all even-positioned terms of bn is -16/3.
If you need math help in anything, go to Super Math Tips, sign up, and you can send any questions you want. They might be answered on this blog!
Hope you like it,
Nadav
nadavs
Friday, June 27, 2008
Geometric Sub-Sequence
Thursday, June 19, 2008
Euler And Series
Yahoo Answers is a goldmine. There are a lot of great questions in there. Today I have one relating to series and Euler. Yes, this guy again. He always comes back for more.
A sequence an is defined as 1/n2(n + 1)2. Find the sum of the series from 1 to infinity.
Remember Euler from before? He proved that the sum of 1/n2 from 1 to infinity is π2/6. Yes, Euler was a smart man. All we need now is to transofrm an to something with 1/n2 which includes sums. But how?
Let's start with the obvious and proceed from there. To create that denominator, we can try to add 1/n2 and 1/(n + 1)2, see what we get, and then decide what else to do. So:
1/n2 + 1/(n + 1)2 = (n + 1)2/n2(n + 1)2 + n2/n2(n + 1)2
Which is:
(n2 + (n + 1)2)/n2(n + 1)2
Open parentheses and add:
(n2 + n2 + 2n + 1)/n2(n + 1)2
(2n2 + 2n + 1)/n2(n + 1)2
To leave only the 1 in the numerator (which will give an, we need to subtract (2n2 + 2n)/n2(n + 1)2:
an = (2n2 + 2n + 1)/n2(n + 1)2 - (2n2 + 2n)/n2(n + 1)2
The second term can be also written as 2n(n + 1), which can be cancelled. At the end, we get:
an = (2n2 + 2n + 1)/n2(n + 1)2 - 2/n(n + 1)
2/n(n + 1) can also be written as (2n + 2 - 2n)/n(n + 1), which can be transformed into:
(2n + 2)/n(n + 1) - 2n/n(n + 1) = 2/n - 2/(n + 1)
Since it's the opposite on an, we can finally conclude that:
an = 1/n2 + 1/(n + 1)2 + 2/(n + 1) - 2/n
Now it's really easy. Notice how 2/(n + 1) and -2/n cancel each other, except for the first term. Their sum is -2/1 + 2/2 - 2/2 + 2/3 - 2/3 + 2/4 - 2/4 + ... . Since 2/∞ is zero, the only remainder of this part of the series is -2.
As Euler said, the sum of the 1/n2 part of the series is π2/6. The 1/(n + 1)2 has the exact same sum, but minus one. Since the sums starts from 1/22 and not 1/12, the 1/12 is left outside.
The total sum of an from 1 to infinity is:
π2/6 + π2/6 - 1 - 2
S = π2/3 - 3
Since the terms of an get very small very quickly, it can be easily approximated for validity.
Hope you like it,
Nadav
nadavs
Thursday, May 29, 2008
Cubic Sequences
Today I have something similar to the cubic function discovery I did exactly two weeks ago. It's somewhat more complex, and therefore more fun. Long? Yes. Fun? Oh, yeah!
Find the nth term: 2, 8, 20, 40, 70, 112, 168, ...
Once again, the genre of "short question, super long answer". Those are the most fun, so let the fun begin.
When you get a sequence and you need to find the nth term, go for differences between numbers. Those differences hide a big secret within them. If the differences are the same, it's a simple arithmetic series. If the differences are not the same, do the differences of the differences. If now they're equal, the relation between the numbers goes through a quadratic expression. If the differences are still not the same, take the differences again and again until they are equal. The number of times you take the differences is the degree of the expression that connects the numbers.
If you see that the differences have a certain ratio to them (for example, try to find the differences for 4, 10, 28, 82, 244, 730), don't bother going more than once or twice. You are dealing with an exponential relation here.
So, let's find the differences of the sequence we're given:
2, 8, 20, 40, 70, 112, 168
6 12 20 30 42 56
6 8 10 12 14
2 2 2 2
Jackpot. We reached equal differences after three times, so the numbers are connected via a cubic expression. To find that expression, we need to set up a system of 4 equations with 4 variables (if you don't know or don't remember, open the link at the top of the post).
Let's call the points that this "function" goes through (1, 2) (2, 8) (3, 20) (4, 40). Now let's set up the system:
a + b + c + d = 2
8a + 4b + 2c + d = 8
27a + 9b + 3c + d = 20
64a + 16b + 4c + d = 40
Subtract the equations from each other to eliminate d:
7a + 3b + c = 6
19a + 5b + c = 12
37a + 7b + c = 20
Subtract them again to eliminate c:
12a + 2b = 6
18a + 2b = 8
And once again:
6a = 2
a = 1/3
Let's start plugging:
12/3 + 2b = 6
4 + 2b = 6
2b = 2
b = 1
Again:
7/3 + 3 + c = 6
c = 2/3
Finally:
1/3 + 1 + 2/3 + d = 2
d = 0
Well, that makes it easier. The nth term is given by:
n3/3 + n2 + 2n/3
Go ahead, plug your number for n and see that it works. For all terms, even the ones which were not included in the calculation.
Enjoy your new knowledge,
Nadav
nadavs
