Alrighty so GOOO SAINTSS.... Im not really a big fan but im happy for them for making history!!! Okay well last week we learned Infinite sequences and series and Sums of Infinite Series.
13-4 Infinite Sequences and Series
*lim
n-infinity: if the degree of the top= the degree of the bottom, then the answer is the coefficients.
Example:
lim n^2+1/2n^2-3n = 1/2
n-infinity
*lim
n-infinity: if the degree of the top is > the degree of the bottom, then your answer is infinity
Example:
lim 7n^3/4n^2-5 = infinity
n-infinity
*lim
n-infinity: if the degree of the top is < the degree of the bottom, then your answer is 0
Example:
lim 5n^2/3n^3+7 = 0
n-infinity
***If no rules apply, then you have to use your calculator to find what the sequence is approaching.
13-5 Sums of Infinite Series
*they can only be found with a geometric serires where /r/<1
Formula: S= t1/1-r
Example: 9-6+4
r= -6/9= -2/3 geometric
/-2/3/<1
S= 9/(1-(-2/3))= 27/5
Example: Write .45 repeating as a fraction
45/100-1= 45/99= 5/11
What i could use some help with is recursive definitions and sigma notation!! Thanks :)
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recursive definitions are not as complicated as they seem
ReplyDeleteyou mainly need to remember that
an-1 or tn-1 is a variable that stands for "previous term"
therefore An-2 or Tn-2 means the term previous to the "previous term"
so when trying to find a sequence take things one at a time
if a problem gives you t1= 3 t2= 6 t3=? t4=?
if tn= Tn-1 + Tn-2
all you have to do is plug in
T3= 6 + 3
therefore t3= 9
the t3 is the term you are looking for
the six came from t2 which is the term previous to the one you were looking for and the three came from the term previous to the term that is previous to the one you were looking for
to find T4 you do the same but remember the plug ins will be different each time you are looking for the next term of the sequence because An-1 or Tn-1 is not a set term it is dependant on what you are looking for and what is before what you are looking for
to evaluate Sigma,
ReplyDeleteFirst, define your limits of summation and your summand.
Then, plug in numbers of the limits into your summand.
Last, generate a series.
So if the limits of summation are 2&5 and the summand is 3h+1, you would plug in the numbers 2,3,4,&5.
It would generate the series
--7,10,13,16
Recursive Definitions are just equations using previous terms. A(n-1) is said PREVIOUS TERM. A(n-2) is the 2prior to the one your looking for.
Terrio also has a good explanation under taylor's blog for sigma notation.
The limit approaches the number but never reaches it and there are a few rules when dealing with limits:
ReplyDelete1. if the degree of the top number=degree of the bottom number then the answer is the coefficients
2. if the degree of the top is > the degree of the bottom then it = infinity
3. if the degree of the top is < the degree of the bottom then it = 0