In today’s class we continued working on the topic we started on Monday, Sigma Notation. We first did our daily POD, which can help us to study for the exam on Friday and the final.
You must multiply every term by 1/x3
After multiplying them you are left with:
Ex1: 3/x (The denominator will go bigger and bigger, making the complete fraction very small)
3
Ex2: -8/x (Always when the variable is as the denominator, the fraction will go to cero)
Another property; its the same of Gauss, but squaring n+1 and dividing by 6.
Practice summation problems with the formula shown:
POD:
Find: Show your work. No calculators.
The first thing you have to do is to eliminate the highest exponent from the denomitator by multiplying the whole equation by the highest power in the denominator, in this case by 3.
You must multiply every term by 1/x3
After multiplying them you are left with:
When you plug in negative infinity in X, the fractions that have the X as the denominator are going to cero, so you must cancel them. The number of this fractions will be so small that they are countless.
Ex1: 3/x (The denominator will go bigger and bigger, making the complete fraction very small)
3
Ex2: -8/x (Always when the variable is as the denominator, the fraction will go to cero)
Therefor you are left with:
*as you notice, when you plugg in negative infinitive to the X value, the answer will be negative infinity.
SOMETIMES PEOPLE MAKE MISTAKES WRITING LIMITS, USUALLY WITH BRAKETS PLACES.
·Never equal the lim of # as n goes to infinitive with the rest of the equation. This is because you are taking the limit of the equation, not telling that this limit equals the equation. For example:
Continuing yesterday’s topic of summations:
You need to remember Gauss’s formula: [n/2(n+1)]
· Steps to do the following problem:
1. Factor the 2
2. Split the summand
3. Apply Gauss’s formula
You need to remember Gauss’s formula: [n/2(n+1)]
· Steps to do the following problem:
1. Factor the 2
2. Split the summand
3. Apply Gauss’s formula
The first step shown is when the summation is divided in two to be able to find the first 20 terms and then subtract 9, because what we are being ask is just for the terms from 10 to 20.
Then, you plug in for the summation of 1 to 20 in Gauss's formula.
This is done as well for the second summation.
Multiply and solve.
Here is another summation problem. This time, we are using the property of the summation of a - b equals summation a - summation b, but backwards. (FROM TWO SIMPLE EQUATIONS TO A MORE COMPLEX ONE)
Another property; its the same of Gauss, but squaring n+1 and dividing by 6.
This is the problem from the last class, the one which we could olny solve the first part.
Practice summation problems with the formula shown:
1 comment:
NICE WORK!!GIRLS!!!
HAHAHA
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