Simplification Rules
The simplification rules are used to simplify expressions. These rules help us to get rid of some of the extra steps that are required when simplifying an expression. The main idea behind these rules is to simplify an expression by combining like terms, grouping terms together and canceling variables. To simplify an expression using addition rule, we add two or more like terms together and then combine them with a common factor. For example: (a + b) + (c + d) = (a + c) + (b + d). Let’s examine our first rule, which is simply an example. Now let’s consider a different problem; one which, at first glance, does not seem to be too different from the previous one. The new problem asks for the sum of i squared from 1 to 4. Now we could work this out directly just as we did before. Just make scratch work with our and add it up. But let’s work through this just enough so that we can see the pattern, see the point of this example. So if we work this out, this equals 3 times (1) squared plus 3 times (2) squared plus 3 times (3) squared plus U times (4) squared. If we apply some elementary mathematics, using the distributive property (the algebraic rule that allows us to multiply a sum of numbers by another number) and multiplication over addition, we can simplify this expression: 3(1^2 + 2^2 + 3^2 + 4^2). That’s 3 equal to 3 times our old friend, which we’re now getting pretty bored of, as we do with old friends, the sum from i=1 to 4 of i squared. sum_(i=1)^4 1^2=30 sum_(i=1)^4 3i^2=3x(1)^2+3(2)^2+3(3)^2+3(4)^2 =3[1^2+2^2+3^2+4^2]=3[sum_(i=1)^4 i^2] We began with an expression, the sum from =1 to something. We then pulled a constant outside of the sigma notation, leaving behind the expression 3i squared with that constant sitting inside the sigma. This is a general rule that can be applied any time you see a constant within an iterated function. Let’s try one more example: Whenever you have an expression that contains a constant, evaluate it by substituting the value of the constant inside the sigma. This can be done because whenever you have an expression with a constant inside the sigma, you can pull it outside, and just then evaluate it. From 1=4 to 25 of 18г cubed – 18 times the sum from 1=4 to 25 of r cubed sum_(r=4)^25 18r^2=18[sum_(r=4)^25 r^3=] Distributive Property A(b+C)=Ab+AC Namely, if you have A times (b + c), that is the same thing as Ab + Ac. And all we are doing is taking this expression and simplifying it by using our knowledge of sigma notation over and over again. Let’s give you a new problem: The sum from =1 to 4, and internally here we’re going to have i squared plus 2i. So let’s just work this out directly. So this is equal to 1 squared + 2 times (1) + 2 squared plus 2 times (2) + 3 squared + 2 times (3) + 4 squared + 2 times (4). And again we could say we’re done, down tools, that’s it. But let’s break this up for our own purposes. Notice in this sum we have terms that look like two different things. One looks like things squared, like these ones. And one looks like two times things, like these. Let’s break those up and write those separately. This is the same thing as being equal to (1 squared + 2 squared + 3 squared + 4 squared) + (2 times (1) + 2 times (2) + 2 times (3) + 2 times (4)). Basically it’s so because we could add numbers in any order we want. That’s equal to our old friend. the sum from i=1 to 4 of i squared. Notice that we have a rule here, that i squared + 2i can be broken up into two summations: i squared and 2i. This is true because summations are additive; one can add them in any order. sum_(i=1)^4 (i^2+2i)=(1^2 +2(1))+(2^2 +2(2))=(3^2+2(3))+(4^2+2(4))=(1^2+2^2+3^2+4^2)+(2(1))+2(2)+2(3)+2(4))=(sum_(i=1)^4 i^2) (sum_(i=1)^4 1^2=30) Let’s do one more rule. This seems really silly and simple, but it makes you think about it a little bit. The sum from k=1 to 10 of 5. Seems like a trick question. Notice that there is no dependence on k. What do we do? This is almost conventional; we’re going to evaluate this as if it were an equality from k=1 to 10, but we don’t do anything with k, we just write down what is in there. This equals 5 plus 5 plus 5 plus, plus, dot, dot, dot, dot, plus 5. How many of them? 10 of them. sum_(K=1)^10 5=(5+5+5+…+5)/10 times=10*5=50 sum_(r=1)^7 8=(8+8+8+…+8)/7 times = 8*7=56 So this is equal to 10 times 5 = 50. The general rule here, whenever you’re summing up a constant, in this case it’s 5, but it could be anything. You just add up that constant the number of times you’re supposed to do. So for example, the sum from =1 to 7 of 8 would be 8 + 8 + itself 7 times. This is the same thing as 8 times 7 or 56.
