Simplification Rules

Simplification Rules The simplification rules are used to simplify expressions. These rules help us to getrid of some of the extra steps that are required when simplifying an expression. Themain 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 togetherand 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 adifferent problem; one which, at first glance, does not seem to be too different fromthe previous one. The new problem asks for the sum of i squared from 1 to 4. Nowwe could work this out directly just as we did before. Just make scratch work with ouri and add it up. But let's work through this just enough so that we can see thepattern, 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 3 times (4) squared.If we apply some elementary mathematics, using the distributive property (thealgebraic rule that allows us to multiply a sum of numbers by another number) andmultiplication over addition, we can simplify this expression: 3(1^2 + 2^2 + 3^2 +4^2). That's 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. We began with an expression, the sum from i=1 to something. We then pulled aconstant outside of the sigma notation, leaving behind the expression 3i squaredwith that constant sitting inside the sigma. This is a general rule that can be appliedany time you see a constant within an iterated function. Let's try one more example: Whenever you have an expression that contains aconstant, evaluate it by substituting the value of the constant inside the sigma. Thiscan 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 r=4 to 25 of 18 r cubed- 18 times the sum from r=4 to 25 of r cubed. Namely, if you have A times (b + c), that is the same thing as Ab + Ac. And all we aredoing is taking this expression and simplifying it by using our knowledge of sigmanotation over and over again. Let's give you a new problem: The sum from i=1 to 4, and internally here we're goingto have i squared plus 2i. So let's just work this out directly. So this is equal to 1 squared + 2 times (1) + 2squared plus 2 times (2) + 3 squared + 2 times (3) + 4 squared + 2 times (4). Andagain we could say we're done, down tools, that's it. But let's break this up for ourown 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 asbeing equal to (1 squared + 2 squared + 3 squared + 4 squared) + (2 times (1) + 2times (2) + 2 times (3) + 2 times (4)). Basically it's so because we could addnumbers 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 twosummations: i squared and 2i. This is true because summations are additive; onecan add them in any order. Let’s do one more rule. This seems really silly and simple, but it makes you thinkabout it a little bit. The sum from k=1 to 10 of 5. Seems like a trick question. Noticethat 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 doanything 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. 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'resupposed to do. So for example, the sum from r=1 to 7 of 8 would be 8 + 8 + itself 7times. This is the same thing as 8 times 7 or 56.