The next in mathematics is series. And here we'll look at arithmetic and geometric progressions and then Taylor and Maclaurin series. So first of all progressions and series. An arithmetic progression is a series of numbers in which there is a constant difference or addition between the terms. And here are the definitions as given in the reference handbook. So here, the last term is a. The difference between the numbers is d. The number of terms is n. And the last term is l. So the relevant equations are given here. The last term is the first term plus number of terms minus 1 multiplied by the difference, or conversely the number of terms is given by this expression. And the summation of the series S is given by either this expression MA plus 1 over 2, or equivalently M multiplied by 2A plus M minus 1D over 2. [SOUND] A geometric progression is a sequence of numbers which increases each term by a constant multiple that we'll denote by r. So, here again are the relevant equations from the handbook. So the first term is a. The common ratio or the multiplicity factor between the terms is r. So the last term l is given by the first term multiplied by r. The ratio multiplied by the raised to the power of n minus 1. And the summation of the series, the sum of the series, is equal to (a- rl) over (1- r) or equivalently a(1- r to the n) divided by (1- r). Some properties of series are given. For the example, the summation of a series, in which each number's a constant equal to C, is just equal to the number of terms multiplied by the constant, and the summation of a series, which is all multiplied by a constant value, C, is equal to C multiplied by the summation of that series. The summation of individual series. Summation of xi plus yi minus for example ci, you just add or subtract the summations of the individual series, etcetera. And a power series whose formula is given here is equal to the summation from i equals 0 to infinity ai. X minus a raised to the power i where i is an integer. And a special case of this is where the constant a in the parenthesis is 0 is equal to the summation from i equals 1 to infinity aix to the i minus 1 is given by this expression here. And finally, another expansion which we'll use is the Taylor series, which is useful for expanding the value of a function about a point, is given by f of x is function of a plus f prime of a over 1 factorial, x minus a, etcetera. And we'll show you an example of that in a minute. So some examples of this. First of all, we have a finite sequences of numbers, 15, 19, 23, 27, et cetera, up to 59. And the question is, the sum of the series is which of these alternatives? So in this case, we see that by inspection, it's an arithmetic series increasing by 4. In other words, each successive number is 4 bigger than the one before it. And, we also see that the value of the first term in the sequence A is 15 and the last term is 59 and the difference D is 4. So, we can use the formula, for example, in this form. The summation is equal to n(a + l) / 2. But we don't yet know how many terms there are, in that sequence, you can probably do it by inspection but that's calculated from this formula that the last term l is a plus the number of terms minus one multiplied by the difference. Or rearranging the number of terms is l minus a over d plus 1. So plugging in, last term is 59, first is 15, difference is 4, so, therefore, there are 12 terms in this series. Now we can go back to our summation formula and evaluate that. So, s is equal to na plus l, which is now equal to 12 times first term plus last term divided by 2 is equal to 444. So, the answer is D. The next example, we have a finite sequence of numbers 20, 30, 45, etc. Up to 227.8. The sum of these series is which of these alternatives? So here we see my inspection that this is a geometric series. Where each term increases by a multiplicity factor of 1.5 compared to the preceding term. And we see that by inspection, the values of the variables are the first term A is 20, the last term L is 227.8 and the multiplicative factor R is 1.5. So we can use the summation equation in this form, S is equal to a minus rl over 1 minus r. We know all the terms in that equation. It is equal to therefore, 20 minus 1.5 times 227.8 divided by 1 minus 1.5. And the answer is 643.4. So, the closest answer is C. And, you can also show either by inspection if you like that there are seven terms here. N is equal to 7 and just convince yourselves that the last expression is indeed given by this expression l is equal to r to the n minus 1. Now we'll finish with an example on the Taylor series and we want to form a Taylor series of the function f of x equals exponential or E to the 2x about the origin about a equals 0, to estimate the value of the function at point B, which is close to A. And the question is, what are the first two terms in the expansion about X equals 0? Which of these alternatives? So, here is the section again from the reference handbook. And, here is the formula for the Taylor series. F of X is equal to function evaluated at A plus the first derivative F prime evaluated at A divided by 1 factorial times X minus A. Plus the second derivative F double prime evaluated at A divided by 2 factorial multiplied by X minus A squared, etcetera. But here, we're asked only for the first two terms, so we just want to evaluate these terms in the expansion. So to do that, first of all we evaluate the function at A, where A is equal to 0. So E to the 2x evaluated at X equals 0 is equal to E to the 0, which is 1. The derivative f prime of x is 2e to the 2x. So the derivative evaluated at x equals 0 is equal to 2e to the 0 which is equal to 2. So, then we can evaluate these to first terms here, is equal to f of a plus f prime of over 1 etc. is therefore equal to f of a is equal to 1. f prime of x is equal to 2 divided by 1 factorial and multiply it and a is equal to 0. So the first terms are 1 plus 2b and the answer is B. Now, let's just look at what we did here. Here's a graph of the function e to the 2x. It looks like this. And what we're doing is finding the Taylor Series about this point 1, so that we can evaluate the function at some other point, say here. So in this case, we're only evaluating the first two terms. So in other words, the value of the function at some nearby place here is equal to, we just take the slope. The value at this point here is equal to the value here which is f of a, plus the derivative, f prime of a, which is the slope of this line multiplied by the horizontal distance from here to here. And that's what the first two terms are. And from this you can see obviously that this is only a good expansion for points which are very close to the original point. So, the first two terms are only good for evaluating functions which are very close. If we wanted to evaluate the function very far away then linearizing this would then become a very poor approximation and we would have to take more terms in the expansion. So this completes our discussion of series.