SUM NOTATION AND LIMIT NOTATION ----------Warren Smith--------- Definition of the sum-symbol: L ---- \ > G(k) = G(F) + G(F+1) + G(F+2) + ... + G(L-2) + G(L-1) + G(L). / ---- k=F G(k) can be ANY function of k. F is the FIRST value of k used in the sum. L is the LAST. k can be ANY LETTER. It is the "running index" in the sum, which varies from the first to the last value in steps of 1. For example 8 ---- \ 2 > k = 4 + 9 + 16 + 25 + 36 + 49 + 64. / ---- k=2 ---------------------------------------------------------- LIMITS lim F(x) x --> v is the limit, as x approaches v, of F(x). F(x) can be ANY function. x can be replaced by any letter. It, like k in the sum notation, is just a placeholder. Normally (if F(x) is "continuous") lim F(x) = F(v) x --> v so that evaluating a limit is normally trivial. For example 2 lim x = 25. x --> 5 However, things can get trickier. First of all, you could have a function such as sign(x) = (+1 if x>0, -1 if x<0, and 0 if x=0) which is discontinuous (jumps in value). Try plotting it to see what I mean. Then lim sign(x) = does not exist. x --> 0 However, at a point where the sign(x) function does not jump, such as x=5, everything is normal: lim sign(x) = +1 x --> 5 Also even in the above jumping example, ONE-SIDED limits still exist: lim sign(x) = -1 ("limit as x approaches 0 from the - side") x --> 0- lim sign(x) = +1 ("limit as x approaches 0 from the + side") x --> 0+ Second, you could have a function which is undefined at the limit point x=v, usually due to a division by 0. Or you could be interested in taking limits as x approaches INFINITY. Examples 5 + 6/x lim ------- = 2 x --> 0 7 + 3/x lim 3/x = does not exist. x --> 0 lim 3/x = +infinity. x --> 0+ lim 3/x = -infinity. x --> 0- 5 + 6x lim ------- = 2 x --> infinity 7 + 3x 2 5 + 6x + 7x lim ------------ = infinity x --> infinity 7 + 3x 2 5 + 6x + 7x lim ------------ = 7/5 x --> infinity 2 7 + 3x + 5x In these cases, the way to evaluate the limit is to simplify the expression for F(x), but during that process you get to use extra simplication rules you would not normally get to use. Namely, A + B, where A is HUGE (limiting value +infinity or -infinity) and B is comparatively negligible, can be replaced by just A. So 5 + 6/x 6/x lim ------- = lim --- = 2 x --> 0 7 + 3/x x --> 0 3/x since 5 is negligible compared to 6/x (which is huge) when x-->0; and 7 is negligible compared to 3/x (which is huge) when x-->0. But the latter expression simplifies to (6/x)/(3/x)=2. And obviously lim 2 = 2. And so 2 5 + 6x + 7x lim ------------ = 7/5 x --> infinity 2 7 + 3x + 5x 2 since 5+6x, although huge if x is huge, still is negligible compared to 7x . 2 and 7+3x, although huge if x is huge, still is negligible compared to 5x . Hugeness rules: B Any monomial A x of degree B is far huger than any monomial(x) of smaller degree than B, when x-->infinity. x Any exponential B for any B>1 is far huger than any polynomial(x) when x-->infinity. A VERY SIMPLE WAY TO ESTIMATE THE NUMERICAL VALUE OF A LIMIT ------------------------------------------------------------ lim F(x) = F(v) x --> v+ is just to use your calculator to evaluate F(v+.01), F(v+.001), F(v+.0001), etc. and hopefully it will be fairly obvious what value these are approaching. Similarly for lim F(x) = F(v) x --> v- just use your calculator to evaluate F(v-.01), F(v-.001), F(v-.0001), etc. If the 2-sided limit exists, these should both appear to be approaching the same value. If no limit exists, then that too will usually be pretty obvious since the values will not seem to be approaching any particular value, or the + and - limits will be obviously different. This calculator-based method will at least give you an estimate of the right answer and is a very useful sanity check on any supposed exact calculation. Plus, you can use this method even if you have completely lost touch with reality. Unfortunately this calculator-based method cannot be used if the limit-value is NOT a number, but instead is FORMULA, which happens if there are other free variables (besides x) in F(x). GEOMETRIC SERIES ---------------- A sum whose consecutive terms are in CONSTANT RATIO (here called k) is a "geometric series." A formula: N ----- N+1 \ n 1 - k ) a k = -------- a / 1 - k ----- n=0 The simplest way to derive the formula: Use N=infinity (and |k|<1) in which case the right hand side is just a/(1-k). To derive this, notice the sum S obeys S - kS = a = the first term in the sum (with n=0) because all the other terms are canceled out ("telescoping"). Then solve for S to get S=a/(1-k). [I personally work by remembering THIS (very simple) formula for an infinite geometric series, only.] Now, notice any FINITE geometric series (N is not infinite) may be thought of as the diffence between two infinite geometric series. Namely the one with initial term a, and the one with initial term N+1 a k . So the final formula is N+1 a a k --- - ------- 1-k 1-k . (If |k|>1, then view the series as going in the other direction, with ratio 1/k, so that is ok, you can verify the formula still works. If k=1 the formula has problems but the sum is so simple you would not be worrying about geometric series at all, so who cares?) ARITHMETIC SERIES ----------------- A sum whose terms have CONSTANT DIFFERENCE (here called k) is an "arithmetic series." A formula: N ----- \ N+1 ) [a + k n] = (2a+kN) --- / 2 ----- n=0 This is best derived by the "pairing trick." You view the N+1 terms in the sum as coming in (N+1)/2 PAIRS. Each pair is a sum of the n=j term and the n=N-j term, and that pair has sum a+kj + a+k(N-j) = 2a+kN. I.e. all the pairs have the SAME sum. (Note, if N+1 is an odd number, then this is N/2 + 1/2 pairs - note the extra half-pair, consisting of the middle term in the sum, alone. This still works out right, but can be a bit confusing.) So, the sum is SUM = (SUM OF EACH PAIR) (#PAIRS) which gives the formula above. I personally operate by just remembering the pairing trick and rederiving the formula each time I need to do one of these sums.