Ok, that's the idea, but try to "spread out" the scaling amongst the a(i)'s. The without loss of generality claim- if it holds for a given sequence, then it holds for any positive multiple of the sequence so the idea is to work with a convenient one. This is essentially the same as my suggestion to adjust your sum so it fits the conditions of 19). You then apply 19) to get an inequality, then try to manipulate it into the form that 20) is asking for. Proving you can do this is essentially justifying the "without loss of generality claim" from that link.
I think "k" is where it should be? ascii math is hard on the eyes though... I mean to include sum(i=0..n,a(i)) = sum(j=1..n+1,a(1-j)) sum(i=0..n,a(i)) = sum(j=2..n+2,a(2-j)) sum(i=0..n,a(i)) = sum(j=3..n+3,a(3-j)) ... all in one go (also negative values of k will work).
Gah. Thanks shmoe. These problems seemed at first to be manageable but now I can hardly do a single one! for b i want to show that m_p < m_p+1 so that i can imply g < m_p from g < m_1 but that is all getting nowhere.
About number 12a. Following the binomial theorem, and extracting the first term (which is one), then subtracting 1 from both sides, you are left with two sums from k=1 to n. What I did was show with induction that the two expressions inside the sum notations were algebriacally equivalent, which in my mind demonstrates that the sums must be the same....but maybe my mind is wrong. Absane mentioned much earlier that you could use combinative reasoning to show these expressions were the same, but I thought induction would be more fitting for a section on....induction. The two algebriac expression are (1/k!)(product:r=0...k-1): ((n-r)/n) and n!/(n^k)(k!)(n-k)! Its easily verified that k=1 gives 1 for both expressions, and considering the case of k+1 for the product side entails to multiplying the induction hypothesis by (n-k)/n(k+1), which, when multiplied by the binomial side, verifies the induction. I suppose my only concern is that showing the algebriac equality of the expressions inside of (the same) summation notation entails to the sums being the same.
alyosha, the problem does not say anything about induction or using induction. it can easily be shown that sum (a(k)) = sum (b(k)) when k=1 to n for both and a(k) = b(k) for all k. You are lagging, by the way. Please Register or Log in to view the hidden image! Don't worry though, I've only been on the same problem for a week now..
shmoe, for 20b on p47, i have proved that G < M_p but I'm at a loss as to how M_q < G? it seems if you substitute x(k)^q for x(k) in part a then the equality is reversed.??? i tried using regular numbers and i got a result which is only satisfied when M_q > G. I don't understand. try for x(1) = 4, x(2) = 3, n = 2, q = -2. lol never mind that. it seems you can't raise both sides of an inequality to a negative power without switching the inequality sign. how about help on 22 though
alyosha, that looks good. Induction to show the terms are equal would be the way to go. As §outh§tar mentioned, it should be no problem to show that two sums are equal when their terms are equal, you can reduce it to a sum of a bunch of zeros being zero by the 'additive property' of the summation notation. rudin is a nice book. One of the best at that level in my opinion. I take it you figured out the full version of 20 then? For #22, did you try to apply the arithmetic/geometric mean inequality from 20)?
for 9d on p57, i proved that every polynomial f(x) of degree n can be written f(x) = (x-a1)(x-a2)...(x-an)*h(x) where h(x) is a polynomial of degree 0 by following an inductive proof here. this implies in particular that f(x) has at most n zeros if f(x) is a polynomial of degree n. Now can I say that because of this, for f(x) to have n+1 distinct zeros, h(x) must equal 0, in which case f(x) = 0 then all the 'coefficients' of this 'polynomial' equal 0? oh, and for e, showing that g(x) = f(x) is sufficient for saying that the two polynomials share the same degree and the same coefficients, correct? also for pg60, 1b, this is what my proof looks like. tell me if it looks right: the set consisting of a finite number of points is the union of each of the finite number of points. from the choice of scale property (and part a) we know that the area of these points is 0. the additive property tells us that the area of the union of all these points is the sum of each of their areas (which is 0). therefore their area is 0. the only problem i fear is that the additive property is good for two sets at a time. i was hoping i could use sigma notation to extend this to a collection of n points but i'm not sure. maybe it's not a problem? also how do i even begin #2, the question on triangular regions? edit: besides which, i thought points and line segments were sets themselves? why is the problem (redundantly) referring to a set of sets then?
Excuse the LOL, but I'm going to refer back to 12 again. For the first inequality, you can show that using the binomial theorem its true for n=2......because the theorem will only get larger with larger n, it is just common sense to say that it must be true for all larger n? For the third inequality, I suppose my concern is showing that the geometric series must be smaller than 2. I showed (using a previously proved formula) that if the series could ever be greater than 2, it would imply that there is some n such that 1/2^n+1 <= 0. Is this an obvious enough absurdity to say that the series could never be greater than 2?
alyosha I presume you're talking about p45. for the 1st inequality, transform (1 + 1/n) into its simplified binomial theorem form. by n > 1, apostol means n >= 2. you can thus further break up your sum into two sums, one from k=1 to 1 and one from k=2 to n. you should now be able to simplify to a 0 < (your sum), which is easy to verify. for the second inequality, it is easy to show that (1-r/n) < 1 for all r>=0 and positive n. then show that the geometic series is less than 1. should be easy from there. for the 3rd inequality, you should incorporate what you learned from #11 after showing that the inequality holds for n=1, 2, and 3 (because 11 holds for n>=4). from there it will be very easy to show that the sum < 2.
I have a question involving number 13 (pg.45) that is actually a rather general question, but it applies to this particular problem. In part b, we consider an inequality with two "unknowns" two positive integers p and n. My concern in these types of inductive proofs would be: must we prove the case for both p+1 AND n+1? or is only the case of p+1 important? Because, incidentally, in part c we focus on the case of n+1 rather than p+1.
sure, you could put this a little more rigorously by substituting the n+1 zero into your expression for f(x) to get 0=(a(n+1)-a1)*...*(a(n+1)-an)*h(a(n+1)). Provided you know that the only way a finite product can be zero is to have one of the terms zero, you can conclude h must be the zero polynomial (since it is already a constant). You can also use induction here, write f(x)=(x-a)*h(x) where h(x) then has degree n-1 and n distinct zeros, conclude h must have zero coefficients. You are asked to prove the coefficients and degree will be identical. You can just apply part (d). Hmm, you want to prove something is true for all natural numbers. What to do... Can you see how to proceed with right triangles? An arbitrary triangle can be broken up into a union of two triangles. You can have a set whose elements are sets themselves, or sets of sets, or..., like { {a}, {a,b}, {a,{b,c}} }
In part b), you don't need induction, applying part a) will handlt all p's and n's simultaneously. In part c) your induction argument on n should be valid for any choice of p, so you can handle all these p's at once. It's conceivable that you would need some sort of 'double induction', though I can't think of a simple case off hand that does. If you had a statement P(n,m) that you wanted to prove true for all positive integers m and n, it would not be enough to prove P(1,1) true and P(n,m) true implies P(n+1,m+1) is true. You'd need something like P(1,1), P(n,m)=>P(n,m+1), and P(n,m)=>P(n+1,m). Something like P(1,1), P(1,m)=>P(1,m+1) and P(n,m)=>P(n+1,m) would do the trick as well (lot's of other possibilities, just think what steps would be needed to cover an integer lattice in the plane).
lol. I already knew that induction was necessary. But that is not what i was having trouble with. the additive property requires that i subtract from the total sum of all points the sum of the areas of the intersections of every combination of two points (such as if points a1 and an intersect, a3 and a7 intersect, a3, a4, and a5 all intersect..). what notation to use for this or how to put it down.. i have no clue. again, my nose tells me maybe it will be easier to show that the area of any such intersection is 0. but again, how to show this. I thought as much but as to what method of proof to use.. i mean, there are no numbers or any symbols to work with. i'll give it another shot after my nap.
