Just start from the bottom and go up. n=1 gives the statement (1+x)>1+x+x^2, so false. n=2 gives... etc. No need to use the binomial theorem either for finding n1 or the rest of the proof.
Shmoe, what is your background in mathematics? and southstar, what is your motivation for doing this?
shmoe's right. I misspoke when I said 'binomial theorem'. I meant a common sense way would be to look at the higher power. Thanks for the help on the absolute value problem; I tried substituting for y earlier but I have no clue why I didn't think of substituting for x instead. One of those dumb things. as for the straightedge thing... oooh. i understand now. i had to ask a friend though and he pointed it out. dont think i would have thought of that since apostol gave me the impression that every solution would somehow follow from previous theorems and so on. this one is an unusual little twist. haha. stupid for not thinking of it. seems rather obvious now. as for the primes one, i don't see how to go about solving with regular induction. even the answer i looked at said regular induction could not be used. there would be no way of determining if a k + 1 was even or odd or prime or what. maybe i'm missing something? for pg43 #4 i began by saying x = - C/B (where C and B are defined in apostol's proof). I'm able to get to AC = B^2. I'm not sure how what I'm doing is different from his proof for AC >= B^2 where he uses x = - B/A; it seems there should be only one such x but here we have two solutions?
my motivation is that i am taking an honors math class (same rigorous proof by proof treatment) after summer is over that begins with vector calculus and assumes everything prior is known by heart. i can do plug and chug calc to a respectable degree but i don't want to start knowing less than half of anyone else. darn, proving the binomial theorem isn't a cakewalk. nevermind, i was able to get it..
Take S(k) as I defined it earlier. Consider k+1. It's either prime or it's not. If it's prime, you're done. If not, you can write (k+1)=ab where a>1 and b>1. Then k>=a and k>=b (can do better than just k of course, but that doesn't matter), so we know a and b are products of primes, hence so is (k+1). You might see someone say regular induction fails here because to say something about k+1 when it's not prime we need to know about integers that will be strictly less than k but you can arrange the S(k) in regular induction to take care of this. Remember regular induction, well ordering, and strong induction are all equivalent, so if you prove something with one of them, you can prove it with all of them, (though you may need to rewrite your S(k) statement). Look at the first sentence of the proof for cauchy-schwarz, the sum over (a(k)x+b(k))^2=0 (specifically for x=-B/A, as this minimizes the sum) if and only if you have equality in cauchy-schwarz. This is a sum of positive things, so equals zero if and only if each term is zero. I'm a graduate student in math.
About his proof of the cauchy shwarz inequality.....maybe I have not looked at it long enough, but what is the significance of proving that it works for x = -b/a ? Or in other words, what is the significance of using the smallest value for the proof?
You started with a sum of non-negative things so you knew it was >=0. It was then rewritten as that quadratic Ax^2+2Bx+C>=0, and this inequality is true for all x. At the specific value of x=-B/A, you ended up with -B^2/A+C>=0, which is rearranged into the inequality you are after. It wasn't really vital that it was a minimum, just that at this value of x it turns into the inequality we are after.
alyosha when you use the quadratic equation to find the 0 of f(x) = Ax^2+2Bx+C, assuming B^2 = AC gives you a zero at x = -B/A. It just so happens that this x value gives us B^2 <= AC. It has something to do with f(x) being positive and therefore its discriminant is negative. I didn't understand why this was so at first but now visually, a negative discriminant reflects the fact that all of f(x) is above the x axis. Sorry shmoe but I was a little confused with what you said but I did a little fidgeting and came up with a new proof so I will post it for you to see if it makes sense. If A = 0 then B^2 = AC is trivial because B = 0. If A is non-zero then assume there is an x such that x = -b(k)/a(k) The sum over a(k)x+b(k) must equal 0: a(k)x+b(k) = 0 from this we have x = -B/A and we also have x = -C/B If -B/A = -C/B then we have B^2 = AC. How do I show that this is the only way the equality sign holds though? alyosha part of Vol 2 may be found here by the way:. Is #11 on pg45 supposed to be proven analytically by the way? And about #12.. Please Register or Log in to view the hidden image!
You don't know all the a(k)'s are non-zero though, so you shouldn't write it like this. The direction you're trying to prove here is straightforward- if you have an x where x*a(k)=b(k) just substitute this into Cauchy-Schwarz to get rid of the b's, and you're done. The other way is a little more work. If you have equality, it means AC-B^2=0. So your quadratic is (assuming A non zero): Ax^2+Bx+C=A(x+B/A)^2 which is zero when x=-B/A. But this quadratic was just another way of writing the sum of (x*a(k)+b(k))^2, k from 1 to n, so we must have: sum(k=1..n, ((-B/A)*a(k)+b(k))^2)=0 and each term must be zero, hence (-B/A)*a(k)+b(k)=0. There is a minor problem here if the a(k)'s are zero and the b(k)'s aren't. You know that you get equality in Cauchy-Shwarz but there is no possible x that will work if there is a non-zero b(k). Not really a big deal, you can swap the a's with the b's and then take x=0. If the a's and b's are all zero, any x at all will work of course. #11 same idea as that question with the minimal n1, just go through n=1,2,3,... until you find the equality holds. The prove using induction that it will hold for each n larger than that. #12a) Try to simplify the product on the RHS, write 1-r/n=(n-r)/n. Write the product using factorials. b)use the binomial thm on the first inequality. use part a) and (1-r/n)<1 on the second inequality. Use #11 and geometric series for the third inequality.
This may seem overly picky, but in the answers for page 35 he lists the general rule of #3 as 2 - 1/2^n and #4 as 1/n, when it seems like they are more accurately 2 - 1/2^(n-1) and 1/(n+1). Could a difference of this kind actually affect the proof in a serious way? Nevermind, you could just define the sum to start counting at n=0 I guess. Shmoe, do you know anything about max rosenschlits intro to analysis? I've heard good things about it.......I also caught a glimpse of that being used at my universitys classes, and I could only smile that they would use a 10 dollar dover book......... Also, I'm trying to determine how to formulate an "inductive" argument for number 8, as you said, to prove that the multiplication of an arbitrary number of inqualities preserves the inequality....
Sorry, not familiar with it. The reviews at amazon looked good though, and it is a cheap second perspective that might be nice to have. I used Pfaffenberger and Johnsonbaugh for my first analysis course (also a Dover book now, ~$23 last I checked) and I liked it. It's on a similar level to Rudin's intro book, so more depth and less user friendly judging from the reviews of max's book. You might want to check them out in the library to see if they are to your liking (both are in googles book search as well). I'd just do it the way Southstar has shown. Or you could take S(k) to be the statement: a(k)*a(k-1)*...*a(2)<=a(k-1)*a(k-2)*...*a(2)*a(1)*c^(k-1) S(k)=>S(k+1) requires multiplying only a pair of inequalities, which you know you can do. This isn't really any different than Southstar's, you'd just be saving the cancellation for the end.
The equation is supposed to be f(x) = Ax^2 + 2Bx + C = A(x+B/A)^2 + C - (B^2)/A. But here x = -B/A is not a zero of the equation. I noticed those errors but I couldn't really follow the rest of your explanation afterwards. I do notice however that because the sum (x*a(k)+b(k))^2 equals 0, f(x) = 0. Substituting x = -B/A into f(x) gives B^2 = AC as desired. In the case that A = 0 however, B = 0 and B^2 = AC = 0. This should take care of both zero and non-zero values for A. Also a(k)x = -b(k) for all x. It is probable that I missed some step anyway because I do not see where the C went over here after completing the square: Ax^2+Bx+C=A(x+B/A)^2 And 12a I was finally able to do. At first I was trying to use induction which, I think, is why I was having trouble. For 12b how do I show in the first inequality that 1 < the sum of (nCk)(n^-k), which is what I simplified the original sum to for k = 1 to n. I am thinking that if I am able to show that (nCk)(n^-k) is positive for all positive integers n and that (nCk)(n^-k) > 1 for n = 1 then I will be on the right path to showing that the sum is also greater than 1 but showing that it is positive is a bit difficult with all those troublesome factorials. alternatively anyway i believe if i can show that (1 + 1/n)^n < (1 + 1/(n+1))^(n+1) for all positive n then i can prove the inequality by showing that it is true for n = 2 and thus must be true for all n >=2. That will also involve some factorial nastiness however. It's too bad we don't have a better way of putting mathematical symbols here so that communication would be more effective. I've seen some other math forum with something called LATEX.. EDIT: I just tried it and after much fiddling came up with 2(n + 1)^(k-1) < n^k, which I can't prove. I give up! Is this part supposed to be done alone by a self learner by the way? I'll feel better if it isn't because this is my first try at real math. ah wait a minute. let's run over this. say we have a sum of a(k) from 1 to n. say i show a(1) > 0. and say i show that an arbitrary a(k) is positive. is that sufficient to say that the sum of a(k) is positive? i think that's safe to take for granted but it's difficult to say what needs to be proven and what needs to be accepted since the order axioms allows for x + y > 0 if x and y are positive except it doesnt specifically mention sigma notation sums). i think it's also safe to say that (1 + n)^-(1+n) > 0 for all positive n (it seems obvious enough that, in the denominator, a positive number raised to a positive number results in a positive number). boy, would i love to play with logarithms right about now. Please Register or Log in to view the hidden image! if these assumptions can be accepted then i can say that (1 + 1/n)^n < (1 + 1/(n+1))^(n+1). of course there's a lot of crunching in between.
You're right about the 2, but remember this is working under the assumption that the Cauchy-Schwarz is an equality, so B^2=AC and the rhs as I wrote it still stands and x=-B/A is a zero. factorial is always positive where you guys have defined it. What are the k=0 and k=1 terms your get from the binomial? (1+1/n)^n is increasing, but you're better off using the binomial to prove the inequality you're after.
scratch that last rambling edit then. i asked: say we have a sum of a(k) from 1 to n. say i show that an arbitrary a(k) is positive. is that sufficient to say that the sum of a(k) is positive? also is it safe to say that (1 + n)^-(1+n) > 0 for all positive n? in the denominator we have u^u for positive u but proving that this is a positive number is beyond me. if we can accept these things then i can show using the binomial theorem that (1+1/n)^n is increasing, in which case i can prove the first inequality. i don't suppose there is a simpler way than this method? for example showing that (nCk)j^-k > 0 for all n>=1. the simplified binomial im using starts from k = 1 so there is no k = 0 but it is equivalent to (1 + 1/n)^n (i confirmed this in 12). (1 + 1/n)^n = 1 + sum[ (nCk)j^-k, for k=1 to n] For n =1 i get 2. i understand what you're saying now about the cauchy schwarz inequality
The sum of a bunch of positive things is positive, yes. u a positive integer then u^u is positive. It's the product of a finite number of positive things. I don't know of a way to show that is increasing that isn't overly complicated. I said k=0 and k=1 and not "n" for a reason Please Register or Log in to view the hidden image! . Just concentrating on the first two terms in the binomial gives: (1+1/n)^n=1+(nC1)*n^(-1)+a bunch of positive stuff this is > 1+(nC1)/n=1+(n!/1!/(n-1)!)/n=1+n/n=2.
They are working from Apostol's "Calculus", volume I. It's a pretty popular book, so shouldn't be hard to find.
Second edition? It seems I have an e-book version of it. Is it Calculus: Volume 1, One-variable calculus, with an introduction to Linear Algebra? Copyright 1967?
I am not sure if you got it or not... did you get #12? If not.. the way I did it was write out (1+1/n)<sup>n</sup> in terms of the binomial theorom... (using the summation notation). I then expanded the binomial and noticed some canceling out... rearranging led to something like: Sum {1/(k!) * [n * (n-1) * (n-2) * ... * (n-k+1)/ n<sup>k</sup>} Where k is from 0 to n. From there you divide out n in all the terms and you should see the product formula. I am not exactly sure if this counts for a "proof" but it's how I derived the formula. I think it can easily be proved by induction.
