Wouldn't that sort of imply that...only the even numbers are being counted? IS that whats being implied?
Yes... look at the values of n. n = 1... 2*n = 2 n = 2... 2*n = 4 n = 3... 2*n = 6 So 2(n+1) is the next even number.
Why couldn't it just be that 2n is the stopping number and that k counts straight through all the natural numbers as usual through 2n? ...though, I can see what the other situation may be, because if I were to choose a new n, it would have to be twice that n....
It does. If n = 1... then k = 1, 2. If n = 2... then k = 1, 2, 3, 4. if n = 3... then k = 1, 2, 3, 4, 5, 6. I think you are overlooking the given and obvious. They want you to prove that the summation formula, for any given n >= 1 equals some constant times n. That is, the sum = constant*n.
hello absane. sorry for the absence. i was struggling with #13 on p45 for a bit but gave up finally. besides which the *ahem* pdf im using has muddled the symbols so its hard for me to read the rest of the problems in that section. whew. im on to functions now and proving polynomial properties, which sounds like a breeze (but it's really not).
For 9b on pg 57, this is my proof: Please Register or Log in to view the hidden image! (each line follows from the previous but i didn't bother to put equal signs) but i don't know if that 'proves' that f(x+a) is still an n-th degree polynomial. my proof for part a was similarly weird. i think this picture method is clearer anyway. what do you think?
for 9b, when you remove (x+a)^n from the sum, you are left with a polynomial of degree n-1 evaluated at x+a. You can use induction on this to say it's also a polynomial of degree n-1. Binomial on (x+a)^n like you did gives a poly of degree n. Poly of degree n + poly of degree n-1 = poly of degree n. Or you can apply the binomial to each term (x+a)^k, getting a double sum. Collect powers of x by reversing the order of summation (which you may want to prove is actually valid). This is good practice. a) should be simpler- show the constant term is zero and pull out an x.
Wait... so what do they want to know? That p(x) = d<sub>0</sub>*x<sup>n</sup> + d<sub>1</sub>*x<sup>n-1</sup> + ... + d<sub>n</sub> for some constants d<sub>k</sub> which are combinations of a and c<sub>k</sub>? And we must do this from the definition of p(x) and f(x)?
Yes, though it doesn't matter where the dk's come from, just that they exist. aside-Didn't it used to be possible to add attatchments here?
You could in the past. However, there is obviously a scipting error somewhere and I posit that Dave won't fix this as he probably wants to converse storage space.
the reason my proof went that way is because i assumed that the addition of the two polynomials would give an n-th degree polynomial but I assumed it was obvious that separating the x^n term from the other terms of lesser power would be enough to show that the polynomial was of degree n. would proving the sum is of degree n-1 go something like: would that be something like if n = 1 then the sum is a monomial, if n = k is a k-th degree polynomial then n = k + 1 is a (k+1) degree polynomial? i don't understand how this would go since it would involve nesting sums. for a i used a g(x) = sum from 1 to n of c(k)x^(k-1). it seemed obvious here too that this was a polynomial of degree n-1 but does that need to be proven too? and c... i tried using f(x) = x(x-2)(x-3) to see if i could get any clues as to what i was supposed to do but.. Please Register or Log in to view the hidden image! Am I supposed to prove such an h(x) exists???
The big problem I see with this is the last line. You should expand this so you can clearly see a polynomial. However, I suspect there is a much easier way to do this. I did my proof of this similar to how you did it. However, I am looking for a quick and cute way to do it.
that's what i was thinking absane. but i don't know of how to simplify it without nesting a sum in order to take care of the (x+a)^k. whatever the simple solution is, it seems that c depends on it as well..
Just expand the c*(x+a)^k term within the summation. Then just "distribute" the summation (sigma) across every term (since you are allowed to do this). Then, you will have one, very long, polynomial that just needs terms collected and simplified. And then you are done. Do you understand? It can be done and it is not too hard to see, it is just too much of a mess for someone like me to do because I am lazy. However I suspect you are not Please Register or Log in to view the hidden image!
Seperating the nth term is fine. Expanding it gives a degree n polynomial, also fine. The problem though is the sum(k=0..n-1,c(k)*(x+a)^k), why is this a polynomial? This is why I suggested induction, on the degree of your polynomial. sum(k=0..n-1,c(k)*(x+a)^k)=g(x+a) where g(x)=sum(k=0..n-1,c(k)*x^k) is a degree n-1 polynomial. By the induction hypothesis, you could conclude g(x+a) is also a degree n-1 polynomial. You then have f(x+a)=your degree n poly from expanding (x+a)^n + g(x+a). Proving the sum of polynomials is a polynomial is simple, just use the definition of polynomial and pg 40, 2(a). That it remains degree n- the only thing that could muck it up is if g(x+a) had a non-zero coefficient for x^n, which we know it doesn't. edit- should mention the base case is for degree 0 polynomials, i.e. constants. That's why it would be good practice Please Register or Log in to view the hidden image!. sum(i=1..n,a(k)) means a(1)+a(2)+...+a(n), B(k,i)=k!/i!/(n-i)! the binomial coefficients. This will look hairy, but translate it carefully to sigmas on paper and it should be nicer: f(x+a)=sum(k=0..n,c(k)*(x+a)^k)=sum(k=0..n,sum(i=0..k,c(k)*B(k,i)*x^i*a^(k-i))) by expanding using the binomial and passing c(k) through the inner sum (inner sum doesn't depend on k). Swapping the order of summation is equivalent to gathering the x^i's. We'll get: sum(i=0..n,sum(k=i..n,c(k)*B(k,i)*x^i*a^(k-i))) which equals: sum(i=0..n,x^i*sum(k=i..n,c(k)*B(k,i)*a^(k-i))) by passing the x^i through the inner sum (inner sum doesn't depend on i). In other words, we get a polynomial, and the coefficient of x^i is just: sum(k=i..n,c(k)*B(k,i)*a^(k-i)) To see why swapping the order of summation works like that, draw a lattice with k on the horizontal axis, i on the vertical. Your double sum adds up all the points in the region bounded by the lines k=i, k=n and i=0. Consider how swapping the sum affects the order you add these points, and the ranges of the variables.
your a) is fine, do a change of variables, j=k-1 say to write it as: sum(j=0..n-1,c(j+1)*x^j) a poly of degree n-1 since the lead coefficient is c((n-1)+1)=c(n) is non-zero since f was degree n. for c) use their hint and define p(x)=f(x+a) (note then we have p(x-a)=f(x)). Part b) tells us p(x) is a poly of degree n. If f(a)=0, then p(0)=0, apply part a) to p(x). use p(x-a)=f(x) and rejoice.
I'm taking a break from b so i can tackle it afresh when i get back but for c i just want to verify that showing f(x+a) = xh(x+a) is enough for the proof. also i'm not sure how to tackle d without using words (if im supposed to show that analytically). it follows from c that an n degree polynomial can have at most n factors (dunno if there needs to be a proof for that). if f(x) = 0 for n + 1 distinct real values then consider the cases where f(x) is not equal to 0 for some real x and where at least one c(k) is non-zero. i figured it would be necessary to show that these cases led to contradiction but.. hmm. i'll come back and look at it.
I'd call that one step shy- sub in (x-a) for x to get the form they're after. Probably easiest to use induction on the degree of f and use part c).
