YAMPT: Yet another math problem thread

[Bullhonkie]

I was given this as an extra credit problem over the weekend but I honestly have no idea where to even begin or what I'm trying to do.

Find the asymptotic for v1+v2+...+vn in the form c?n^a
Identify c and a

Can anyone shed some light by explaining this and start me off in the right direction?

 

[olds]

Describing a reaction or activity that approaches, but never reaches, a given value - the asymptote.

I suck at math, that was from google.

 

[Q]

I've heard of asymptote but not asymptotic.

Asymptote has something to do with putting it in the TI83 and graphing it

 

[blinky8225]

Out of curiosity, what math class is this for?

Are you familiar with harmonic numbers? It's similar to that series.

 

[Bullhonkie]

Originally posted by: blinky8225
Out of curiosity, what math class is this for?

Are you familiar with harmonic numbers? It's similar to that series.

Whoops I probably should've mentioned that. It's for integral calculus. Nothing fancy, just the 2nd quarter in the calculus sequence. We haven't started doing integration yet.

I have no idea what harmonic numbers are.

Originally posted by: Quintox
I've heard of asymptote but not asymptotic.

Asymptote has something to do with putting it in the TI83 and graphing it

Maybe my prof meant asymptote. He's russian and not the greatest with english grammar. Either way though, I still have no idea what to do.

Help me someone?

 

[Inferno0032]

Originally posted by: Bullhonkie

Originally posted by: blinky8225
Out of curiosity, what math class is this for?

Are you familiar with harmonic numbers? It's similar to that series.

Whoops I probably should've mentioned that. It's for integral calculus. Nothing fancy, just the 2nd quarter in the calculus sequence. We haven't started doing integration yet.

I have no idea what harmonic numbers are.

Originally posted by: Quintox
I've heard of asymptote but not asymptotic.

Asymptote has something to do with putting it in the TI83 and graphing it

Maybe my prof meant asymptote. He's russian and not the greatest with english grammar. Either way though, I still have no idea what to do.

Help me someone?

You are in your 2nd quarter in calculus in college, and you haven't started integration yet? I'm a senior in high school and I've already done integration. And then 3D integration when finding functions' values when spun around certain points...

Anyways, asymptotic line I believe he was referring to, that's what you call the x=?? line that the function approaches, but never equals.

 

[sponge008]

I think it's referring to an asymptotic series expansion (http://en.wikipedia.org/wiki/Asymptotic_expansion), how to figure it out I have no idea.

 

[RESmonkey]

Hrmm...I'm in BC (equivalent to Calc II) and I've never seen asymptotic expansion before.

 

[blinky8225]

Well, the harmonic series is the series 1/1+1/2+1/3+1/4+...+1/n.
Its asymptotic limit can be found by taking the integral of 1/n to get ln[n]+C.

Likewise, by taking the integral of the square root of n, one gets (2/3)n^(3/2).
Thus, c=2/3 and a=3/2.

That's my answer, and I'm fairly certain that it is right, but the question seems a little vague, and (2/3)x^(3/2) isn't a true asymptote. Rather, it's a very good approximation; the series actually diverges from (2/3)x^(3/2) as it gets very large.

I'm only a freshman in college myself, so I hope someone else on this forum could provide a second opinion.

 

[Q]

Originally posted by: blinky8225
Well, the harmonic series is the series 1/1+1/2+1/3+1/4+...+1/n.
It can be approximated by taking the integral of 1/x to get the ln[x].

Likewise, by taking the integral of the square root of x, one gets (2/3)x^(3/2).
Thus, c=2/3 and a=3/2.

That's my answer, and I'm fairly certain it is right, but the question seems a little vague and (2/3)x^(3/2) isn't a true asymptote. Rather, it's a very good approximation; the series actually diverges from (2/3)x^(3/2) as it gets very large.

I'm only a freshman in college myself, so I hope someone else on this forum could provide a second opinion.

Sounds good to me. I'm a freshman in college too and I have no idea what you said so that sounds right. At least I know I'm not going to major in Math

 

[maziwanka]

i do not remember math anymore

 

[Bullhonkie]

Originally posted by: blinky8225
Well, the harmonic series is the series 1/1+1/2+1/3+1/4+...+1/n.
It can be approximated by taking the integral of 1/n to get the ln[n].

Likewise, by taking the integral of the square root of n, one gets (2/3)n^(3/2).
Thus, c=2/3 and a=3/2.

That's my answer, and I'm fairly certain that it is right, but the question seems a little vague, and (2/3)x^(3/2) isn't a true asymptote. Rather, it's a very good approximation; the series actually diverges from (2/3)x^(3/2) as it gets very large.

I'm only a freshman in college myself, so I hope someone else on this forum could provide a second opinion.

Thank you. The question seems a bit vague to me as well but I have absolutely nothing to go by.

Is there a way of finding those values without taking the integral? We're going to start integration this week, but since he gave us this over the weekend I'm assuming there's some other way to do this.

 

[blinky8225]

Have you done Riemann sums, yet? You could probably do it by taking the limit of a Riemann sum as n approaches infinity.

 

[Bullhonkie]

Originally posted by: blinky8225
Have you done Riemann sums, yet? You could probably do it by taking the limit of a Riemann sum as n approaches infinity.

Yeah we have, but I don't think I understand them very well. Would you mind explaining how that would work?

 

[QED]

Originally posted by: Bullhonkie

Originally posted by: blinky8225
Well, the harmonic series is the series 1/1+1/2+1/3+1/4+...+1/n.
It can be approximated by taking the integral of 1/n to get the ln[n].

Likewise, by taking the integral of the square root of n, one gets (2/3)n^(3/2).
Thus, c=2/3 and a=3/2.

That's my answer, and I'm fairly certain that it is right, but the question seems a little vague, and (2/3)x^(3/2) isn't a true asymptote. Rather, it's a very good approximation; the series actually diverges from (2/3)x^(3/2) as it gets very large.

I'm only a freshman in college myself, so I hope someone else on this forum could provide a second opinion.

Thank you. The question seems a bit vague to me as well but I have absolutely nothing to go by.

Is there a way of finding those values without taking the integral? We're going to start integration this week, but since he gave us this over the weekend I'm assuming there's some other way to do this.

Perhaps he wanted you to think in reverse order.

Define F(n) to be the partial sum of the square roots of 1 through n. You can approximate the "derivative" of F(n) using as follows:

F'(n) = ( F(n) - F(n-1) ) / ( n - ( n - 1)) = F(n) - F(n-1) = sqrt(n).

Now you know he wants the asymptote of F(n) in the form of c * n^a for some c and some a. Let A(n) be the asymptote of F(n). Hence, A(n) = c * n^a for some c and some a. The derivative of A(n) is A'(n) = c * a * n^(a-1).

By matching up the derivative of the asymptote with the "derivative" of F(n), you must have:

c* a = 1 and (a - 1) = 1/2. Solving for c and a gives you a=3/2, c=2/3.

 

[frostedflakes]

I wonder if your professor is just using the wrong wording. It sounds like he wants you to find the general form of the series, which would be summation 1*n^(1/2) from n=1 to n=n. But the word asymptotic thrown in there completely throws me off, heh.

 

[blinky8225]

Another method would be to using the fact that 1+2^2+...+n^2=n(2n+1)(n+1)/6. That would approximately be the area under f(n)=n^2. f(n)=n^(1/2) would be its inverse. Thus, the area under n^(1/2) would be n^3-n(2n+1)(n+1)/6. Then, you would just take the first term as the others terms become less significant as n approaches infinity and you also get a=3/2 and c=2/3.

This method is probably a bit confusing in words. It helps if you draw it out.

Oh, I forgot to add, you would also have to make a substitution of n=x^(1/2)

 

[sponge008]

Oh, on second thought, this is exactly equivalent to a Riemann sum, that's probably the way your prof wanted you to do it. See how the two are equivalent? Then once you have the sum, you should be able to find the end behavior.