Question

Help? http://i.imgur.com/U8EuJ.png

Here is my work on the first one so far: http://i.imgur.com/B4cOp.jpg

Here's a similar solution, but I still can't figure out all the steps: http://i.imgur.com/1ghgd.png

Answer 1

webassign?
I hate that place.

Answer 2

It's alright. We're stuck with it. It's better than CalcPortal.

Answer 3

Is this for a calc class or a linear algebra class?

Answer 4

Calc I

Answer 5

I'm not really following this
\[R_n\] notation, do you know what this is called? Perhaps I've known it with different variables.

Answer 6

It's just the Riemann sum evaluated at the right endpoints...

Answer 7

Your first part is good so far, are you having trouble with the limit?

Answer 8

I'm just not sure where to go from there. I guess I need to simplify it somehow, but I'm not sure how.

Answer 9

Alright let me type it in here so I don't have to keep flipping

Answer 10

\[\lim_{N\rightarrow \infty} \frac{32}{N}\sum_{j=1}^{N} 2(-2+\frac{4j}{N})^3 + (-2+ \frac{4j}{N})^2\]
Does that look about right?

Answer 11

That's what I have so far, but do I need to do something after that?

Answer 12

That cant be right... that limit goes to 0 right in that first term... constant/infty = 0
But the integral of the function over the interval would give a numeric, positive, answer...

Answer 13

@ agreene
what they have is correct

Answer 14

*shrug*

This is probably why I never do approximate sums by hand... now I'm curious to see how this is evaluated though.

Answer 15

it is pretty ugly...multiply it out and use the formulas for

then sum of 1,sum of j, sum of j^2 and sum j^3

Answer 16

You're still going to need to hide that constant of summation somewhere so it will be annihilated and not make the limit 0...

Answer 17

?

Answer 18

the 32/n would be out front multiplying everything in the sum, in the way you described and then it would still be 0.

Answer 19

no it wont..try it

Answer 20

You have to change the summation somehow and cancel out N's, leaving separate terms, one of which will simplify to zero and the other will be a positive number.

Answer 21

(In the example I posted, the whole number was 2032 and the other terms have N in the denominator, so I guess they're zero.) But I still don't know how to get that far.

Answer 22

\[\lim_{n\rightarrow\infty}\frac{32}{N}\sum_{j=1}^{N}(\frac{128j^3}{N}-\frac{176j^2}{N}+\frac{80j}{N}-12)\]
Thats what I get when I expand... this limit is still 0.

Answer 23

How do I get rid of the summation signs? Where do the j's go? Ugh. I'm gonna need to find my book and read this section again.

Answer 24

no it is not

Answer 25

That limit is emphatically not zero.

Answer 26

It looks like zero to me too, but I know it's not.

Answer 27

\[\frac{32}{\infty}\sum = 0*\sum=0\]

Answer 28

Thats what I see, atleast... I know it cant be the case though.

Answer 29

You must be more careful with your infinities. Look, even just the last bit:
\[\frac{1}{N} \sum_{j=1}^{N} 12 = 12 \]

Answer 30

@agreene
you can't do the limit that way

Answer 31

Furthermore,
\[\sum_{j=1}^{N} j = \frac{N(N+1)}{2}\]
\[\sum_{j=1}^{N} j^2 = \frac{N(N+1)(2N+1)}{6} \]
and
\[\sum_{j=1}^{N} j^3 = \frac{N^2(N+1)^2}{4} \]

Answer 32

Those formulae are handy though... and going to make this rather lame.

Answer 33

for example
\[\frac{1}{n}\lim_{n\rightarrow \infty}\sum \frac{k}{n}=\frac{1}{n^2}\lim_{n\rightarrow \infty}\frac{n(n+1)}{2}\]
\[=\lim_{n \rightarrow \infty}\frac{n^2+n}{2n^2}=\frac{1}{2}\]

Answer 34

You know that
\[\sum_{j=1}^{N} 12 = 12N \]
right? Can't rely on Wolfram for everything it seems.

Answer 35

Can we finish the answer to the first question? Isn't it supposed to be in the form "Number plus or minus a number over N"?

Answer 36

be careful with the algebra here.
\[\sum_{k=1}^nk\] is a polynomial of degree 2, so you should have
\[\frac{1}{n^2}\] out front to make the limit finite. likewise
\[\sum_{k=1}^n k^2\] is a polynomial of degree 3, so it requires a
\[\frac{1}{n^3}\] out front

Answer 37

Yes, though it'll certainly take a little bit to simplify it to that point. Use the formulas I listed up there.

Answer 38

I'm doing this on paper, I've had enough LaTeX for today.

Answer 39

Scanning now...

Answer 40

I'm trying. No idea if I'm doing it right, though. http://i.imgur.com/047c1.jpg

Answer 41

While I don't really feel like going through the algebra to make sure, the idea seems right. Here's my

Answer 42

Here's my work *

Answer 43

Wait, so... I see how the limit as N approaches infinity of RN is 128/3, but what does RN equal? It's not 128/2 + 128/N + 256/3N^2, is it?

Answer 44

Oops, that should be
\[\frac{128}{3} + \frac{512}{N} + \frac{256}{N^2} \]
but yeah.

Answer 45

There's no 3 in the last denominator? (WebAssign won't accept that answer.)

Answer 46

damn it I can't type. Yeah there should be a 3.

Answer 47

Haha, I thought so.

Answer 48

Excellent. The fruit of our efforts: http://i.imgur.com/YbYmm.png

Thanks so much. I still need to work through the next two problems, but I'm just starting on them now. I'll hack away at them until I get frustrated and come back asking for help. :-P

Answer 49

Those little green keys just light up my heart. The concept is fairly simple, it's just tedious algebra. If it starts to wear on you too hard, type the summation (without the limit) into wolfram alpha and it may make your life a little easier. But make sure you know what's going on.. :)