Mr.Stewart is confusing me. Estimating sums

[flyingpig]

Homework Statement

[PLAIN]http://img842.imageshack.us/img842/1404/unledro.png Now what I don't understand is this part

[PLAIN]http://img821.imageshack.us/img821/7479/unledoq.png

Is he trying to say that $$\frac{1}{2n^3} = R_n$$? Is that how he deduced that $$\frac{1}{2n^3} < 0.0005$$?

 

[Mark44]

No, it's saying that R_n <= 1/(2n²), and if he can find an n for which 1/(2n²) < 0.0005, then R_n will be < 0.0005.

 

[HallsofIvy]

Given the sum [itex]\sum a_n[itex]with $a_n$ <b>decreasing</b> (and, in order that the sum be convergent, they must be decreasing to 0), mark those points $(n, a_n)$ on a graph, draw vertical lines from each such point, and a horizontal line from each to the left so that you get a sequence of rectangles, each having width 1, height $a_n$ and so area $a_n$. The total area in all those is the sum $\sum a_n$. <br /> <br /> Now, draw a curve through the upper <b>left</b> corner of each rectangle. Call that function f(x). Then $f(n)= a_n$ for each positive integer n. Also, notice that the curve passes <b>beneath</b> the horizontal line on each rectangle. That is, the area beneath the curve, in each rectangle, is <b>less</b> than the area in that rectangle and the total area under the curve, from x= 1, is <b>less</b> than the total area of the rectangles.<br /> <br /> That is, <br /> $$\int_1^\infty f(x)dx< \sum_{n=1}^\infty a_n$$<br /> <br /> Taking f(x)= $1/x^3$ when $f_n= 1/n^3$ does exactly that.[/itex][/itex]

 

[flyingpig]

No, it's saying that R_n <= 1/(2n²), and if he can find an n for which 1/(2n²) < 0.0005, then R_n will be < 0.0005.

So then $$R_n\leq \frac{1}{2n^3}$$, but he also says $$R_n\leq 0.0005$$, so how could he turned it into

$$\frac{1}{2n^3}<0.0005$$

 

[SammyS]

If he chooses n so that $$\frac{1}{2n^3}<0.0005\,,$$ then $$R_n\leq \frac{1}{2n^3}<0.0005\quad\to\quad R_n<0.0005$$

 

[Mark44]

I'm looking at what you posted, and it says

$$R_n\leq \frac{1}{2n^2}$$
Note the exponent of 2, not 3.

Stewart doesn't say that 1/(2n²) < 0.0005. He says he wants to find n so that this is true.

If he can find such a number n, then he will have R_n < 0.0005.

 

[flyingpig]

Thanks for catching that Mark44 lol, I didn't realize I had a 3 instead of a 2. I don't know why I keep thinking it is a 3.

I just want to ask, when he solved it, you should get $$n>\sqrt{1000}$$

$$\sqrt{1000}\approx 31.6$$ and you round up to 32 because there is no such thing as "half a term", but what if n $$\sqrt{1000}\approx 31.2$$ (31.2 I made up randomlly), would we still choose n to be 32 or 31? At first I thought it would be 32, but my book chooses 31.

It was an another example in my book.

 

[Mark44]

Thanks for catching that Mark44 lol, I didn't realize I had a 3 instead of a 2. I don't know why I keep thinking it is a 3.

I just want to ask, when he solved it, you should get $$n>\sqrt{1000}$$

$$\sqrt{1000}\approx 31.6$$ and you round up to 32 because there is no such thing as "half a term", but what if n $$\sqrt{1000}\approx 31.2$$ (31.2 I made up randomlly), would we still choose n to be 32 or 31? At first I thought it would be 32, but my book chooses 31.

It was an another example in my book.

For 1/(2n²) < .0005, you need n > sqrt(1000) ~ 31.6, but you also want n to be an integer, so you take n = 32. If you chose n = 31, then 1/(2n²) ~ .00052, which is larger than .0005.

If you needed to find n > sqrt(973) ~31.2, I would take n = 32. If your book chooses a different number, I would need to see more information about this other problem.

 

[flyingpig]

[PLAIN]http://img35.imageshack.us/img35/1369/unledlh.png

For part c), they did this

$$\frac{1}{3n^3}<\frac{1}{10^5}$$

$$\frac{1}{n^3}<\frac{3}{10^5}$$

Now here is where it gets confusing, I did some algebra and I found you eventually have to flip over the inequality sign, but if I were to just invert both sides, does that mean I also have to change inequality? Anyways

[PLAIN]http://img64.imageshack.us/img64/9656/unledif.png

 

[Mark44]

[PLAIN]http://img35.imageshack.us/img35/1369/unledlh.png

For part c), they did this

$$\frac{1}{3n^3}<\frac{1}{10^5}$$

$$\frac{1}{n^3}<\frac{3}{10^5}$$

Now here is where it gets confusing, I did some algebra and I found you eventually have to flip over the inequality sign, but if I were to just invert both sides, does that mean I also have to change inequality?

Yes. Assuming that a and b are both positive, a < b ==> 1/a > 1/b.
A geometric explanation is that if b is to the right of a, then 1/b will be a smaller number than 1/a.

Anyways

[PLAIN]http://img64.imageshack.us/img64/9656/unledif.png[/QUOTE]

They found n $\approx$ 32.2, so they're taking n > 32. Since n is an integer, this means that n >= 33.

 

[flyingpig]

But why n > 32? The approxixmate is 32.2, which means that it should be n > 32.2

Why choosen n > 32? If it was 32.1, wouldn't the error be no longer within 10^-5?

 

[Mark44]

Keep in mind that n is an index, so it's an integer[/color] value. The inequalities n > 32, n > 32.2, n > 32.1 are all equivalent to n >= 33, for n an integer.