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 $a_n$ decreasing (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 $\sum a_n$.

Now, draw a curve through the upper left 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 beneath the horizontal line on each rectangle. That is, the area beneath the curve, in each rectangle, is less than the area in that rectangle and the total area under the curve, from x= 1, is less than the total area of the rectangles.

That is,
$$\int_1^\infty f(x)dx< \sum_{n=1}^\infty a_n$$

Taking f(x)= $1/x^3$ when $f_n= 1/n^3$ does exactly that.

 

[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 value. The inequalities n > 32, n > 32.2, n > 32.1 are all equivalent to n >= 33, for n an integer.