## 8.8 #54

Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --John Mason

It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing $\sqrt{x^4-1}$ and $\sqrt{x^4}$ It was easy from there. --Josh Visigothsandwich

Thanks. I'm not really sure that I needed to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --John Mason

It does not converge for me either, but Josh, be careful with your comparison. It's not really valid to use $1/\sqrt{x^4}$ as the function for comparison, because $x/\sqrt{x^4} < x/\sqrt{x^4-1}$, albeit infinitesimally less. I used $(g(x) = 1/x^{0.99}) > (f(x) = x/\sqrt{x^4-1})$, in which case P < 1 and diverges. --Randy Eckman 15:44, 26 October 2008 (UTC)

That is true. But if you start out with that as your comparison:

$\sqrt{x^4} > \sqrt{x^4 - 1}$

$\text{For } x > 0$

$x\sqrt{x^4} > x\sqrt{x^4 - 1}$

$\frac{x}{\sqrt{x^4 - 1}} > \frac{x}{\sqrt{x^4}} = \frac{x}{x^2} = \frac{1}{x}$

$\int^\infty_4 \frac{dx}{x} = \infty$

Certainly much cleaner than working with decimal powers. --John Mason

Also, remember, $\int_1^{\infty}\frac{1}{x^p}dx$ diverges as long as $p\le 1$. So if p = 1, it still diverges. That's why it's okay to use the comparison I used. I did what John did, only I multiplied the inequality by x after taking the inverse of both sides and switching the inequality sign.His Awesomeness, Josh Hunsberger

## Pg. 329, #16

How accurate do we need to make our answers for the roots? After four iterations, I have the first point accurate to four digits. The text doesn't specify a number of correct digits, and out of curiosity I found the precise roots on Mathematica. I don't know how the textbook could expect us to calculate exactly this:

{x -> 0.630115} {x -> 2.57327}

--Randy Eckman 17:43, 26 October 2008 (UTC)

I went to 10 digits, as that was all my calculator could show. And for the record "Reckman" is a very cool name. --John Mason

And I went to five digits because that's all my calculator would show me (I think I can change that, but i wasn't sure how). His Awesomeness, Josh Hunsberger

## Pg.329, #5

I don't really understand how to do #5. It seems like there isn't an actual function. Are we supposed to use maple? Can someone help get me started?

--Klosekam 16:19, 27 October 2008 (UTC)

Since you need to know where

$e^{-x} = 2x + 1$

you can just subtract out one side and solve for the roots (aka at what value of x the function takes zero). So you can use

$f(x) = 2x + 1 -e^{-x}$