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:

Mathematicapg329num16 MA181Fall2008bell.PNG

{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} $

--John Mason

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