Preserved Topic: lil' calc siggie

Total rip off weadah (originally), eye, and silence, but I have my calc ap test tomorrow, and I needed a break from studying. Just a quickie...err not so quick actually.



Again, I know this is completely unoriginal so don't rag on me for it.

[This message has been edited by mbridge (edited 05-10-2001).]

[This message has been edited by mbridge (edited 05-10-2001).]

I think I'm just gonna make a sig like that with Weadah's name on it.

wouldnt Jestahs name be more origianl Taxon?.. *grin*

i love it, mbridge.

Reminds me of my TI-92+

yea, looks like my '89...

[This message has been edited by mbridge (edited 05-10-2001).]

reminds me of 4U HSC.... good ol' aussie exams...


___________________
b u n c h a p i x e l s

good gosh. i cant believe this.

that is really cool.

yeah.

and gee whizz all these are good. =)

like it alot mbridge, really tight and clean man.

cool

[This message has been edited by Weadah (edited 05-10-2001).]

Good news, my exam turned out to be extremely easy. There was only one multiple choice problem that I couldn't do. I guess they've watered all the AP exams down just like every other exam.



[This message has been edited by mbridge (edited 05-10-2001).]

Cool sig. I thought the exam was fairly easy also, although the second set of long answer questions almost got me, I had a tough time with them, couldn't do one of the integration things (integration isn't my strength), and the taylor's theorem question almost caught me, but I managed to get it right anyway, I think.

- Slime

I have to agree with you on that one. Question 6 Free Response was by far the hardest. There was also a tricky U-substitution one on the no calculator free response part.

The question I ommitted was a power series expansion...we never covered that in class. It was x^2/(1-x^2), I think. Any insight on that one?


[This message has been edited by mbridge (edited 05-10-2001).]

Oh, I remember that one. It was sort of a trick question. They said this:

"the taylor series for 1/(1-x) is 1 + x + x^2 + x^3 + x^4 + ..."

then they asked you to find the expansion for x^2/(1-x^2)

Well, you don't actually have to find the taylor series expansion for that, although you could do it, but it'd take you a long time. You have to think of it like this:

let f(x) = 1/(1-x)
we also know that f(x) = 1 + x + x^2 + x^3 + ...

now you must realize that f(x^2) = 1/(1-x^2)
therefore f(x^2) = 1 + x^2 + x^4 + x^6 + ...

and x^2 * f(x^2) = x^2/(1-x^2)
so x^2 * f(x^2) = x^2 + x^4 + x^6 + ...

And that's the answer. Well, I'm not sure that was the exact question they asked, but that was the general logic that you had to use.

I think I missed the u-substitution one you mentioned. =) I actually did part C of that problem (solving the differential equation y' = -3xy) to find the integral they asked for. Then I got to part C and realized I had done part A wrong. =) Did you get -4?

- Slime

Yea, that's how I was going to do the power series expansion, substituted in x^2 for x and then multiplied by x^2 but the answer wasn't there.....

For question 6?

I know the limit one (definition of a derivative) was 2/9,
D was 3/2, I don't remember what the other questions ask. What were they?

Was question C the interval of convergence one? I got something like -4 <= x < 4

I don't remember an answer of -4....


[This message has been edited by mbridge (edited 05-11-2001).]