Conquering Calculus

[StupidGuy]

I have decided to study Calculus independently (that is, not as part of a course or program--purely on my own).

I like to spend as much time on a problem as I can before I ask for help, because I believe that I will remember the problem-solving method better if I go through the frustration of failure before being told where I went wrong.

With that in mind, I will use this thread to post questions whenever I get stuck. Hopefully someone will be nice enough to help me out.

I'm only on section 1.2, and I'm already stuck. Here's the problem (by vsubj, fsubj, et cetera, I mean a function letter followed by a subscript):

Find the velocities vsub1, vsub2, vsub3 and formulas for vsubj and fsubj.

f = 0, 1, 0, 1
v = 1, -1, 1
vsubj = -1^j+1

So far so good. The formula for vsubj works (the first element in the series is vsub1, not vsub0).

Now, I've been trying to come up with a formula for fsubj for over half an hour now(the first element of f is 0, not 1).

I'm a verbal thinker, so my thought process has gone like this:

"I need a function that will generate 0 for odd inputs and 1 for even inputs or vice versa, because the elements of the series follow an alternating pattern and the only relevant difference between element numbers (j) is that some are odd and some are even.

I know that I can generate -1 and 1 given odd or even inputs by raising -1 to exponential powers. So now I need to find a formula that will generate 0 for -1 and 1 for 1, or vice versa."

At this point I came up with fsubj = -1^j-1 - (-1^j-1), but it doesn't quite work. However, I didn't see any other way to generate 0 or 1 depending on whether or not the input is odd or even. So I spent five minutes banging my head against my desk, and then I came here.

I'm sure that this is a very simple problem and that I'm just missing something obvious or not thinking about the problem in the right way.

 

[Unco]

You found \(\displaystyle \mbox{v_j = {(-1)}^{j + 1}}\) generated \(\displaystyle \mbox{ v_j = 1, -1, 1}\) where \(\displaystyle \mbox{j}\) began at 1.

We want something similar for \(\displaystyle \mbox{f_j = 0, 1, 0 , 1}\) where \(\displaystyle \mbox{j}\) begins at 0.

Notice that if we add 1 to each term of \(\displaystyle \mbox{v_j}\) and divide by 2 we get \(\displaystyle \mbox{1, 0, 1}\). Also note that if \(\displaystyle \mbox{j}\) began at 0, \(\displaystyle \mbox{v_j = -1, 1, -1}\).

 

[Gene]

Just some style comments.
Vsubj can be typed as
V_j = V_j
You can also do
V^j = V^j

Don't spare the ()'s
typing V^j+1 needs them.
v^(j)+1 is different than v^(j+1) and we can't be sure which you mean without them. Even more with functions.
E.g sin x+5 as sin(x)+5 or sin(x+5)
At the top of the screen are 4 buttons. One of them will have a drop-down menu that covers LaTex which does all sorts of math-magic. (I can't get there on my ISP so I can't be too specific.) Most of the posts use that.

 

[StupidGuy]

Thanks for your help, Unco. And Gene, thanks for the style suggestions.

Okay, here's another one:

Suppose v_j = r^j. Show that f_j = (r^{j + 1} - 1)/(r - 1) starts from f₀ = 1 and has f_j - f_{j - 1} = v_j. (Then this is the correct f_j = 1 + 4 + ... + r^j = sum of a geometric series.)

Okay, showing that f₀ = 1 is easy. I had no problems with that. But then when I try to go about proving that f_j - f_{j - 1} = v_j = r^j I run into problems.

So this is my thought process so far: all I have to do is set f_j - f_{j - 1} equal to r^j and then do some algebraic maniuplation to reach an obvious identity. Right? Okay:

(r^j+1 - 1)/(r - 1) - (r^j - 1)/(r - 1) = r^j

Now, I'm not sure if I even set this up correctly. If f_j is (r^{j + 1} - 1)/(r - 1) then f_{j - 1} should be (r^j - 1)/(r - 1) right? If not, then that coudl be where I screwed up. But if the equation is set up correctly, then:

[(r^{j + 1} - 1) - (r^j - 1)]/(r-1) = r^j

I'm not sure how to manipulate this equation to reach an identity. I think that the js are throwing me a bit. I mean, there's no rule of exponents for subtracting a^b - a^b-1. I guess I'm just lost here.

I probably don't even have it set up right. -_- In fact, since I don't see to able to figure this out on my own, I probably shouldn't even be studying Calculus...

Edit: I think the numbers inside the brackets changed my font size, but I'm not sure what the HTML tag is to insert a bracket without the parser interpeting it as code.

Edit by TKHunny: Repaired missing HTML tag.

 

[Gene]

You're making too much of it.
[(r^{j + 1} - 1) - (r^j - 1)]/(r-1)=
[r^j+1 - r^j]/(r-1) =
[r^j(r-1)]/(r-1) =
r<sup>j

(Looks well typed to me)

 

[tkhunny]

Why would a guy named "StupidGuy" be trying to study calculus on his own?

I'm just asking.

 

[StupidGuy]

You know, I knew that I had to end up with something multiplied by r - 1 in the numerator, and that that something had to be r^j, so I probably should have just tried it and then checked to see if it worked. Duh...sometimes you have to start with an ideal solution and then see if you can find a way to get there without cheating! Well, I'm glad that I failed to solve this, because now I've learned something important.

And tkhunny, I'm not stupid, but I'm not particularly talented when it comes to math--hence "StupidGuy". In fact, if I took an IQ test I'd probably score in the 140-150 range (on a 15 SD Stanford-Binet test), which would lead one to believe that I'm better at math than I really am. My "verbal intelligence" is probably significantly higher than 150, and my "math intelligence" is certainly significantly lower, so it would average out.

But yeah, I'm a smart guy who's not so great at math.

 

[tkhunny]

Seems like an attitude problem or a self-fulfilling prophecy, nonetheless.

Pull through.

 

[StupidGuy]

Here's a problem that doesn't even make sense to me. It's not that it's difficult, I just have no idea what the expression is supposed to mean.

U(t) = step from 0 to 1 at t=0.

...Huh? How can a variable equal a "step"?

Graph U(t-1).

Okay, so the slope is...what? One, because t has no coeffecient, or "step from 0 to 1"? If it's a slope of 1, how can it jump from 0 to 1 at t=0? Is "step from 0 to 1" another way of just adding 1 to the entire function, giving it a y-intercept of 1?

Uh, yeah...this problem doesn't really make sense to me. Maybe I'm not familiar with the letter U's usage in mathematics...?