Recent content by calcnd

As for your other point about solving it in another fashion, I could add n 1's to the right of the last 1 on the string, then evenly partition them. But that still requires keeping track of how many 1's have been encountered.

Yes, but I need to be able to go back and fourth keeping track of what I encounter. So, for example, if the input string is 1111 I need to get an output of 1110111. But I can't "count" the three ones and then tell it to write three, as the machine can't remember what it's encountered other...

I'm so very confused on how to go about these problems. Define a TM that for every n duplicates a string of the form 1^n, creating 1^n 0 1^n. Does the machine calculate any function? We're using the notion that a string of n+1 n's represents n. Basically, I've surmised that I need a...

Not really. The professors themselves were the students who excelled and took an interest, whereas the general decline in the student body should probably be attributed to attitude more than anything. Mean IQ is not going down, as far as I'm aware. Speaking as an eighteen-year-old who is to...

Well, are there 100 cm per km or 100,000? I'd set it up like this, anyways. \frac{change in distance}{change in time} =\frac{\Delta d}{\Delta t} \frac{3 cm}{year} = \frac{\Delta d}{\Delta t} {\Delta t} = \frac{\Delta d}{3 \frac{cm}{year}}

I did it quickly, so my answer might be off, but keep doing L'Hopital's rule until you get a proper answer. (I got \frac{1}{6}, by the way. Basically, just keep differentiating until you get rid of the 1 in the e^{x}-1 bracket, as then you'll get a proper answer.

I can. But how does that help me determine the value of X at which G(X) equals zero? :(

As stated above, L'Hopital's rule, or, if that's not allowed, rationalize the numerator.

Is a mathematical markup language called LaTeX. Here's some more info for this forum's LaTeX engine: https://www.physicsforums.com/showthread.php?t=8997

Heh... I noticed I did my derivative entirely wrong. I guess that's why you shouldn't use the computer when you're tired. Anyways, G(0)=0^2-e^(\frac{1}{1+0}) =-e Which is < 0 G(2)=2^2-e^{\frac{1}{1+2}} =4-e^{\frac{1}{3}} Which is > 0 Meaning G(x)=0 is contained somewhere between x=0 and...

Hrm, that's what I'm not sure of. I know the x value is within the interval, due to the intermediate value theorem, as you stated. I could suggest that I solve G'(x)=0, but I could imagine that would be useless as there's no reason to think that G(c)=0 is a critical point. :S

Use the Intermediate Value Theorem and/or the Mean Value Theorem and/or properties of G'(x) to show that the function G(x) = x^2 - e^{\frac{1}{1+x}} assumes a value of 0 for exactly one real number x such that 0 < x < 2 . Hint: You may assume that e^{\frac{1}{3}} < 2 . So I'm completely lost...

I have no idea how to do permutations properly, so my answer is obviously incorrect. I go a simplified expression of 16[2!(4!)] = 768 I'd be interested to learn how to do this properly.

Thank-you for clarifying that.

Covalent bonds involve the sharing of electrons to gain a stable electron configuration. They do not become noble gases. (You are correct to think that the group 8 gases are inert, though.) I'm unsure if you've done electronegativity, but a covalent bond occurs when the electronegativity...