So...Should I have waited until after people had read the problem (and thus be thinking about it) to mention it was homework?

Or is this question actually that bad?

## Search found 847 matches

- Sat Sep 26, 2009 8:13 pm UTC
- Forum: Science
- Topic: Standard addition involving HPLC
- Replies:
**3** - Views:
**1068**

- Sat Sep 26, 2009 2:12 am UTC
- Forum: Science
- Topic: very basic physics question (thats driving me nuts)
- Replies:
**9** - Views:
**4755**

### Re: very basic physics question (thats driving me nuts)

u as in the coefficient of static friction? What's acting on the box on the left? From your diagram, it dosen't look like there would be any forces acting on it (other than normal and gravitational, which would cancel) :? Are the boxes attached by a rope or something to that effect? I presume that 7...

- Thu Sep 24, 2009 10:06 pm UTC
- Forum: Science
- Topic: Standard addition involving HPLC
- Replies:
**3** - Views:
**1068**

### Standard addition involving HPLC

Yes, this is homework. So, those of you still with me, the question goes as follows: "The concentration of retinol (vitamin A) in serum was determined by HPLC with UV absorbance detection. A 1.00 mL sample of serum was placed in a 10 mL test tube. To precipitate the proteins, 800.0 microL of et...

- Sun Sep 20, 2009 9:15 pm UTC
- Forum: Mathematics
- Topic: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?
- Replies:
**8** - Views:
**1223**

### Re: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?

I think the thing with showing the coefficients are equal must have been what my book was refering too.

It did say "...easy to show by multiplying the series...", and the proof using the derivative dosen't really multiply.

It did say "...easy to show by multiplying the series...", and the proof using the derivative dosen't really multiply.

- Sun Sep 20, 2009 3:33 pm UTC
- Forum: Mathematics
- Topic: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?
- Replies:
**8** - Views:
**1223**

### Re: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?

So, I've had some sleep, and am now making significantly less typos. Binomial expansion for (i+j) isn't fun. Also not something I've done for awhile. so... (a+b)^(i+j)= summation from k=0 to k=i+j of (i+j)!/(k!*(i+j-k)!) * a^(i+j-k)*b^k. Each of those terms is gonna be over (i+j)!, so the coefficien...

- Sun Sep 20, 2009 6:37 am UTC
- Forum: Mathematics
- Topic: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?
- Replies:
**8** - Views:
**1223**

### Re: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?

So I want to come up with some f(i,j)=coefficient for each side, and I'll discover that both sides would result in the same coefficient therefore equal? Sounds like a good idea, except so far I've had a tendency to get completely muddled after the terms get past the power of 2. Also, I have no idea ...

- Sun Sep 20, 2009 5:57 am UTC
- Forum: Mathematics
- Topic: Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?
- Replies:
**8** - Views:
**1223**

### Proof that e^(z_1+z_2)=e^(z_1)*e^(z_2) using power series?

Hello xkcd people who all seem to act like everything below a masters degree equivalent in math is childs play! :P I'm looking for help in proving that e^(z_1+z_2)=e^(z_1)*e^(z_2) using the power series for e^z as the basis, where z is a complex number. The numbers are intended as subscripts, not ad...