Messages

1

Quiz-B / Re: Quiz-B P2

« on: April 08, 2018, 07:04:47 PM »

So at the last part, I let f(x)=x^n and take n times derivatives to get (*).
The first half is aimed at finding a general conclusion.

2

Quiz-B / Re: Quiz-B P2

« on: April 08, 2018, 01:51:31 PM »

I got an idea to prove the formula * shown in the reply part.

3

Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6

« on: April 08, 2018, 12:50:30 PM »

I think there is a type of question c). Should be [ln(x+epsilon i)]', here the power of v is 1 and should not be shown.
Also for the second half part of this question, should we use the equality from b) rather than from a)?

4

Term Test 1 / Re: P1

« on: April 07, 2018, 09:41:44 PM »

Answer for this question

5

Quiz-7 / Re: Thursday's quiz

« on: April 02, 2018, 07:28:17 PM »

u=xyz-P(x, y, z)(x2+y2+z2-R2)
u=-[Pxx(x2+y2+z2-R2)+2xPx+2xPx+2P+Pyy(x2+y2+z2-R2)+2yPy+2yPy+2P+ Pzz(x2+y2+z2-R2)+2zPz+2zPz+2P]=0
Rearranging it we can get:
P(x2+y2+z2-R2)+4(xPx+yPy+zPz)+6P=0
As P is rotational symmetric, we first consider for P(x, x, x)=mx+n
Then 3Pxx(3x2-R2)+12Pxx+6P=0
Plug in P=mx+n, Px=m, Pxx=0, we can get m=n=0
So P(x, y, z)=0
Then u=xyz