Legendre Form. for Pn(0), if Pn(x)=1/(2^n n!)(d^n)/(dx^n)(x^

[danger9918]

my professor gave us a sheet for bonus marks and this came in :

Legendre formulas are defined by :
Pn(x)= (1)/(2^n n!) (d^n)/(dx^n)(x^2-1)^2 , n = 1,2,3...
find the formula for Pn(0)

please either post an explanation or the way to solving it.. thanks
p.s. if ur gna post explanation, explain the whole thing because i never studied legendre formula's before :S

 

[danger9918]

hard ?

10 views and no replies ??
is it that hard ?

 

[stapel]

Re: hard ?

10 views and no replies ??
is it that hard ?

Since "views" may come from fellow students taking a gander, and since not all tutors are qualified in the same area, one should not feel insulted or otherwise injured if "views" do not always convert into "replies".

Also, the ad in your signature, for some sort of get-rich-quick scheme, makes your post look less than legitimate, and could very likely be scaring off many viewers (who don't what their machines infected, etc).

Note: Since you need extensive lessons, and since we cannot reasonably-feasibly provide those hours of instruction within this environment, your requirement that you be taught the topic is unlikely to result in anything other than links to other pages of possible utility.

Eliz.

 

[galactus]

This looks like Rodrigues' formula. Try googling it. It's been a long while since I looked at this. (Mathematical Physics).

\(\displaystyle \L\\P_{n}(x)=\frac{1}{2^{n}n!}\cdot\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}\)

Since you have a 2 as the exponent of (x^2-1)^n, then perhaps n=2 in the rest?.

\(\displaystyle \L\\P_{2}(x)=\frac{1}{8}\cdot\frac{d^{2}}{dx^{2}}(x^{2}-1)^{2}\)

I believe \(\displaystyle \L\\P_{2}(x)=\frac{1}{2}(3x^{2}-1)\)

Then \(\displaystyle \L\\P_{2}(0)=\frac{1}{2}(3(0)^{2}-1)=\frac{-1}{2}\)

Don;t take my word for it. Do a google or something to be sure.

 

[danger9918]

This looks like Rodrigues' formula. Try googling it. It's been a long while since I looked at this. (Mathematical Physics).

\(\displaystyle \L\\P_{n}(x)=\frac{1}{2^{n}n!}\cdot\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}\)

Since you have a 2 as the exponent of (x^2-1)^n, then perhaps n=2 in the rest?.

\(\displaystyle \L\\P_{2}(x)=\frac{1}{8}\cdot\frac{d^{2}}{dx^{2}}(x^{2}-1)^{2}\)

I believe \(\displaystyle \L\\P_{2}(x)=\frac{1}{2}(3x^{2}-1)\)

Then \(\displaystyle \L\\P_{2}(0)=\frac{1}{2}(3(0)^{2}-1)=\frac{-1}{2}\)

Don;t take my word for it. Do a google or something to be sure.

ummm... sorry about that its actually (x^2-1)^n ... btw i removed the ad, hopefully that will help in getting more replies..