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A question to solve

by therector24
Tags: solve
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therector24
#1
Oct17-07, 10:53 AM
P: 24
hey
hi,
iam new here and one of my friends told me about this site
it amazing and nice.

lets start the real things;

iam stuk in finding a solution 4 this question

can u solve this
http://www.physicsforums.com/attachm...4&d=1192636318

and where n=0,1,2,3,4,5,.................
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CompuChip
#2
Oct17-07, 11:01 AM
Sci Advisor
HW Helper
P: 4,300
Welcome to PF therector24.
Policy here is that we don't just give away the answer, but rather a hint on how to obtain it. Therefore, please show us some work, what you got so far. Then we will be able to point you in the right direction.

If you didn't do anything yet: start by finding the n'th derivative of the polynomial in that function.
therector24
#3
Oct18-07, 01:59 AM
P: 24
thank u for your interesting
well, i tried to substitute with finding the derivative
till n= 6 and all of the answers were 0
except when n=0,2
the answer was 1 & .5
so here iam stuk in finding the formula
if all of the answers were zeros it was going to be nothing
i just want an idea and i will solve it
and if i knew the answer i will post it as fast as i can
thank u.

therector24
#4
Oct19-07, 02:22 PM
P: 24
A question to solve

hey, i found the formula but the problem that it is not right when we say n=0,1,2 but after that its correct so can i make an exception on that waiting for reply!!!
inferi
#5
Oct20-07, 03:23 AM
P: 16
I am also stuck in this formula ,I tried to derive seven times by only substituting the n untill n=7,I did that because i wanted to find something that i can build my formula but i could not.
anyone can help?
sennyk
#6
Oct20-07, 10:30 AM
P: 73
I found the solution this morning. There is a recursive relationship.

[tex] f(x) = (x^2-1)^n[/tex]

[tex]\frac{d^n}{dx^n}(f(0))= (\frac{d^(^n^-^2^)}{dx^(^n^-^2^)}f(0))(M_n)[/tex]

[tex]\frac{d^0}{dx^0}(f(0))= 1[/tex]

[tex]\frac{d^1}{dx^1}(f(0))= 0[/tex]

[tex]M_2 = -4[/tex]

[tex]M_n = M_n_-_2 - 32(\frac{n}{2} - 1)[/tex]

[tex]M_o_d_d[/tex]

is not necessary, because odd n values of the derivative are zero.

You still have to multiply by the other term. [tex]\frac{1}{(2^n)(n!)}[/tex]
therector24
#7
Oct20-07, 12:55 PM
P: 24
so about the formula
there is a problem that u have to derive in order to have m
the porpuse of the formula is that u can get the answer just by substituting but here i have to find m after finding the derivative.
u mean m is the last number after deriving
but how to find m with out finding the derivative???????
sennyk
#8
Oct20-07, 01:29 PM
P: 73
You don't have to find the derivative. This function recurses back to the 0th derivative, which means that you don't find the derivative. You just plug in 0 for x. Here's an example of the iteration.

[tex] n = 0 [/tex]

[tex] \frac{d^0}{dx^0}[f(0)] = 1 [/tex]


[tex] n = 2 [/tex]

[tex] M_2 = -4 [/tex]

[tex] \frac{d^2}{dx^2}[f(0)] = 1(-4) = -4 [/tex]


[tex] n = 4 [/tex]

[tex] M_4 = -4 - 32(4/2 - 1) = -4 - 32 = -36 [/tex]

[tex] \frac{d^4}{dx^4}[f(0)] = -4(-36) = 144 [/tex]

etc...

You don't have to compute any derivatives; to derive this I had to compute many derivatives (Maple helps here). Then I found the recursive relationship (aka pattern) that works when x = 0.
therector24
#9
Oct20-07, 01:53 PM
P: 24
here we don't want to use maple we want a formula, that means when u substitute you get the finall asnwer directly (by hand)
sennyk
#10
Oct20-07, 02:02 PM
P: 73
You don't use maple. You use the recursive formula.

Here's an easier example of recursion.

[tex] H(x) = x[H(x-1)] [/tex]
[tex] H(0) = 1 [/tex]

There is another representation of this formula. Can you figure out what it is?

If this is too difficult for you, ask someone what recursion is. I'm sure some one on one help will help you understand this.
therector24
#11
Oct20-07, 03:01 PM
P: 24
thank u so much u helped me alot
inferi
#12
Oct20-07, 03:10 PM
P: 16
thank u sennyk u were a great help


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