Newton's identities in Sage

jtaa
17 ●2 ●4 ●11

EDIT
I actually need: $s_k=[c_1s_{k-1}+...+c_{k-1}s_1-kc_k]$ ? could somebody help me to change tobias welch's answer so that it computes $s_k$ instead of $c_k$ ?
END EDIT

I'm combining netwon's identities with le verrier's algorithm

I need some help coding the following on python.

$c_k=\frac{-1}{k}(s_k+c_1s_{k-1}+c_2s_{k-2}+\dots+c_{k-1}s_1)$
where $s_k=Tr(A^k)$ , for some square matrix A, $\forall k=1,2,3,\dots,n$

So, i'd like to type in $c(k)$ and python spits out the value for $c_k$ as defined above.

answered 12 years ago

twch
368 ●3 ●13 ●21

An recursive algorithm would be quite inefficient for this problem as yout would have to recalculate all c_k coefficients in each step.

The following few lines should work for your problem (if I hopefully didn't make any mistakes with the formulas):

def c(k,A):
    c_arr=[-A.trace()]
    for j in range(k-1):
        c_arr=c_incr(c_arr,A)
        print c_arr
    return c_arr[k-1]

def c_incr(c_arr, A):
    k=len(c_arr)+1
    c_new=(A^k).trace()
    for c_k in c_arr:
        k-=1
        c_new+=c_k*(A^k).trace()
    c_new=-1/(len(c_arr)+1)*c_new
    c_arr.append(c_new)
    return c_arr

link

Comments

it worked thanks!

jtaa ( 12 years ago )

what if i wanted a new formula for s_k=[c_1s_{k-1}+...+c_{k-1}s_1-kc_k] ?

jtaa ( 12 years ago )

You should consider learning some programming. It will save you a lot of time.

ppurka ( 12 years ago )

add a comment

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Newton's identities in Sage

1 Answer

Comments

Your Answer

Question Tools

Stats

Related questions

Newton's identities in Sage savecancel

1 Answer

Comments

Your Answer

Question Tools

Stats

Related questions

Newton's identities in Sage