Question

Proof with induction..

Answer 1

\[f ^{(n)}(x)=(-1)^{n}n!\left[ \frac{ 1 }{ x ^{n+1} }+\frac{ 1 }{ (x+1)^{n+1} }+\frac{ 1 }{ (x+2)^{n+1} } \right]\]

Answer 2

I do this but I have no idea what's the proof here or if this is correct. Do I just leave it like this or add something? Not really sure how to prove it.

Answer 3

ok does (n) note in f means derevative right ?

Answer 4

@rational

Answer 5

yes n as in nth derivative

Answer 6

so f^0 should give main function right ?

Answer 7

\(f (x)= \left[ \frac{ 1 }{ x }+\frac{ 1 }{ (x+1) }+\frac{ 1 }{ (x+2) } \right] \)

Answer 8

yes, that's the main one

Answer 9

now for k=1 , check from the given statment and check from derivative

Answer 10

then go through induction ,
assume its true for k
\(\large f ^{(k)}(x)=(-1)^{k}k!\left[ \frac{ 1 }{ x ^{k+1} }+\frac{ 1 }{ (x+1)^{k+1} }+\frac{ 1 }{ (x+2)^{k+1} } \right] \)


then show for k+1
\(\large f ^{(k+1)}(x)=(-1)^{k+1}({k+1} )!\left[ \frac{ 1 }{ x ^{k+2} }+\frac{ 1 }{ (x+1)^{k+2} }+\frac{ 1 }{ (x+2)^{k+2} } \right] \)
lets see if we could seperate some terms

Answer 11

What do you mean by separate terms? I got until this point but I don't know what else I need to do to prove it. @BSwan

Answer 12

well , f^k+1 is the first derevative of f^k , right ?

Answer 13

if you have this function , how do you derevative it
\(\large f ^{(k)}(x)=(-1)^{k}k!\left[ \frac{ 1 }{ x ^{k+1} }+\frac{ 1 }{ (x+1)^{k+1} }+\frac{ 1 }{ (x+2)^{k+1} } \right] \)

Answer 14

I just make k bigger to find the nth derivative that I want.

Answer 15

this is the induction step , i would helpyou in detali but im bze :(
but i guess this hint is enough for you to continue ,
hope maybe @rational ( if you have time ) , incase you need something farther

Answer 16

just imagin
\(\large g(x) =(-1)^{k}k!\left[ \frac{ 1 }{ x ^{k+1} }+\frac{ 1 }{ (x+1)^{k+1} }+\frac{ 1 }{ (x+2)^{k+1} } \right]\)

find g'(x) , it will be the same to f^(k+1) (x)
got it now ?

Answer 17

http://openstudy.com/study#/updates/540314f7e4b0e197b3e2e06c

Answer 18

yep kinda, it's a messy derivation tho