Question

Prove using induction:

Answer 1

For every n element of all positive integers
\[D ^{n} (\ln x) = \left( -1 \right)^{n-1}\left[ \frac{ \left( n-1 \right)! }{ x ^{n} } \right]\]

Answer 2

Begin by proving it is true for 1

Answer 3

done...

it ended in 1/x both sides...

Answer 4

I'm just confused as to how I'm going to prove K+1...

Answer 5

What is the K+1 term for the left side

Answer 6

-1^( (k+1)-1) * ( ( (k+1)-1)!/ (x^(k+1) )

Answer 7

Then It'll be just (-1)^k ( (k!)/ (x^(k+1)) )

Answer 8

@shubhamsrg Can you help me again? If it's okay with you...

Answer 9

@hartnn please guide me again in this problem...

Answer 10

\[LHS= (-1)^{((k+1)-1)}*\frac {[(k+1)-1]!}{x^{k+1}}\]

Answer 11

\[\huge =(-1)^k*\frac{k!}{x^{k+1}}\]

Answer 12

\[\huge ~that's~ what~ you ~have ~to~ prove~ \\\huge = D^{k+1}ln(k+1)\]

Answer 13

your assumption is

D(k) lnx = (-1)^(k-1) (k-1)!/(x^k) to be true

for k+1,
LHS = D(k+1) lnx = d/dx (D(k) lnx)
= d/dx ( (-1)^(k-1) (k-1)!/(x^k) ) = (-1)^(k-1) (k-1)! (-k)/(x^(k+1))

when you simplify this, you'll see its just the RHS for n=k+1

Answer 14

@Loser66 Why did you change the x into K+1? I mean, what gives us the ground to do that?

Answer 15

OOPS. I 'm sorry, the RHS still be ln x

Answer 16

however, when proving the induction step, you must divide it into 2 cases, n odd and n even.

Answer 17

no enuff! its proven! go watch happy feet

Answer 18

oh. Ahm, can you help me at the LHS? I don't know how to differentiate it. Te factorial scares me...

Answer 19

I am not sure, let me look up at my note. You may have to find out the closed form of it. Not sure I can help or not.

Answer 20

@jim_thompson5910

Answer 21

oh yea, my friend , dan said that it's proven!! look @shubhamsrg posted above. that's the proof.

Answer 22

I saw it too. But I didn't quite get how he differentiated the one with a factorial..

Answer 23

@dan815 come here to explain him/her. he doesn't understand the proof.daaaannnn, pleaaase

Answer 24

Sorry for troubling you...

Answer 25

hi, this proof by shubash works
your assumption is

D(k) lnx = (-1)^(k-1) (k-1)!/(x^k) to be true

for k+1,
LHS = D(k+1) lnx = d/dx (D(k) lnx)
= d/dx ( (-1)^(k-1) (k-1)!/(x^k) ) = (-1)^(k-1) (k-1)! (-k)/(x^(k+1))

when you simplify this, you'll see its just the RHS for n=k+1