Interesting Question (1 Viewer)

[bored.of.u]

Found this question fun and interesting =D:
Prove d<SUP>n</SUP>/dx<SUP>n </SUP>[1/(1-x)] = n!/(1-x)<SUP>n+1 </SUP>for all positive integers n.

 

[drowning_pocoyo]

For n = 1

LHS = d/dx (1/1-x)
LHS = -1.(1-x)^2.-1
LHS = 1/(1-x)^2
RHS = 1!/(1-x)^2
RHS = 1/(1-x)^2
Hence the statement is true for n = 1

Assuming the statement is true for n = k
d^k/dx^k (1/1-x) = k!/[(1-x)^k]

For n = k + 1
d^k+1/dx^k+1 (1/1-x) = (k+1)!/[(1-x)^k+1]

LHS = d/dx [d^k/dx^k (1/1-x)]
LHS = d/dx {k!/[(1-x)^k+1]} (inductive assumption)
Here k is a constant so d/dx k! = 0
LHS = [-k!.(k+1).(1-x)^k.-1]/[(1-x)^(2k+2)]
LHS = [(k+1)!.(1-x)^k]/(1-x)^(2k+2)]
LHS = (k+1)!/[(1-x)^(k+2)]
LHS = RHS

Hence the statement holds for n=k+1 when n=k holds and by mathematical induction it is true for all positive integers.

 

[untouchablecuz]

For n = 1

LHS = d/dx (1/1-x)
LHS = -1.(1-x)^2.-1
LHS = 1/(1-x)^2
RHS = 1!/(1-x)^2
RHS = 1/(1-x)^2
Hence the statement is true for n = 1

Assuming the statement is true for n = k
d^k/dx^k (1/1-x) = k!/[(1-x)^k]

For n = k + 1
d^k+1/dx^k+1 (1/1-x) = (k+1)!/[(1-x)^k+1]

LHS = d/dx [d^k/dx^k (1/1-x)]
LHS = d/dx {k!/[(1-x)^k+1]} (inductive assumption)
Here k is a constant so d/dx k! = 0
LHS = [-k!.(k+1).(1-x)^k.-1]/[(1-x)^(2k+2)]
LHS = [(k+1)!.(1-x)^k]/(1-x)^(2k+2)]
LHS = (k+1)!/[(1-x)^(k+2)]
LHS = RHS

Hence the statement holds for n=k+1 when n=k holds and by mathematical induction it is true for all positive integers.

no need for induction, i think

D_(0)=(1-x)^-1
D_(1)=0!(-1)(-1)(1-x)^-2=1!(1-x)^-2
D_(2)=1!(-2)(-1)(1-x)^-3=2!(1-x)^-3
D_(3)=2!(-3)(-1)(1-x)^-4=3!(1-x)^-4
D_(4)=3!(-4)(-1)(1-x)^-5=4!(1-x)^-5

...

D_(n)=(n-1)!(-n)(-1)(1-x)^-(n+1)=n!(1-x)^-(n+1)

does this constitute a proof?

 

[drowning_pocoyo]

no need for induction, i think

D_(0)=(1-x)^-1
D_(1)=0!(-1)(-1)(1-x)^-2=1!(1-x)^-2
D_(2)=1!(-2)(-1)(1-x)^-3=2!(1-x)^-3
D_(3)=2!(-3)(-1)(1-x)^-4=3!(1-x)^-4
D_(4)=3!(-4)(-1)(1-x)^-5=4!(1-x)^-5

...

D_(n)=(n-1)!(-n)(-1)(1-x)^-(n+1)=n!(1-x)^-(n+1)

does this constitute a proof?

thatss nice (Y) i dunno for proofs i thought it common to do LHS = RHS

 

[Iruka]

no need for induction, i think

D_(0)=(1-x)^-1
D_(1)=0!(-1)(-1)(1-x)^-2=1!(1-x)^-2
D_(2)=1!(-2)(-1)(1-x)^-3=2!(1-x)^-3
D_(3)=2!(-3)(-1)(1-x)^-4=3!(1-x)^-4
D_(4)=3!(-4)(-1)(1-x)^-5=4!(1-x)^-5

...

D_(n)=(n-1)!(-n)(-1)(1-x)^-(n+1)=n!(1-x)^-(n+1)

does this constitute a proof?

Not really.

What you've done is just a non-rigorous form of induction.

 

[untouchablecuz]

Not really.

What you've done is just a non-rigorous form of induction.

yer it seems that way

any other way to prove it besides induction?

 

[lychnobity]

yer it seems that way

any other way to prove it besides induction?

The Lych Method:

It's obvious.

Can't think of any other way than induction or what you did

 

[untouchablecuz]

The Lych Method:

It's obvious.

Can't think of any other way than induction or what you did

hahahaha, i googled "the lych method" before i saw your username

also, did you get my email? (essay)

 

[lychnobity]

hahahaha, i googled "the lych method" before i saw your username

also, did you get my email? (essay)

Yeah, I finished about 10 mins ago, feel free to send us another message, I'm test free for 2 months

 

[untouchablecuz]

Found this question fun and interesting =D:
Prove dⁿ/dxⁿ[1/(1-x)] = n!/(1-x)ⁿ⁺¹for all positive integers n.

hence find the half derivative of 1/(1-x)