1. ## Induction with derivates

Show by induction that the nth derivative f^n(x) of

f(x) = sqrt(1-x)

is

f^n(x) = -(2n)!/((4^n)n!(2n-1)) * (1-x)^((1/2)-n)

for n>= 1

Thanks

2. Originally Posted by Gigabyte

Show by induction that the nth derivative f^n(x) of

f(x) = sqrt(1-x)

is

f^n(x) = -(2n)!/((4^n)n!(2n-1)) * (1-x)^((1/2)-n)

for n>= 1

Thanks
no surprise here. follow the general pattern of proving things by induction. show that the formula works for the first derivative. then assume it works for the nth derivative, so the equation you posted holds. then differentiate both sides of the equation to show the formula holds for the (n + 1)st derivative. you should end up with the same formula, but all the n's replaced by (n + 1)'s

3. Would you differentiate the equation with respect to x?? Would this just not make the first part of the equation a constant??

SO pretty much how do you differentiate it?

P.s. All other induction that i have done has involved a sequence, is this not necessary for all cases.

Thanks

4. Originally Posted by Gigabyte
Would you differentiate the equation with respect to x?? Would this just not make the first part of the equation a constant??
yes, with respect to x, what elese would you do it with respect to? differentiating with respect to x is implied.

SO pretty much how do you differentiate it?
good ol' chain rule

P.s. All other induction that i have done has involved a sequence, is this not necessary for all cases.

Thanks
doesn't matter. the method of prove by induction is the same, no matter what you are working with

5. Sweet, thats what i suspected but i am unsure on how to deal with the factorial.

i got

-(1/2 - n) (- (2n)!)/((4^k) k! (2k-1)) * (1-x)^((-1/2)-k)

thanks

6. Originally Posted by Gigabyte
Sweet, thats what i suspected but i am unsure on how to deal with the factorial.

i got

-(1/2 - n) (- (2n)!)/((4^k) k! (2k-1)) * (1-x)^((-1/2)-k)

thanks
why are you mixing n's and k's? also, if you take the derivative, the power decreases by -1

7. Sorry was just using k. all k's where suppose to be n's.

-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1)) * (1-x)^((-1/2)-n)

-1/2 is found by 1/2 - 1!!!

8. Originally Posted by Gigabyte
Sorry was just using k. all k's where suppose to be n's.

-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1)) * (1-x)^((-1/2)-n)

-1/2 is found by 1/2 - 1!!!
the new power is 1/2 - n - 1 = 1/2 - (n + 1)

9. thanks for that i get that part now.

-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1)) * (1-x)^((1/2)-(n+1))

How would you expand the red part of the equation or do you not have to do that??

10. Originally Posted by Gigabyte
thanks for that i get that part now.

-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1)) * (1-x)^((1/2)-(n+1))

How would you expand the red part of the equation or do you not have to do that??
the trick here is to multiply by $1 = \frac {4(n + 1)(2n + 1)}{4(n + 1)(2n + 1)}$

simplify and you will get the desired result

11. Awesome that was really helpful thanks.

I've tried to simplify as far as possible but i got.

2(2n)!/4^(n+1)n! * (1-x)^((1/2)-(n+1)) Equation 1

Does this seem kind of right??
It's obviously not what i expected, as it should equal the original equation just with (n+1) in place of the n's.

Intermediate steps for the red part of the equation i just got where:

2(4n^3 + 4n^2 - n - 1)(2!)/ 4(4n^3 + 4n^2 - n -1)4^n(n!)

canceling the orange parts, gave me the equation 1.

12. Originally Posted by Gigabyte
Awesome that was really helpful thanks.

I've tried to simplify as far as possible but i got.

2(2n)!/4^(n+1)n! * (1-x)^((1/2)-(n+1)) Equation 1

Does this seem kind of right??
It's obviously not what i expected, as it should equal the original equation just with (n+1) in place of the n's.

Intermediate steps for the red part of the equation i just got where:

2(4n^3 + 4n^2 - n - 1)(2!)/ 4(4n^3 + 4n^2 - n -1)4^n(n!)

canceling the orange parts, gave me the equation 1.
nope, that's not completely what you want. you want (2(n + 1))! in the top right? and there should be a (2n + 1) in the denominator, long with an (n + 1)!

a bit more of a nudge in the right direction: change the top this way. split the first 4 into 2*2, use one two to multiply the (n + 1) to get (2n + 2) and the other 2 to multiply the (n - 1/2) to get (2n - 1)

so you have: $\frac {(2n + 2)(2n + 1)}{4(n + 1)(2n + 1)} \cdot (2n - 1) \cdots$

13. Originally Posted by Jhevon
nope, that's not completely what you want. you want (2(n + 1))! in the top right? and there should be a (2n + 1) in the denominator, long with an (n + 1)!

a bit more of a nudge in the right direction: change the top this way. split the first 4 into 2*2, use one two to multiply the (n + 1) to get (2n + 2) and the other 2 to multiply the (n - 1/2) to get (2n - 1)

so you have: $\frac {(2n + 2)(2n + 1)}{4(n + 1)(2n + 1)} \cdot (2n - 1) \cdots$
I thought that i was multiplying
-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1))

Where does the (n+1) on the top line of what i have from differentiating the original equation come from??

14. Originally Posted by Gigabyte
I thought that i was multiplying
-(1/2 - n) (- (2n)!)/((4^n) n! (2n-1))