Can you get the answer to a really hard math problem?

[Zero Plasma]

Problem---k and n are positive integers. Let f(x) = 1/(xk - 1). Let p(x) = (xk - 1)n+1 fn(x), where fn is the nth derivative. Find p(1).

 

[Cogman]

Link I dont think homework is allowed here (this smells very much like homework)

 

[eigen]

Originally posted by: Smoke0
Problem---k and n are positive integers. Let f(x) = 1/(xk - 1). Let p(x) = (xk - 1)n+1 fn(x), where fn is the nth derivative. Find p(1).

Yeah I smell homework...and this problem is not that hard.Let me guess Cs major in Discrete math or maybe a second semester DE class.

 

[Mik3y]

Originally posted by: Smoke0
Problem---k and n are positive integers. Let f(x) = 1/(xk - 1). Let p(x) = (xk - 1)n+1 fn(x), where fn is the nth derivative. Find p(1).

that problem is not hard dude. if you want to know how to do it, open up your mathbook and look at the examples.

 

[TuxDave]

Ok... I figured it out. So what do I win?

 

[Zero Plasma]

It's not homework.Heres the answer.

Solution

Answer: (-1)n+1 n! kn.

A routine differentiation.

Let pn(x) = (xk - 1)n+1fn(x). So fn = pn/(xk - 1)n+1. Differentiating, fn+1 = (pn' (xk - 1) - (n+1)kxk-1pn)/(xk - 1)n+2. Hence pn+1(1) = - (n+1)kpn(1). Also p1(1) = -k. Hence pn(1) = (-1)n n! kn.

 

[Gamingphreek]

That doesn't sound like discrete math. Sounds like Pre Calc.

-Kevin

 

[TuxDave]

Originally posted by: Smoke0
Problem---k and n are positive integers. Let f(x) = 1/(xk - 1). Let p(x) = (xk - 1)n+1 fn(x), where fn is the nth derivative. Find p(1).

Originally posted by: Smoke0
It's not homework.Heres the answer.
Solution
Answer: (-1)n+1 n! kn.
A routine differentiation.
Let pn(x) = (xk - 1)n+1fn(x). So fn = pn/(xk - 1)n+1. Differentiating, fn+1 = (pn' (xk - 1) - (n+1)kxk-1pn)/(xk - 1)n+2. Hence pn+1(1) = - (n+1)kpn(1). Also p1(1) = -k. Hence pn(1) = (-1)n n! kn.

wtf mate? PN or P?

 

[tak3shiro]

Doesn't look like pre-calc. I'm in pre-calc right now, looks more like Calc I.

 

[imported_IKnowNothing]

Math...I love it but it is to damn simple.

English...now there is something a bit harder to predict. Unless you study the emotional and reactional factors....never mind....did I say I love math

Dish out a question about quantum theories and relativity. Love debating!

Math is a chain of events leading to the creation of Universal understanding.

 

[CycloWizard]

Originally posted by: IKnowNothing
Math...I love it but it is to damn simple.

English...now there is something a bit harder to predict. Unless you study the emotional and reactional factors....never mind....did I say I love math

Dish out a question about quantum theories and relativity. Love debating!

Math is a chain of events leading to the creation of Universal understanding.

Can I call you during my test tomorrow?

 

[NewBlackDak]

I loved math until I hit Calc in college. I'm a visual learner, and some problems are to difficult to visualize until you solve them. Drives me nuts.

 

[Turkish]

Heh, that really is NOT a hard math problem

<< -- Ex-engineer but is proud to say he got A's in all math classes (7 of them)

 

[itachi]

try proving euler's equation and laplace's theorem (better known as laplace transform) then use that proof to prove the trig identities, laws, rules, and half angle formulas.. (not concurrently).. now thats a b1tch. not that hard, but extremely time consuming.

 

[Tiamat]

Originally posted by: itachi
try proving euler's equation and laplace's theorem (better known as laplace transform) then use that proof to prove the trig identities, laws, rules, and half angle formulas.. (not concurrently).. now thats a b1tch. not that hard, but extremely time consuming.

heeh, I dipped into that in my partial diff eq's class. Definately took a bit of working at. Let's just say one sign error kept me stunned for 2 hrs and 3 sheets of paper :/

 

[eigen]

Originally posted by: Tiamat

Originally posted by: itachi
try proving euler's equation and laplace's theorem (better known as laplace transform) then use that proof to prove the trig identities, laws, rules, and half angle formulas.. (not concurrently).. now thats a b1tch. not that hard, but extremely time consuming.

heeh, I dipped into that in my partial diff eq's class. Definately took a bit of working at. Let's just say one sign error kept me stunned for 2 hrs and 3 sheets of paper :/

bah, there is so much more interesting stuff than wasting time with that crap.
Graph Colorouing on the other hand......

 

[DrPizza]

Originally posted by: itachi
try proving euler's equation and laplace's theorem (better known as laplace transform) then use that proof to prove the trig identities, laws, rules, and half angle formulas.. (not concurrently).. now thats a b1tch. not that hard, but extremely time consuming.

Hmmmm
Wouldn't you be using the Laplace transform to demonstrate the trig identities, rather than prove them?
It's been a lonnnnnnng time for me, but working backwards, integral came from the derivative, definition of the derivative,... when you're doing lim h->0 (f(x+h)-f(x)) / h, you end up using trig identities to simplify the derivatives of trig functions. So, the tools you're using to prove the trig identities were actually derived from trig identities.

 

[itachi]

Originally posted by: DrPizza
Hmmmm
Wouldn't you be using the Laplace transform to demonstrate the trig identities, rather than prove them?
It's been a lonnnnnnng time for me, but working backwards, integral came from the derivative, definition of the derivative,... when you're doing lim h->0 (f(x+h)-f(x)) / h, you end up using trig identities to simplify the derivatives of trig functions. So, the tools you're using to prove the trig identities were actually derived from trig identities.

ok you just hurt my head there..
i'm not sure what you're getting at by stating that you use trig identities to simplify the derivatives of trig functions? you don't need trig identities to prove it.. you do need to use euler's equation though. both euler's and laplace's methods are complex number based solutions.. euler's formula is the basis for trig identities, etc.. and laplace transforms can simplify complex equations. convert cos and sin back into their complex forms.. and you'll have exponential functions. no need for trig simplification.

heeh, I dipped into that in my partial diff eq's class. Definately took a bit of working at. Let's just say one sign error kept me stunned for 2 hrs and 3 sheets of paper :/

can't say that i know exactly what you're talking about.. but i got an idea. series solutions in ode kept me going for a couple sheets of paper before i realized that i forgot something.

oh.. and this is the question i got on my exam for ordinary differential equations a couple years ago. this is what happens when you oversleep and miss the exam.. then beg the professor to let you make it up, giving a lame and obviously bs excuse for why you missed it.

 

[helpme]

Originally posted by: itachi
try proving euler's equation and laplace's theorem (better known as laplace transform) then use that proof to prove the trig identities, laws, rules, and half angle formulas.. (not concurrently).. now thats a b1tch. not that hard, but extremely time consuming.

I don't careeeeeeeeee as long as t is replaced with s

 

[syconub]

maybe I should just post my algebra homework every day