Factorial Notation (1 Viewer)

[STx]

Hey, i just saw these q's in the Cambridge 3U(yr12), and couldnt figure them out, im sure their supposed to be simple though:

1a) If f(x)=xⁿ,
find:
(i) f⁽ⁿ⁾(x)
(ii) f^(k)(x), where k≤n

And Identities on the Binomial Coefficients
2) Consider the identity (1+x)²ⁿ= _r=0∑^{2n 2n}C_r*x^r

a) show that _r=0∑^{2n 2n}C_r=2²ⁿ {Done}

(b) ²ⁿC₁+²ⁿC₃+²ⁿC₅ +...+²ⁿC_2n-1=2^2n-1

thanks

 

[Riviet]

For 1, it's just asking for what you get after differentiating f(x) n times and k times (imagine n number of ' instead of the letter n).

f⁽ⁿ⁾(x)=n(n-1)(n-2)...3.2.1(x⁰)
=n!


f^(k)(x)=n(n-1)(n-2)...(n-(k-2))(n-(k-1))(x^n-k)
=n(n-1)(n-2)...(n-k+2)(n-k+1))(x^n-k)

For 2, try something similar to what is done in excercise 5F Q2 b) (i) and (ii).

Hope that helps.

edit: fixed error identified by sasquatch.

 

[STx]

How did you get these parts:

fⁿ(x)=n(n-1)(n-2)...3.2.1(x⁰)
=n!

f^k(x)=n(n-1)(n-2)...(n-(k-2))(n-(k-1))(x^n-k)

=n(n-1)(n-2)...(n-k+2)(n-k+1))(x^n-k)

 

[alcalder]

OK, so if
f(x) = xⁿ
Then
f⁽¹⁾(x) = n x^n-1 (differentiate once)
f⁽²⁾(x) = n(n-1) x^n-2 (differentiate again)
f⁽³⁾(x) = n(n-1)(n-2)x^n-3 (differentiate a third time)

and so on.

If we want to find f^(k)(x), look at the pattern above and generalise. (this is differentiating k times)

f^(k)(x) = n(n-1)(n-2)(n-3)...(n-(k-1)) x^n-k

To find f⁽ⁿ⁾(x), put in k=n (this is differentiating n times until x has the power 0)


f⁽ⁿ⁾(x) = n(n-1)(n-2)(n-3)...(n-(n-1)) x^n-n
=n(n-1)(n-2)(n-3)...(1) x⁰

And this is just n!

EDIT: OOPS! Fixed my brackets too.

 

[Riviet]

How did you get these parts

Sorry to be a bit too brief with my working. The last terms in f⁽ⁿ⁾(x) are just the last terms in n! After differentiating n times the index of x would decreasing by -1 each time and multiplying this by n gives -n which gives n-n=0, hence x⁰.

Similarly for f^(k)(x), after differentiating k times you are subtracting from the index of x k times and each time you are subtracting 1 from the index so multiplying this by k gives -k which gives n-k, hence x^n-k.

edit: thanks for the reminder, sasquatch, I was being a little lazy with the brackets.

 

[sasquatch]

(b) 2nC1+2nC3+2nC5 +...+2nC2n-1=22n-1

Ew thats ugly...

(1+x)²ⁿ = ²ⁿC₀ + ²ⁿC₁x + ²ⁿC₂x² +... + ²ⁿC_2nx²ⁿ

If x = 1

[1] 2²ⁿ = ²ⁿC₀ + ²ⁿC₁ + ²ⁿC₂ +... + ²ⁿC_2n

(1-x)²ⁿ = ²ⁿC₀ - ²ⁿC₁x + ²ⁿC₂x² -... - ²ⁿC_2nx²ⁿ

If x = 1,

[2] 0 = ²ⁿC₀ - ²ⁿC₁ + ²ⁿC₂ - ... - ²ⁿC_2n

[1] - [2]:

2²ⁿ - 0 = (²ⁿC₀ + ²ⁿC₁ + ²ⁿC₂ +... + ²ⁿC_2n) - (²ⁿC₀ - ²ⁿC₁ + ²ⁿC₂ - ... - ²ⁿC_2n)

2²ⁿ = ²ⁿC₀ + ²ⁿC₁ + ²ⁿC₂ +... + ²ⁿC_2n)
- ²ⁿC₀ + ²ⁿC₁ - ²ⁿC₂ + ... + ²ⁿC_2n

2²ⁿ = 2²ⁿC₁ + 2²ⁿC₃ + 2²ⁿC₅ + ... + 2²ⁿC_2n

2²ⁿ = 2(²ⁿC₁ + ²ⁿC₃ + ²ⁿC₅ + ... + ²ⁿC_2n)

2²ⁿ/2 = ²ⁿC₁ + ²ⁿC₃ + ²ⁿC₅ + ... + ²ⁿC_2n)

2^2n-1 = ²ⁿC₁ + ²ⁿC₃ + ²ⁿC₅ + ... + ²ⁿC_2n)



Also Riviet, make sure you bracket the n when using the f⁽ⁿ⁾(x) notation for the nth derivative. fⁿ(x) refers to something different: the function of the function of the function ....(n times) of x, or

fⁿ(x) = f(...f(...f(x)...)...) (n times)

 

[STx]

(b)

(1-x)²ⁿ = ²ⁿC₀ - ²ⁿC₁x + ²ⁿC₂x² -... - ²ⁿC_2nx²ⁿ

Why are you using that ? Thats not part of the question right?

 

[SoulSearcher]

(1-x)²ⁿ = ²ⁿC₀ - ²ⁿC₁x + ²ⁿC₂x² -... - ²ⁿC_2nx²ⁿ

If x = 1,

[2] 0 = ²ⁿC₀ - ²ⁿC₁ + ²ⁿC₂ - ... - ²ⁿC_2n

This line is supposed to be (1-x)²ⁿ = ²ⁿC₀ - ²ⁿC₁x + ²ⁿC₂x² - ... - ²ⁿC_2n-1x^2n-1 + ²ⁿC_2nx²ⁿ, so then you will be able to prove that ²ⁿC₁ + ²ⁿC₃ + ²ⁿC₅ + ... + ²ⁿC_2n-1 = 2^2n-1

You can also use the expansion of (1+x)²ⁿ = ²ⁿC₀ + ²ⁿC₁x + ²ⁿC₂x² + ... + ²ⁿC_2n-1x^2n-1 + ²ⁿC_2nx²ⁿ, and then substitute in x = -1, and then follow out the rest of sasquatch's steps.
<SUP></SUP>

 

[STx]

thanks, heres another one i cant quite complete yet:

By comparing coefficients of xⁿ⁺¹ on both sides of (1+x)ⁿ.(1+x)ⁿ = (1+x)²ⁿ, show that:

 

[STx]

thanks, heres another one i cant quite complete yet:

By comparing coefficients of xⁿ⁺¹ on both sides of (1+x)ⁿ.(1+x)ⁿ = (1+x)²ⁿ, show that:

Help pls kthx , okay i expanded both sides but i cant campare the coefficients properly as they wont match up like the answer requires, also what would give a coefficient of xⁿ⁺¹ on the RHS?

 

[sasquatch]

i tried this question yesterday and i got the RHS fine.. but on the LHS i got something like

LHS = (n 1)(n n) + (n 2)(n n-1) + (n 3)(n n-2) + .... + (n n)(n 1)

On the RHS i got:

RHS = (2n 0) + (2n 1)x + (2n 2)x^2 + ... + (2n n+1)x^(n+1) + ... + (2n 2n)x^2n

so the coeff of x^(n+1) is (2n n+1)

And by using the fact that (n r) = n!/[r!(n-r)!]

(2n n+1) = (2n)!/[(n+1)!(2n - n - 1)!]
= (2n)!/(n+1)!(n-1)!

So yeah it seems as if the questions is screwed, or involves a weird manipulation of the LHS terms that neither of us can see. So yeah im not sure...

So weird..

 

[pLuvia]

Yeh same I tried the LHS but couldn't get their preferred answer, but when I just multplied the wanted terms and let x=1 I got their answer

 

[sasquatch]

hmm that question is weird then. why would they give you incorrect guidance..

 

[STx]

but when I just multplied the wanted terms and let x=1 I got their answer

yes its screwed. So that method works?

 

[pLuvia]

But only for the LHS, the RHS you use sasquatch's method, but then the question doesn't make sense

 

[sasquatch]

hehe thats so wierd..

 

[airie]

For LHS, basically you make all the possible "combos" for a term of xⁿ⁺¹.

From sasquatch's working,

on the LHS i got something like

LHS = (n 1)(n n) + (n 2)(n n-1) + (n 3)(n n-2) + .... + (n n)(n 1)

and from the question,

thanks, heres another one i cant quite complete yet:

By comparing coefficients of xⁿ⁺¹ on both sides of (1+x)ⁿ.(1+x)ⁿ = (1+x)²ⁿ

a term of xⁿ⁺¹ may be obtained by x.xⁿ, or x².x^n-1, x³.x^n-2 etc, where

the coefficient of x from the first bracket in (1+x)ⁿ.(1+x)ⁿ, ie. (1+x)ⁿ is ⁿC₁, and that of xⁿ from the second bracket of (1+x)ⁿ is ⁿC_n, multiplying these two terms yields ⁿC₁.ⁿC_n.xⁿ⁺¹ --> the coefficient is ⁿC₁.ⁿC_n;

similarly for all other "combos" of taking terms from each of the two brackets ie. for x² from the first and x^n-1 from the second, then for x³from the first and for x^n-2 from the second etc, the coefficient of the resulting terms of xⁿ⁺¹ will each match with the terms that sasquatch worked out above,

on the LHS i got something like

LHS = (n 1)(n n) + (n 2)(n n-1) + (n 3)(n n-2) + .... + (n n)(n 1)

Therefore the coefficient of xⁿ⁺¹ on LHS will equal to what sasquatch worked out here, and therefore LHS = RHS for the question as they are both the coefficient of xⁿ⁺¹ when (1+x)ⁿ.(1+x)ⁿ and (1+x)²ⁿ is expanded.

 

[STx]

^Yeah i know how to do those types of questions, but couldnt get the required answer as below, nor did sasquatch

Anyway, i just noticed this property: ⁿC_r= ⁿC_n-r. So if we use it on: LHS = (n 1)(n n) + (n 2)(n n-1) + (n 3)(n n-2) + .... + (n n)(n 1)
For example: (n n)=(n 0), and if its applied to the other terms it yields the required answer. Thanks anyway

 

[airie]

^Yeah i know how to do those types of questions, but couldnt get the required answer as below, nor did sasquatch

Anyway, i just noticed this property: ⁿC_r= ⁿC_n-r. So if we use it on: LHS = (n 1)(n n) + (n 2)(n n-1) + (n 3)(n n-2) + .... + (n n)(n 1)
For example: (n n)=(n 0), and if its applied to the other terms it yields the required answer. Thanks anyway

o.0 ... It was done above, the LHS. Maybe my explanation of "combos" wasn't clear?

 

[sasquatch]

Oh shit... is that how easy it was to get from my line to the required answer.. god damn..haha oh well good work STx!!!!