# Q15 (1 Viewer)

#### kenkap

##### Member
Re: General Thoughts: Mathematics Ext 2

Hell there was an integration by parts question in question 15,
Am i missing something????

#### Alkanes

##### Active Member
Re: General Thoughts: Mathematics Ext 2

Am i missing something????
Recurrence formula one.

#### lolcakes52

##### Member
Re: General Thoughts: Mathematics Ext 2

I put my solutions to 16 here http://community.boredofstudies.org/showthread.php?t=293623
for 15
ai) simple expansion of the difference of two roots being greater than zero
aii) (x-1)>=0 , y-x>=0 multiply and your done
aiii) the left inequality is part ii, the right is part i.
iv) didn't attempt

bi) conjugate root theorem, mention that the conjugate of ia is -i*the conjugate of a.
ii) split the third term of the polynomial into two parts and factorise the two parts
iii) didn't do
iv) product of the roots, then say something about how conjugate alpha times alpha is the modulus of alpha
v) sum of roots
vi)didnt do

#### v1

##### Active Member
Q15
a) (iv) using iii) 1<=j<=n
sqrt(n) <= sqrt(j(n-j+1)) <= (n+1)/2
let j=1,2,3...n and multiply them all
sqrt(n^n) <= sqrt([1*2*3....*n][n*(n-1)*(n-2)...1] <= [(n+1)/2]^n
sqrt(n^n) <= sqrt([n!][n!]) <= [(n+1)/2]^n
sqrt(n^n) <= n! <= [(n+1)/2]^n

Last edited:

#### SoresuMakashi

##### Member
Q15
a) (iv) using iii) 1<=j<=n
sqrt(n) <= sqrt(j(n-j+1)) <= (n+1)/2
let j=1,2,3...n and multiply them all
sqrt(n^n) <= sqrt([1*2*3....*n][n*(n-1)*(n-2)...1] <= [(n+1)/2]^n
sqrt(n^n) <= sqrt([n!][n!]) <= [(n+1)/2]^n
sqrt(n^n) <= n! <= [(n+1)/2]^n
Very nice.