what is the general equation for the binomial expansion of (1 + x)^n when n is negative or a fraction
1 + nx + (n(n-1)x^2 / 12) + (n(n-1)(n-2)x^3 / 12*3) + ….
how would you explain the expansion
- the one is always there (refer later into flashcards)
- for the fractions just multiply n by the next -ve number
- and multiply the consecutive numbers from 1 for the denominator
- and remember the increasing power of x each time
for (p + qx)^n, what is the condition for the binomial expansion to be valid if n is negative
|x| < |p/q|
how would you binomially expand 1/(1 + x)^2 up to x^3
- it can be written as (1 + x)^-2
- first term is 1
- second term is -2x as n= -2
- next term is [(-2)(-3)]x^2 / 12
- last term is [(-2)(-3)(-4)]x^3 / 123
- making 1 - 2x + 3x^2 - 4x^3
what is the domain of x for which the expansion is valid
- |x| < |p/q|
- p = 1 and q = 1
- so |x| < 1
what would be the method for binomially expanding a fraction, such as (1+2x)^3 / (1-x)^2
- write it out as a multiplication, so the denominator bracket would have a -ve power
- binomially expand both up to a necessary point (up to the power it asks for)
- multiply both expansions together and tidy up
how would you work out the domain of x when two brackets are binomially expanded
- work out the conditions for both brackets
- then combine them together
- if one of the brackets has a +ve power i has no condition, so the total condition would be by the -ve one
what is the form you need to rearrange every (p + qx)^n bracket into if you want to binomially expand it
- p^n(1 + qx/p)^n
- this is why the 1 is always at the beginning lol
what would you rewrite 1 / root of (4 - x) into
- (4 - x)^-1/2
- 4^-1/2 * (1 - x/4)^-1/2
