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OpenStudy (anonymous):

use binomial expansion to find the the 1st three terms of (16-3x)^-1/4

OpenStudy (anonymous):

do you know pascal triangle?

OpenStudy (anonymous):

yeah i do but i am not so familiar with it

OpenStudy (anonymous):

does that help?

OpenStudy (turingtest):

the \(k\)th term of \((a+b)^n\)is given by\[\binom nka^{n-k}b^k\]where\[\binom nk={n!\over k!(n-k)!}\]

OpenStudy (anonymous):

and how do i go from here turing

OpenStudy (anonymous):

n = -1/4 k is the term you wanna find

OpenStudy (turingtest):

but pascalls triangle and the binomial series won't fork for fractions, so you need to convert it I think

OpenStudy (anonymous):

(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + ... + an−1xyn−1 + anyn, i think you should consider using this method since you are asked to find 3 first terms

OpenStudy (anonymous):

convert it into a decimal 0.25?

OpenStudy (turingtest):

no because factorial is not defined for non-integers unfortunately :( I'm having to look at notes, sorry

OpenStudy (turingtest):

Here's something you can read along with me if you like... http://mathcentral.uregina.ca/QQ/database/QQ.09.98/evans2.html

OpenStudy (anonymous):

oh ok turing thanks

OpenStudy (turingtest):

hm... I'm having trouble with how they seem to suggest using the fractions in the binomial theorem now I'm interested, I'll get back to you hopefully

OpenStudy (accessdenied):

FFM provided me a link on "Generating Functions" not long ago, which coincidentally features the use of binomial theorem to noninteger powers: http://users.softlab.ntua.gr/~nidal/discrete/notes/examples/gen2.pdf Just scroll down a ways to the "Binomial Theorem Revisited"

OpenStudy (anonymous):

the power of a fraction makes this question a tussle

OpenStudy (accessdenied):

16^(-1/4) -1/4 16^(-5/4) (-3x) -1/4(-5/4) 16^(-9/4) (-3x)^2 -------- + --------------------- + ---------------------------- 0! 1! 2! Which at least seems to match up with what I'm getting on wolfram. D:

OpenStudy (anonymous):

accessdenied in these notes~(hyperlink)there is no fractional power

OpenStudy (turingtest):

@FoolForMath @KingGeorge fractional binomial expansion help

OpenStudy (anonymous):

i had come with something like that too access but doubted it i dont know why. it then worked out like: term 1: 2^4(-1/4) term 2: (-1/4)(2)^4(-5/4) term 3: ((-1/4)(-5/4))/2! 16^-1/4.9x^2

OpenStudy (anonymous):

\[\large (16-3x)^{-\frac 14 }=16\left(1-\frac3{16} x\right)^ {-\frac 14 } \] Now apply this formula:

OpenStudy (anonymous):

fool m not getting it still

OpenStudy (accessdenied):

\[ (16(1 - \frac{3}{16}x))^{\neg 1/4} = 16^{\neg 1/4} (1 - \frac{3}{16}x)^{\neg 1/4}\]

OpenStudy (anonymous):

You shouldn't because I made a mistake, EDIT: \[\large (16-3x)^{-\frac 14 }=16^ {-\frac 14 }\left(1-\frac3{16} x\right)^ {-\frac 14 } \] Now apply this formula:

OpenStudy (anonymous):

would this be the first term or what @AccessDenied

OpenStudy (accessdenied):

We're basically trying to rewrite the original expression to fit FFM's formula.

OpenStudy (anonymous):

oh ok let me try it out

OpenStudy (anonymous):

\[\large \left(1-\frac3{16} x\right)^ {-\frac 14 } = 1 +\frac 3{48}x + \frac{45 }{8192} x^2 +\cdots \]

OpenStudy (anonymous):

how did we conclude to the above answer @foolformath sorry i am bit slow tonight(south african time)

OpenStudy (anonymous):

Did you see the formula I attached? @TuringTest he/she is all yours ;)

OpenStudy (kinggeorge):

I'm pretty sure the second term should have a coefficient of \[-\frac{3}{16}\cdot-\frac{1}{4}=\frac{3}{64}\]

OpenStudy (anonymous):

Yes you are right, I am making silly mistakes, EDIT 2: \[ \large \left(1-\frac3{16} x\right)^ {-\frac 14 } = 1 +\frac 3{64}x + \frac{45 }{8192} x^2 +\cdots \]

OpenStudy (anonymous):

I need to be caffeinated! bbl

OpenStudy (accessdenied):

and then once you have that, you just have to bring back the 16^(-1/4) factored out and multiply it back to the terms. 16^(-1/4) = 1/2 so all terms would be multiplied by 1/2, and you'd have the first three terms there.

OpenStudy (anonymous):

so my 1st term would be: 1/2? 2nd: -8? 3rd:5/4*9^2?

OpenStudy (accessdenied):

\[ \large \frac{1}{2} \left(1-\frac3{16} x\right)^ {-\frac 14 } = \frac{1}{2}\left(1 +\frac 3{64}x + \frac{45 }{8192} x^2 +\cdots \right)\]Like that.

OpenStudy (accessdenied):

Just distribute the 1/2 to each term (we only need to worry about first three). 1/2 is the first term. (1/2)(3x/64) is the second. (1/2)(45x^2 / 8192) is the third.

OpenStudy (anonymous):

thank you guys for the great help

OpenStudy (anonymous):

Not really, I am full of mistakes in this thread :/

OpenStudy (anonymous):

lol we learn from our mistakes remember @FoolForMath

OpenStudy (anonymous):

We fail exams due to your mistakes :P

OpenStudy (anonymous):

very true but nonetheless thanks for participating

OpenStudy (accessdenied):

You're welcome. :) I just wanted to show the way I did it (using the formula in that link), since it seems it worked also. It's just not as nice as FFM's formula makes it. :P \[ \large{(a + b)^n = \frac{a^n}{0!} + \frac{n a^{n-1} b}{1!} + \frac{n(n-1) a^{n-2} b^2}{2!} + \cdots} \\ \\ \frac{16^{-1/4}}{0!} + \frac{(-1/4) 16^{-1/4 - 1} (-3x)}{1!} + \frac{(-1/4)(-1/4 - 1) 16^{-1/4 - 2} (-3x)^2}{2!} \\ \frac{16^{-1/4}}{0!} + \frac{(-1/4) 16^{-5/4} (-3x)}{1!} + \frac{(-1/4)(-5/4) 16^{-9/4} (-3x)^2}{2!} \\ \frac{1}{2} + \frac{3x}{4*2^5} + \frac{5*9x^2}{2!*2^9 } \\ \frac{1}{2} + \frac{3x}{4*32} + \frac{45x^2}{16*2^10} \\ \frac{1}{2} + \frac{3x}{128} + \frac{45x^2}{16384} \\ \]

OpenStudy (anonymous):

Both are same thing.

OpenStudy (anonymous):

thanks alot mates

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