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

do you know pascal triangle?

yeah i do but i am not so familiar with it

http://upload.wikimedia.org/wikipedia/en/math/d/f/a/dfab2268985c30e5404a3accfc87da07.png

does that help?

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

and how do i go from here turing

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

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

(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

convert it into a decimal 0.25?

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

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

oh ok turing thanks

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

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"

the power of a fraction makes this question a tussle

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:

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

@FoolForMath @KingGeorge fractional binomial expansion help

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

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

fool m not getting it still

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

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:

would this be the first term or what @AccessDenied

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

oh ok let me try it out

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

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

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

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

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

I need to be caffeinated! bbl

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.

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

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

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.

thank you guys for the great help

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

lol we learn from our mistakes remember @FoolForMath

We fail exams due to your mistakes :P

very true but nonetheless thanks for participating

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 \[ \large{(a + b)^n = \frac{a^n}{0!} + \frac{n a^{n-1} b}{1!} + \frac{n(n-1) a^{n-2} b^2}{2!} + \cdots} \\ \\ \frac{16^{-1/4}}{0!} + \frac{(-1/4) 16^{-1/4 - 1} (-3x)}{1!} + \frac{(-1/4)(-1/4 - 1) 16^{-1/4 - 2} (-3x)^2}{2!} \\ \frac{16^{-1/4}}{0!} + \frac{(-1/4) 16^{-5/4} (-3x)}{1!} + \frac{(-1/4)(-5/4) 16^{-9/4} (-3x)^2}{2!} \\ \frac{1}{2} + \frac{3x}{4*2^5} + \frac{5*9x^2}{2!*2^9 } \\ \frac{1}{2} + \frac{3x}{4*32} + \frac{45x^2}{16*2^10} \\ \frac{1}{2} + \frac{3x}{128} + \frac{45x^2}{16384} \\ \]

Both are same thing.

thanks alot mates

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