Question

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

Answer 1

do you know pascal triangle?

Answer 2

yeah i do
but i am not so familiar with it

Answer 3

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

Answer 4

does that help?

Answer 5

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

Answer 6

and how do i go from here turing

Answer 7

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

Answer 8

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

Answer 9

(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

Answer 10

convert it into a decimal 0.25?

Answer 11

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

Answer 12

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

Answer 13

oh ok turing thanks

Answer 14

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

Answer 15

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"

Answer 16

the power of a fraction makes this question a tussle

Answer 17

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:

Answer 18

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

Answer 19

@FoolForMath @KingGeorge fractional binomial expansion help

Answer 20

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

Answer 21

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

Now apply this formula:

Answer 22

fool m not getting it still

Answer 23

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

Answer 24

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:

Answer 25

would this be the first term or what @AccessDenied

Answer 26

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

Answer 27

oh ok let me try it out

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

Yes you are right, I am making silly mistakes,

EDIT 2:

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

Answer 33

I need to be caffeinated! bbl

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

thank you guys for the great help

Answer 39

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

Answer 40

lol we learn from our mistakes remember @FoolForMath

Answer 41

We fail exams due to your mistakes :P

Answer 42

very true but nonetheless thanks for participating

Answer 43

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} \\
\]

Answer 44

Both are same thing.

Answer 45

thanks alot mates