Question

(1) Find binomial expansion for:

(2) Find the expansion for the given function using Maclaurin series for common functions:

sqrt( 1+ sinx)

Answer 1

\[\frac{ x^2 + 16x + 2 }{ x^3 + 23x^2 + 20x - 4}\]

Answer 2

you need to factor it

Answer 3

@SkepticGod , the roots aren't integers.

Answer 4

Look at it on Wolfram:
http://www.wolframalpha.com/input/?i=+%28x%5E2+%2B+16x+%2B+2%29%2F%28x%5E3+%2B23x%5E2+%2B+20x+-4%29

Answer 5

I think I remeber it now so isn't it 1+sin^2=cos^2x ?

Answer 6

I think the first thing you have to do is solve for x and then plug them in for the "x"

Answer 7

sqrt(1+sinx)

sinx = sqrt(1+cos^2x)

sqrt(1+ sqrt(1+cos^2x)) = useless, If I understood you well.

Answer 8

We need to get rid of the root. I tried using the common expansion for

(1+x)^m than plugging x = sinx

Got it, but long solution.

Answer 9

sorry nevermind lol

Answer 10

I mean there's nothing I could really expand further because both denominator and nominator are prime

Answer 11

@SkepticGod , thanks at least you're thinking with me.

Answer 12

no problem! :D seem like I knew it but I haven't learn that Taylor/Maclaurin polynomial and series so..

Answer 13

@SkepticGod , did you learn binomial ?

Answer 14

yea i did when I was in High school

Answer 15

@SkepticGod , Do you know that when the power is negative , it's infinite series ?

Answer 16

what i know is expanding ding binomial thats all, wish i could help but too bad i haven't expert on series

Answer 17

Trojo: Have you done similar problems requiring binomial expansions? At first glance, this one is a bear.

Answer 18

@mathmale , I 've been doing binomial expansions for about 1 year ( + ve real number power) and for 0.5 year in expansions with negative , rational numbers.

Answer 19

Have you tried every approach you can think of to factor numerator and denominator?

Answer 20

@mathmale , Tried grouping (failed), finding roots using calc ( none integers) long division doesn't give any beneficial results.

Answer 21

@mathmale , I even thought of partial fractions xD

Answer 22

Have you tried dividing the numerator into the denominator?

Answer 23

@mathmale I got 1- something

by adding , subtracting.

which was of no use.

divinding ? you will get reminder in the denominator.

Answer 24

Do you have your calculus book with y ou?
I'll go get mine. Need a quick review of the binom. thm.

Answer 25

Excuse me, binomial expansions. Sorry.

Answer 26

Binomial Theorem: just in case
(x+y)^n = \[\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right) x^{n-k} y ^{k}\] where,
\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{ n! }{ k!(n-k)! }\]

Answer 27

@SkepticGod , Your formula won't work with a negative power like in our case.

Answer 28

Replace n with -n

Answer 29

thanks, Skeptic.

Just realized something. Binomial expansions involve expanding powers of binomials, such as in \[(1+x)^m\]

Answer 30

The complicated rational function you're working on certainly does not look like a binomil at first glance. Am I missing something?

Answer 31

it's an infinite series... not a binomial expansion. he meant that.

Answer 32

Totally, correct. We have to do some magic and turn it into good thing.

Answer 33

the binomial theorem for any index...

Answer 34

(1) Find binomial expansion for:

(2) Find the expansion for the given function using Maclaurin series for common functions:

sqrt( 1+ sinx)

Trojan: what say you?

Answer 35

Where does your "sqrt(1+sin x) come from?

Answer 36

\[(1+y)^n = 1 + ny + \frac{n(n+1)}{2}y^2 + \cdots \]

Answer 37

I don't know mate. That's why I am asking.

the second one we have to get rid of the square root or use this method:
using the common expansion for

(1+x)^m than plugging x = sinx

Got it, but long solution.

Answer 38

The sqrt(1+sinx) is another question.

If one is solved and I will take the other and post a new question.

Answer 39

Did you notice that the question had been here for 4 hours ?

Answer 40

Trojan, do you agree or disagree with Parth's comment, in which he says you meant an "infinite series" (which could be either a Taylor or Maclaurin series?

Answer 41

@mathmale , I totally agree with parthkohli , he understand what I mean.

The binomial is infinite as the power is negative (x^3+..)^-1

Answer 42

See this:

http://mathworld.wolfram.com/NegativeBinomialSeries.html

Answer 43

But does the given denominator have the format (a+b)?

Answer 44

Nope.

Answer 45

Maybe if you can find the roots as 1+ sqrt(3)/3 or something I can play around with partial fractions.

Answer 46

\[(a+b)^3=a^3+3a^2b+3ab^2 + b^3.\] and I don't see that your denom has that form. Thus, as interesting as that Wolfram excerpt is, it may not aply hear.

Answer 47

apply here, I meant.

Answer 48

Partial fractions would work, probably, if and only if y ou can factor the denominator.

Answer 49

I've begun trying to factor it using synth. div., but have not finished. Have you tried factoring by this method?

Answer 50

I may use the formula if I want ..

((x^3 + 23x^2) + (20x - 4))^-1

Let x^3 + 23x^2 be z
Let 20x -4 be y

(z + y)^-1 use the formula replace and keep expanding each one considering an infinite series I will die before archiving my aim.

Answer 51

Actually, Trojo, you need only write out the first three or so terms to illustrate the pattern. I'd much rather that you didn't die.

Answer 52

Again: Have you tried factoring the den. using synth. div? I've tried the possible roots 1, -1, 2 and -2, and have found that none lead to factors of the den.

Answer 53

Could we temporarily abandon binom. expansions and focus on Maclaurin expansins?

Have you tried using the def'n of the Mac expansion to come up with the first few terms of the Mac expansion of the given rational function?

Answer 54

Imagine :

(x^2 + 16x + 2) ( z+y)^-1

expand (z+y)^-1:

(z+y)^-1 = 1/z * (1 + y/z)^-1

= 1/z * [ 1 + y/z + (y/z)^2 + (y/z)^3 + .......]

= [1/z + y/z^2 + y^2/z^3 +y^3/z^4 + ....]
[1/z + y/z^2 + y^2/z^3 +y^3/z^4 + ....](x^2 + 16x + 2)

[ 1/z x^2 + x^2 * y/z^2 + y^2/z^3 * x^2 + .... ] +[ 16x * 1/z * 16 x * y/z^2 + ....] + [ 2 * 1/z + y/z^2 * 2 + 2 * y^2/z^3 + ...]

Then replace z, y with actual values and simplify

Answer 55

@mathmale , Hahaha I used the fking damn calculator and got this:

root 1 : -22.08625958
root 2: 0.1675002457
root 3: -1.0871240664

Answer 56

I see what you've done at the beginning: (x^2 + 16x + 2) ( z+y)^-1
but lose you as you expand this idea further.

So you are saying that your den has 3 real roots?
If that's the case, you can convert each into a factor and then use Partial Fractions.

Answer 57

But what a messy bit o' work this will turn into!

Answer 58

\[\frac{ x^2 + 16x + 2 }{ x^3 + 23x^2 + 20x - 4} \\~\\=-\frac{1}{4} (x^2 + 16x + 2 ) (1-g)^{-1} \\~\\=-\frac{1}{4} (x^2 + 16x + 2 ) (1+g+g^2+\cdots )\]

where \(g = \frac{1}{4}(x^3+23x^2+20x) \)

Answer 59

just write first few terms in the series and move to next problem :)

Answer 60

is that an OK solution? i thought exactly the same thing but... looks too "unsimplified"?

Answer 61

I would write at least upto the x^4 term and put \(\cdots \)

Answer 62

Quite an interesting substitution. Would you separate the given rational function into 3, all with the same denom? Then mult. the binomial expansion by x^2, by 16x and by 2, one by one?

Answer 63

@ganeshie8 , How did you get rid of the 4 and got the 1/4. the question require simplification.

Answer 64

@mathmale , Same as finding the expansion and multiplying times (x^2 +16x + 2)

Answer 65

@ganeshie8 , true, i need the first 3 terms.

Answer 66

Yes, that's familiar and makes sense.

Answer 67

@ParthKohli , We need to simplify , I would 've solved it years ago.

Answer 68

\[\frac{ x^2 + 16x + 2 }{ x^3 + 23x^2 + 20x - 4} \\~\\
\frac{ x^2 + 16x + 2 }{-4[ 1 - \frac{1}{4}(x^3 + 23x^2 + 20x )]} \\~\\
=-\frac{1}{4} (x^2 + 16x + 2 ) (1-g)^{-1} \\~\\=-\frac{1}{4} (x^2 + 16x + 2 ) (1+g+g^2+\cdots )\]

where \(g = \frac{1}{4}(x^3+23x^2+20x) \)

Answer 69

So, approaching this problem that way, we do have a binomail to expand:\[(1-g)^{-1}\]

Answer 70

@ganeshie8 , nice trick indeed. Now simplify please xD

Answer 71

@ParthKohli , your turn. find the 3 terms. Alexander isn't here to multiply.

Answer 72

@mathmale , yeah and a lot of inner binomials to expand too.

Answer 73

I didn't understand this at all... D:

Answer 74

Seems that we'd get a sum \[x^2(1-g)^{-1}+16x(1-g)^{-1}+2(1-g)^{-1}\]

Answer 75

how would you find the degree zero term there?

Answer 76

1, g, g^2, ... all generate degree zero terms.

Answer 77

Hmm theoretically. maybe just 2

Answer 78

@ParthKohli , remember that g= x^3 + 23x^2 + 20x

which will give degree 1 +

Answer 79

Parth: Same comment from me on that. g(0)=0; (1-g) is not zero.

Answer 80

sorry :(

Answer 81

@ParthKohli , Go multiply it's easy lol

you will get expansion of (x^3 + 23x^2 + 20x)^2 < Ok ?

((x^3+23x^2) + 20x)^2 After expanding you will get

(x^3 + 23x^2)^n (n differs) Now expand again and you've finished the first expansion go to the next.

Answer 82

why do you want to expand (x^3 + 23x^2 + 20x)^2

Answer 83

unless your prof is a psycho

Answer 84

there has to be a simpler way

Answer 85

@ganeshie8 to get power 20 x ,
Maybe he got a trick.

By the way do you know simpler solution for the second ?

Answer 86

@ParthKohli , Correct.

Answer 87

Trojo: I'm still wondering whether you've looked thru your text, notes, etc., for concrete examples of how to expand rational functions such as yours into a binomial series.

Answer 88

\[\frac{ x^2 + 16x + 2 }{ x^3 + 23x^2 + 20x - 4} \\~\\
\frac{ x^2 + 16x + 2 }{-4[ 1 - \frac{1}{4}(x^3 + 23x^2 + 20x )]} \\~\\
=-\frac{1}{4} (x^2 + 16x + 2 ) (1-g)^{-1} \\~\\=-\frac{1}{4} (x^2 + 16x + 2 ) (1+g+g^2+\cdots )\\~\\
=-\frac{1}{2}-\frac{13 x}{2}-\frac{285 x^2}{8}+\mathcal{O}(x^3)
\]

where \(g = \frac{1}{4}(x^3+23x^2+20x) \)

Answer 89

I am satisfied with above work... but i would like to know if there is any other pleasant way to expand...

Answer 90

@mathmale , The professor was meant to solve this e.g ( + other e.g) with us. Unluckily the lecture was canceled.

Answer 91

@ganeshie8 , I am sure we will do some simplification before getting on.

see this:

sqrt(1+sinx) -> I solved it like :using the common expansion for

(1+x)^m than plugging x = sinx

Got it, but long solution.

any shorter one ?

Answer 92

Thats pretty much how I would do it..
Have you also plugged in the series of sinx ?

Answer 93

Yeah :((

and simplified

Answer 94

looks good, i can't think of anything better...

Answer 95

I also thought of

sqrt ( 1 + cos(pi/2 - x)) = sqrt(1 +2cos^2(pi/4 - x/2) - 1) = sqrt(2) cos(pi/4 - x/2)

cos(pi/4 - x/2) isn't mac :/