Question

Expand to first 3 terms \[\huge \frac{3+4x+x^2}{\sqrt[3]{1+\frac{x}{2}}}\]

Answer 1

\[3 + 7/2 +\] to lazy to calculate 3º derivative in 0.

Answer 2

Taylor series?
oh man...

Answer 3

calculate the second derivative and use taylor series

Answer 4

I don't think so,
Use expansion?

Answer 5

\[(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2\]

Answer 6

What i predicted,
\[(3+4x+x^2)(1+x/2)^{-1/3}\]
then n=-1/3

Answer 7

u are using binomial formula for not natural exponents....

Answer 8

i made the taylor expansion in 0

Answer 9

u can make it anywhere just avoid -2

Answer 10

I've never expanded a negative fractional exponent befor, I think I'lll have to get back to you...
curious to see how to do it though, maybe factoring the top can help somehow

Answer 11

I tried rationalizing the denominator and simplifying as much as possible, but that was to no avail...
still thinking though

Answer 12

ok

Answer 13

@JamesJ

Answer 14

There's an easier way to do this. Find the Taylor series of
\[ \frac{1}{(1+y)^{1/3}} \]

Once you have the first three or so terms of that, substitute y = x/2 and multiply that series by (3 + 4x + x^2)

Answer 15

My book didn't teach me about taylor series (even my syllabus does not include taylor series)
Is it possible to use \[(1+x)^{n}=1+nx+\frac{n(n-1)}{2}x^2\]?

Answer 16

Yes

Answer 17

when i use n=-1/3 x=x/2 (bit confusing for x=x/2 part)

(3+4x+x^2)(1+x/2)^(−1/3)
Expand-----------
\[(3+4x+x^2)[1-\frac{x}{6}+\frac{-\frac{1}{3}(-\frac{1}{3}-1)}{2}(\frac{x^2}{4})]\]

Answer 18

My answer is
\[\frac{x^4}{18}+\frac{2x^3}{9}-\frac{x^2}{6}+7x\] but thats not the correct answer

Answer 19

So the terms of 1/(1+(x/2))^(1/3) are
1 - x/6 + x^2/18
as you have it.

Double check your multiplication. Notice both have a constant term, but there's no constant in your product of the two.

Answer 20

Oh whops multiplied wrongly, lol so this is the answer
\[\frac{x ^{4}}{18}+\frac{x ^{3}}{18}+\frac{x^2}{2}+\frac{7x}{2}+3\]
thanks @JamesJ :D