Question

find the maclaurin series for the function: (x(5x+1))/(1-x)^3

Answer 1

http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries_files/eq0025M.gif

I know it looks complex, but you just need to find f(0), and then find a few derivatives of f

Answer 2

it is odd that this function seems to have been cooked up to give all nice integer coefficients
i am wondering if there is some sort of gimmick to get the answer
perhaps partial fractions, but i am not sure if that is quicker

Answer 3

yea i have no idea how to do it im lost

Answer 4

@agent0smith do u know how to start it?

Answer 5

Take it one step at a time. First find f(0) (ie plug in 0 for x)

Then find the first derivative, and f'(0) by plugging in x=0

Answer 6

ok so f(0) is 0

Answer 7

then i take the first deriv of the function and plug in 0 again?

Answer 8

Yes

Answer 9

i was given the series for 1/(1-3)^3 does that help?

Answer 10

could i take that and find the maclaurin series for the top and then just multiply it by the 1/1-x)3 answer? or no

Answer 11

Hmm, i don't think you can do that, since the top of this function is x(5x+1)... not just some constant.

Answer 12

ok cause i was given that 1/(1-x)^3 = 1/2 summattion n(n-1)x^n-2 with n=2

Answer 13

but idk how to use that to help me

Answer 14

I'm not sure you can... your function is more complex than just 1/(1-x)^3

Answer 15

hmm im lost...

Answer 16

partial fractions gives it to you if you want to use that

Answer 17

too bad she is gone
if you decompose in to partial fractions you get three that are easy to find the expansion of, although it is still kind of a pain, but perhaps easier than taking successive derivatives

Answer 18

That's what i considered... but it seems it's going to give us more than just the simple series... you'll have the other fractions too.

Answer 19

it turns out that if you use partial fractions you get nice integer for the numerators

Answer 20

of course i did it with wolfram, not by hand but you get
\[\frac{5}{1-x}-\frac{11}{(1-x)^2}+\frac{6}{(1-x)^3}\] and she said she was given the expansion for the last one

Answer 21

the first one is just the well known geometric series, and the second one is the derivative of that , except of course for the 11 up top
then you combine them but it will still be hard to find a general formula
question is rather a pain, but extra derivative will make it almost impossible quick

Answer 22

@satellite73 sorry im back i had to get wifi

Answer 23

oh ok, i hope you understood what i wrote
it is still going to be a pain, but lots less pain than taking successive derivatives ad infinitum

Answer 24

i kinda understand what u have done..what is the next step?

Answer 25

compute the series for each
the first one is the well known geometric series
\[\frac{1}{1-x}=\sum_{n=1}^{\infty}x^k\] although here you have to multiply it by 5 and get\[\frac{5}{1-x}=5\sum_{n=1}^{\infty}x^k\]

Answer 26

but what about the rest of the numerator?

Answer 27

like i said, it is a pain
the next one is almost \(-\frac{1}{(1-x)^2}\) which is the derivative of the first one

Answer 28

the derivative of \(1+x+x^2+x^3+...\) is \(1+2x+3x^2+4x^3+...\) so it expands that way, although you still have to multiply by 11

Answer 29

and you have to multiply it by \(-11\) giving \[-11\sum_{n=1}^{\infty}nx^{n-1}\]

Answer 30

so far we have
\[5\sum_{n=1}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}\]

Answer 31

oops typo there, should start the first one at zero \[5\sum_{n=0}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}\]

Answer 32

1/(1-x)^3= \[\sum_{n=2}^{\infty} n(n-1)x^n-2\]

Answer 33

and lastly you said you were given the expansion for \(\frac{1}{(1-x)^3}\)

Answer 34

thats what i was given

Answer 35

i assume you mean
\[\sum_{n=2}^{\infty}n(n-1)x^{n-2}\]

Answer 36

yea i couldnt get it to go in the exponent lol

Answer 37

ok so your last task it so combine them in to one series term by term
i am not sure how you are going to get a nice formula out of this, but i am sure it is possible

Answer 38

yea idk how to combine them into one term

Answer 39

oh and you also have to remember to multiply the last one by \(-6\)

Answer 40

\[5\sum_{n=0}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}-\sum_{n=2}^{\infty}n(n-1)x^{n-2}\]

Answer 41

what about the 6?

Answer 42

another damn typo, it is hard to write all this out
\[5\sum_{n=0}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}-6\sum_{n=2}^{\infty}n(n-1)x^{n-2}\]

Answer 43

its ok i am so grateful for u helping me with this

Answer 44

now how about a general term? lets start with \(n=0\) and see what we get

Answer 45

damn something is wrong

Answer 46

oooh i see it!

Answer 47

it should be
\[5\sum_{n=0}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}+6\sum_{n=2}^{\infty}n(n-1)x^{n-2}\]

Answer 48

now it should work better

Answer 49

im lost as to what we do now? im sorry i missed the lesson on maclaurin serioes so im lost

Answer 50

if \(n=0\) you get \(0\) which is what we want

Answer 51

lets start with this line \[5\sum_{n=0}^{\infty}x^n-11\sum_{n=1}^{\infty}nx^{n-1}+6\sum_{n=2}^{\infty}n(n-1)x^{n-2}\]

Answer 52

you want the index to start at \(n=0\) for all of them

Answer 53

so we have to modify the last two correct?

Answer 54

yes

Answer 55

5∑n=0∞xn−11∑n=1∞nxn+6∑n=2∞n(n−1)xn

Answer 56

lol

Answer 57

sorry idk how to use these buttons well lol

Answer 58

lets change them all to \(k\)
now the first one we can leave alonoe
the second we want to start at 0, so we make \(k=n-1\) which makes \(n=k+1\)

Answer 59

each \(n\) gets replaced by \(k+1\) and we get for the second one
\[-11\sum_{k=0}^{\infty}(k+1)x^k\]

Answer 60

third one we want to start at 0, so make \(k=n-2\) making \(n=k+2\) so each \(n\) gets replaced by \(k+2\)

Answer 61

so then its n+2 (n+1)x^n

Answer 62

oops another typo!

Answer 63

\[6\sum_{k=0}^{\infty} (k+2)(k+1)x^k\]

Answer 64

now they all start at the same place, and we can combine like terms as we say in the math biz

Answer 65

so now we can just add them?

Answer 66

\[5\sum_{n=0}^{\infty}x^n-11\sum_{k=0}^{\infty}(k+1)x^k+6\sum_{k=0}^{\infty}(k+2)(k+1)x^k\]

Answer 67

yeah add them

Answer 68

\[5\sum_{k=0}^{\infty}x^n-11\sum_{k=0}^{\infty}(k+1)x^k+6\sum_{k=0}^{\infty}(k+2)(k+1)x^k\]

Answer 69

i guess we still have to compute
\[5-11(k+1)+6(k+2)(k+1)\]

Answer 70

something is bothering me here
when \(k=0\) we should get \(0\)

Answer 71

what happened to the x^k in the first two?

Answer 72

\[5\sum_{k=0}^{\infty}x^k-11\sum_{k=0}^{\infty}(k+1)x^k+6\sum_{k=0}^{\infty}(k+2)(k+1)x^k\]

Answer 73

oo so we factor out x^k right

Answer 74

give me a second, let me see if i can figure out the mistake

Answer 75

ok

Answer 76

damn i can't see it
we should be getting 0 if \(k=0\) i think right? we know the first term of the expansion is 0

Answer 77

hmm idk

Answer 78

is it possible one of the coefficient s are in correct?

Answer 79

idk i am not getting it right

Answer 80

is works if the last coefficient is 3

Answer 81

what do you mean?

Answer 82

i mean constant

Answer 83

if instead of 6.. if its 3

Answer 84

really? maybe we made a mistake in the expansion

Answer 85

5−11(k+1)+3(k+2)(k+1)

Answer 86

that gives us zero correct?

Answer 87

how did u get the numbers 5 11 and 6?

Answer 88

partial fraction decomposition

Answer 89

from the derivative by partial fractions?

Answer 90

and it ought to work, because \(5-11+6=0\)

Answer 91

is that part correct?

Answer 92

yes, i did it with wolfram
maybe i am being stupid somewhere

Answer 93

i really dont understand the steps of the maclaurin series so idk about the parts before..but i understand the changing the series to n=0 and all that looks correct to me

Answer 94

lets try \(-\frac{6}{(x-1)^3}\) again
we should get \(-6\) as the first term

Answer 95

oh i have a idea
what did you say the expansion of \(\frac{1}{(x-1)^3}\) was?

Answer 96

\[\sum_{n=0}^{\infty} n(n-1)x^n-2\]

Answer 97

with the -2 in the exponent

Answer 98

starting at 2 right?