Question

please helpppppppppppppppppp...

Answer 1

\(\rm Expand~log(1+x+x^2+x^3)~in ~powers~of~x~upto~x^8\)

Answer 2

@ganeshie8 @Hero @zepdrix @AravindG pls help

Answer 3

@SithsAndGiggles

Answer 4

sounds like you are wanting to find taylor series at a=0

Answer 5

m not so familiar with these problems :(

Answer 6

\[f(x) \approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+ \cdots + \frac{f^{(8)}(x)}{8!}x^8\]

Answer 7

yep

Answer 8

but finding the 8th derivative is bit tedious

Answer 9

well do you already know a series formula for say log(x)?

Answer 10

hey @freckles i'll be right was waiting 4 a long time
i'll be back in 15mins
got a urgent work
please excuse me for sometimee
ill be back asap

Answer 11

well if you know this:
\[\log(1+u)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{u^n}{n} \\ \text{ then you can replace } u \text{ with } x+x^2+x^3 \\ \log(1+x+x^2+x^3)=\sum_{n=1}^{\infty } (-1)^{n+1} \frac{(x+x^2+x^3)^n}{n}\]
and at least you would haven't to find derivatives

Answer 12

still looks a bit tough

Answer 13

m back

Answer 14

thanks for your patience @freckles :)

Answer 15

\[1+x+x^2+x^3=\frac{1-x^4}{1-x} \\ \log(1+x+x^2+x^3)= \log(1-x^4)-\log(1-x)\]
I wonder if this form will be easier to look at

Answer 16

wow
wonderful!!!
looks great :)

Answer 17

\[\log(1-u)=-\sum_{n=1}^{\infty} \frac{u^n}{n} \]

Answer 18

I think that does look easier
i'm gonna try it

Answer 19

yep looks great in my opinion

Answer 20

so \[\rm \log(1-x)=-x-\frac{ x^2 }{ 2 }-\frac{ x^3 }{ 3 }-\frac{ x^4 }{ 4 }-\frac{ x^5 }{ 5 }-.......\infty\]

Answer 21

yeah and you would only need to go to x^8

Answer 22

and then you do the same thing for log(1-x^4) only going up til you gt x^8

Answer 23

hint: you only need two terms for that one :p

Answer 24

need to substitute x-->x^4
in that series

Answer 25

yah

Answer 26

two terms?

Answer 27

yeah :)

Answer 28

\[\log(1-u)=-(u+\frac{u^2}{2}) \text{ where } u=x^4 \]

Answer 29

4*2=8

Answer 30

lol oh like that hahaha

Answer 31

so the final answer would look like?

Answer 32

you write both series until you reach x^8 for each...
\[\log(1-x^4)-\log(1-x) \\ =-(x^4+\frac{x^8}{2})+(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^8}{8})\]

Answer 33

oh so for each term the series must reach to x^8

Answer 34

right since your question says go to x^8

Answer 35

or up to x^8 whatever

Answer 36

oh okay...

Answer 37

shall i post it here or a new post?

Answer 38

\[\rm~If~y^{1/m}+y^{-1/m}=2x~\\~Prove~that~:~(x^2-1)y_{n+2}+(2n+1)xy_{n+1}+(n^2-m^2)y_n=0\]

Answer 39

doesn't matter

Answer 40

what does y_i mean?

Answer 41

well its successive differentiation

Answer 42

ok

Answer 43

i'm used to \[y^{(i)}\] as notation for that

Answer 44

y^n
y_n
dy/dx
f'(x)

Answer 45

the first thing I would try is to see if I could find y'...
\[y^\frac{1}{m}+y^{\frac{-1}{m}}=2x \\ \text{ multiply both sides by } y^{\frac{1}{m}} \\ y^{\frac{2}{m}}+1=2x \cdot y^{\frac{1}{m}} \\ y^{\frac{2}{m}}-2x \cdot y^{\frac{1}{m}}+1=0 \\ \text{ \let } u=y^{\frac{1}{m}} \\ u^2-2x \cdot u+1=0 \\ u=\frac{2x \pm \sqrt{(-2x)^2-4(1)(1)}}{2(1)} \\ \]

my first thought is to solve for y... from some reason
let me think

Answer 46

okay

Answer 47

\[y=x \pm \sqrt{x^2-1} \\ \text{ \let's look at one of these for now } \\ y=x+\sqrt{x^2-1}... y'=1+\frac{1}{2}(x^2-1)^{\frac{-1}{2}}(2x) =1+x(x^2-1)^\frac{-1}{2} \\ \]
I don't know that might look crazy after awhile

Answer 48

hmm.. :/

Answer 49

do you know of any ideas or like where the problem comes from at least
so i can look up the topic

Answer 50

hey u actually solved it
\[y^{1/m}=(x+\sqrt{x^2-1})\\y=(x+\sqrt{x^2-1})^{m}\]

Answer 51

I didn't prove anything though

Answer 52

now diff y once

Answer 53

\(\sqrt{x^2-1}~y_1=m~y\)

Answer 54

oops I replace u with y earlier
I meant to put y^(1/m)

Answer 55

since we need y_2
so we will differentiate again

Answer 56

do you know Lebnitz theorem?

Answer 57

@freckles u der?

Answer 58

yeah

Answer 59

i look up theorem
\[\frac{d}{dx}(\int\limits_{y_0}^{y(1)} f(x,y) dy)=\int\limits_{y_0}^{y_1} f_x(x,y) dy\]

Answer 60

that is what is says leibniz's rule is

Answer 61

ihk how to use lebnitz

Answer 62

for this problem?

Answer 63

wait a sec

Answer 64

\[\rm y_n=(uv)_n=u_nv+nC_1~\cdot ~u_{n-1}~\cdot~v_1+nC_2~\cdot u_{n-2}v_2\]

Answer 65

it is like this
lol

idk how to choose u n v

Answer 66

i don't recall that :p

Answer 67

I would have to do some more research

Answer 68

but @rvc you are able to solve your problem now?

Answer 69

or you still stuck and we should call @ganeshie8 ?

Answer 70

yep

Answer 71

call him

Answer 72

I did
let's see if he comes running :p

Answer 73

@ganeshie8 @zepdrix

Answer 74

http://math.stackexchange.com/questions/906642/if-y-frac1m-y-frac-1m-2x-show-that-x2y-n22n1xy-n1
I don't know if I get this solution.
Do you understand it at all?

Answer 75

well I might get parts of it but not the whole thing

Answer 76

thats what i was saying
after differentiating twice if we apply the theorem we will get the ans
but idk how to apply it :(

Answer 77

\[\frac{d^n}{dx^n}[(x^2-1)y^{(2)}] \\=[x^2-1] \cdot \frac{d^{n}}{dx^n}(y^{(2)})+ \frac{d}{dx}[x^2-1] \cdot \frac{d^{n-1}}{dx^{n-1}}(y^{(2)})+\frac{d^2}{dx^2}[x^2-1] \cdot \frac{d^{n-2}}{dx^{n-2}} (y^{(2)}) \\ \\ \text{ but } \frac{d^3}{dx^3}[x^2-1]=0 \text{ and so } \frac{d^k}{dx^k}[x^2-1]=0 \text{ for } k \ge 3 \\ \text{ so we have } \frac{d^n}{dx^n}[(x^2-1)y^{(2)}] \\ =[x^2-1]y^{(n+2)}+2x \cdot y^{(n+1)}+2 \cdot y^{(n)}\]

Answer 78

oops forgot to throw in the other coefficient action going on

Answer 79

\[[x^2-1]y^{(n+2)}+n \cdot 2x \cdot y^{(n+1)}+\frac{n(n-1)}{2!} \cdot 2 \cdot y^{(n)}\]

Answer 80

http://wdjoyner.com/teach/calc1-sage/html/node106.html
using the formula on this page

Answer 81

i c

Answer 82

\[\frac{d^n}{dx^n}(xy^{(1)})=x \cdot \frac{d^n}{dx^n}y^{(1)}+ n \cdot \frac{d}{dx}(x) \cdot \frac{d^{n-1}}{dx^{n-1}} y^{(1)} \\ \text{ we don't need to go any further because } \frac{d^{2}}{dx^2}(x)=0 \\ \text{ and so } \frac{d^k}{dx^k}(x)=0 \text{ for } k \ge 2\]

Answer 83

\[\frac{d^n}{dx^n}(xy^{(1)})=x y^{(n+1)}+n y^{(n)}\]

Answer 84

and then we can do the last term easily

Answer 85

actually if only i understood how to use the theorem i would have solved it further n would not have wasted your precious time :(
im extremely sorry for that :(

Answer 86

I was using the formula above...
http://wdjoyner.com/teach/calc1-sage/html/node106.html
the exact one on this page

Answer 87

that is the theorem im talking about :)

Answer 88

one sec it got cut off

Answer 89

\[\frac{d^n}{dx^n}[(x^2-1)y^{(2)}] \\=[x^2-1] \cdot \frac{d^{n}}{dx^n}(y^{(2)})+ n\frac{d}{dx}[x^2-1] \cdot \frac{d^{n-1}}{dx^{n-1}}(y^{(2)})+ \\\frac{n(n-1)}{2}\frac{d^2}{dx^2}[x^2-1] \cdot \frac{d^{n-2}}{dx^{n-2}} (y^{(2)}) \\ \\ \text{ but } \frac{d^3}{dx^3}[x^2-1]=0 \text{ and so } \frac{d^k}{dx^k}[x^2-1]=0 \text{ for } k \ge 3 \\ \text{ so we have } \frac{d^n}{dx^n}[(x^2-1)y^{(2)}] \\ =[x^2-1]y^{(n+2)}+n \cdot 2x \cdot y^{(n+1)}+\frac{n(n-1)}{2} \cdot 2 \cdot y^{(n)}\]

Answer 90

did you understand this?

Answer 91

like I chose v to be x^2-1
and u to be y^(2)

Answer 92

I plugged them into that formula but as you will start to see when we get to the third derivative of v we will 0
and 0 after that so the rest of the formula doesn't matter

Answer 93

yep
understanding a bit

Answer 94

\[v=x^2-1 \\ \frac{dv}{dx}=2x \\ \frac{d^2v}{dx^2}=2 \\ \frac{d^3v}{dx^3}=0 \\ \frac{d^4v}{dx^4}=0 \\ \cdots \frac{d^nv}{dx^{n}}=0 \text{ for } n \ge 3 \]
so you only need the first three terms from that formula on that page

Answer 95

yes
i m getting it now
that makes sense to me lol

Answer 96

\[\frac{d^n}{dx^n}(uv)=\frac{d^n u}{dx^{n}} \cdot v+n \frac{d^{n-1}u}{dx^{n-1}} \cdot \frac{dv}{dx} +\frac{n(n-1)}{2} \frac{d^{n-2}u}{dx^{n-2}} \frac{d^2v}{dx^2} \text{ since } v=x^2-1\]