Question

\[f(x)=\frac{x}{2x^2 +1}=\]
\[x*\frac{1}{2x^2+1}\]
I don't know how to write this as a sum

Answer 1

\[\sum_{n=0}^{\infty}\]

Answer 2

n is varying or x is varying?

Answer 3

I'm not sure, It's a power series

Answer 4

here is an example from my book
\[\frac{x^3}{(x+2)}= x^3*\frac{1}{x+2}=x^3\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^n=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^{n+3} \]

Answer 5

firstly tell me do you know what is a power series?

Answer 6

here is what I under stand...
we have a function like \[\frac{x^3}{(x+2)}\]
and we are trying to express it as a sum of infinitely many terms.
so we're trying to trace the line : \[\frac{x^3}{x+2}\] with partial sums...

Answer 7

I think something just clicked

Answer 8

\[\sum c_n(x-a)^n=c_0 +c_1(x-a)^2+....\]

Answer 9

\[x=c_n\]
\[(x-a)=\]

Answer 10

a=-1

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
http://mathworld.wolfram.com/PowerSeries.html

Answer 12

check them they will help you

Answer 13

\[x=2x^2\]?

Answer 14

Are you looking for the Taylor series of the function you wrote?

Answer 15

no, Power series

Answer 16

\[\sum c_n(x-a)^n=c_0 +c_1(x-a)^2+....\]

Answer 17

you gave an example from a book.
which book is it?

Answer 18

stewart

Answer 19

6th edition I believe

Answer 20

calculus?

Answer 21

yes

Answer 22

I think \[c_n =x\] in this problem

Answer 23

would that make any sense? I'm new to power series

Answer 24

I can write the whole theorem if you'd like

Answer 25

yes that might help

Answer 26

is it one-variable calculus or multivariable?
the book.

Answer 27

early transcendentals

Answer 28

i gues that's one variable

Answer 29

does it have violins on the cover?

Answer 30

yes

Answer 31

If the power series \[\sum c_n(x-a)^n\] has a radius of convergence R>0 then the function f defined by \[f(x)=c_0 +c_1(x-a)+c_2(x-a)^2+....=\sum_{n=0}^{\infty}c_n(x-a)^n\]
is differentiable ( and therefore continuous) on the interval (a-R, a+R) and
(i) \[f'(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+....=\sum_{n=1}^{\infty} nc_n(x-a)^{n-1}\]
(ii) \[\int f(x)dx=C+c_0(x-a)+c_1\frac{(x-a)^2}{2}....=C+\sum{n=0}^{\infty}c_n\frac{(x-a)^{n+1}}{n+1}\]

Answer 32

the radii of convergence of the power series in equations (i) and (ii) are both R

Answer 33

that was quite tedious

Answer 34

i bet

Answer 35

so.
are you looking for a series that converges to
\[ \frac{x}{2x^2+1} \]
i.e.
\[ \frac{x}{2x^2+1}=\sum_{n=0}^\infty c_n(x-a)^n \]

Answer 36

the question reads
Find a power series representation for the function and determine the interval of convergence

Answer 37

give a minute ok?

Answer 38

sure, I've been trying to figure this out for 2 days now

Answer 39

does it give a point for the convergence?
i mean, you can't have a power series to cenverge to the function "everywhere"

Answer 40

In the example that I gave, the interval of convergence is (-2,2)

Answer 41

\[\frac{x^3}{(x+2)}= x^3*\frac{1}{x+2}=x^3\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^n=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^{n+3}\]

Answer 42

I will be right back

Answer 43

are you back?

Answer 44

an

Answer 45

anyway

Answer 46

in the example you just gave me if
\[ f(x)=\frac{1}{x+2} \]
then
\[ f^{(n)}=\frac{(-1)^{n+1}\cdot n!}{(x+2)^{n+1}} \]
do you agree?

Answer 47

I'm back

Answer 48

not quite...why is it n+1?

Answer 49

is it because it's even

Answer 50

you are right. it should be only n

Answer 51

well the book has \[(x+2)^{n+1}\] why is it n+1?

Answer 52

ok

Answer 53

the consecutive derivatives would be
\[ f'(x)=\frac{-1}{(x+2)^2} \]
\[ f''(x)=\frac{2}{(x+2)^3} \]
\[ f'''(x)=\frac{-6}{(x+2)^4} \]
do you agree?

Answer 54

yes

Answer 55

ooh hence n+1?

Answer 56

so the n-th derivative is
\[ f^{(n)}(x)=\frac{(-1)^n\cdot n!}{(x+2)^{n+1}} \]
agree?

Answer 57

yes sir! one more thing...why n! ?

Answer 58

notice de numerators of the fractions.
they are 1=1!, 2=2! and 6=3!
ok?

Answer 59

yes

Answer 60

got it

Answer 61

for instance
\[ f^{(4)}(x)=\frac{24}{(x+2)^5} \]