Question

Power series question.

Answer 1

\[\sum_{k=0}^{\infty} (-1)^n*(2x)^{2n}\]In my book it says that we dan wrive this as
\[\sum_{k=0}^{\infty}c_{k}x ^{k}\] with \[c_{k}= (-1)^{k/2}2^k\] if k is even. and\[c_{k}= 0\] if k is not even.

Why can we set \[c_{k}\] equal to these values?

Answer 2

check your sigma please

Answer 3

wrive should be write.*

Answer 4

I figured, and k should be n, or the other way around

Answer 5

yes, the first sigma should be n=0 instead of k=0.

Answer 6

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ (-1)^n\cdot (2x)^{2n}\right]}\)

and you just need to covert this to a form of: \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ c_n\cdot x^n\right]}\) ?

Answer 7

yes indeed, I need to the values for c_{k} to calculate the convergence radius. The values are given but I don't know how these were formed.

Answer 8

you just need a bunch of manipulations...

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ (-1)^n\cdot (2x)^{2n}\right]}\)

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ (-1)^n\cdot x^{2n}\cdot 2^{2n}\right]}\)

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ (-1)^n\cdot x^{2n}\cdot 4^{n}\right]}\)

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty }\left[ (-4)^n\cdot x^{2n}\right]}\)

or something like this....

Answer 9

but for the radius of convergence you can use the ratio test, without doing any conversions. Ratio test directly

Answer 10

Which course is this, calculus, right?

Answer 11

My native language is dutch so I'm not to sure what it is called in English but it's where you see all the things about convergence, divergence of series.

Answer 12

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|}\)

Answer 13

this is the ratio test

Answer 14

Yes I know about that method, but I'm in a chapter explaining the use of the convergence radius. Which uses \[R = \lim_{k \rightarrow \infty}(c_{k}/c_{k+1})\] to find values for |x| that makes the serie convergent.

Answer 15

i think you are using the ratio test, but confusing something

Answer 16

and it is the a(n+1)/a(n) ...

Answer 17

Thats why I need to find the value for \[c_{k}\], which is given. But I don't know how to get these.

Answer 18

In your case, (if we use it directly from were you started)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|}\)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|\frac{(-1)^{n+1}(2x)^{2n+1}}{(-1)^{n}(2x)^{2n}}\right|}\)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|\frac{(-1)(-1)^{n}(2x)(2x)^{2n}}{(-1)^{n}(2x)^{2n}}\right|}\)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|\frac{(-1)\cancel{(-1)^{n}}(2x)\cancel{(2x)^{2n}}}{\cancel{(-1)^{n}}\cancel{(2x)^{2n}}}\right|}\)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|(-1)2x\right|}\)

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|-2x\right|}\)

Answer 19

Now, for the series to covert, this limit of the ratio has to be less than 1

Answer 20

\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|-2x\right|}\)
there is no variables here, (as we treat x as a constant in the limit)
so this limit just equals....
\(\large\color{black}{ \displaystyle \lim_{n \rightarrow \infty}\left|-2x\right|=|-2x|}\)

For series to converge this limit must be equal less than 1, therefore you set.

\(\large\color{black}{ \displaystyle\left|-2x\right|<1}\)

Answer 21

then solve for x.

Answer 22

The only thing to do, is that you would need to check for interval boundaries by plugging in the x value into the series and seeing if it converges or not.

Answer 23

The ratio test you give makes use of the d'Alembert test. This is without the convergence radius. As I'm now learning about the convergence radius I need to find these values for \[c_{k}\]. My equation was correct. So radius = \[= \lim_{k \rightarrow \infty}\left| C_{k}/C_{k+1} \right|\]

Answer 24

Then to see if it converges, I can easily find values for |x| so that |x| < Radius; and thus is convergent for these values |x|.

Answer 25

@SolomonZelman Thanks for the input but I have already seen this method and understand it. The only thing I can't figure out is how they got to these values for \[c_{k}\] I gave earlier.

Answer 26

Maybe you understand me better by reading this; http://en.wikipedia.org/wiki/Power_series#Radius_of_convergence.
This is the method I intend to use, as your see it says \[r^{-1}\], by making that r you get the equation I gave earlier.

Answer 27

but it is \(\displaystyle \frac{a_{n+1}}{a_n}\)

Answer 28

r^(-1)=...

Answer 29

So r gives \[c_{k}/c_{k+1}\]

Answer 30

yes, limit rules allow you

\( {\rm r} =\displaystyle \frac{a_n}{a_{n+1}}\)

Answer 31

so you would have
|1/(-2x)|=r

Answer 32

You're correct in all you are saying but it is not quiete the answer I'm looking for. I would like to know why \[c_{k}\] equals the formulas I gave earlier, both if k is even and if k is not even.

Answer 33

I believe that for odd k's you get an imaginary output (you would be getting some square root of -1, and that is \(i\) )

Answer 34

so they give you that this imaginary value isn't at all a value - that imaginary is zero.
((I can see this, if you look at imaginary numbers as at rotations))

Answer 35

I can see why it is zero for k = odd, it says the series will only have terms for k is even, so the coefficient should be zero.

Still the formula for k is even is still bothering me.

Answer 36

So basically I found out that for the values of Ck they set n equal to k/2. Which is where these different powers come from.

But I don't see why they would do this. If you substitute n back into k, it would give the same result. Why would they want to to change these powers (n=k/2) instead of simply leaving the original as it is and say n =k. @SolomonZelman

Answer 37

with (-1)^n you would only have positives for even n's.
so they want alternative series for even n's.

but I honestly don't know why they give you that \(c_k\) formula in the first place.

Answer 38

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \left[(-1)^n(2x)^{2n}\right]}\)

negative for n=1, n=3, n=5, and for all odd n's
positive for n=0, n=2, n=4 and for all even n's

Answer 39

I didn't, or don't usually see convergence using \(c_k\) ... ratio test is the most frequent and convenient to me thing..... sorry that i can't really help you more.

Answer 40

True, they eventually conclude the same thing but it is merely used to show the use of convergence radius. I understand it more or less now, thanks for the time spent helping me!

Answer 41

yeah, shortly:

RADIUS OF CONVERGENCE
1) Ratio test, and set the value you get for limit less than 1 and solve for x.
Why less than 1? Because we are taking it as a geometric series, and geometric series will converge when |common ratio| is less than 1.

INTERVAL OF CONVERGENCE
2) Then, the x solutions, say, you got B>x>A
have to be checked, by plugging in x=A and x+B into the series and seeing if it converges and solution includes this value, or with the plugged value series diverges and thus this value is excluded from the solution.