Question

PLEASE HELP! I need help with these two questions urgently =( http://puu.sh/d3nyh/a06a169609.png
http://puu.sh/d3nyS/ff5327ac75.png

Answer 1

It may not be as hard as you are thinking.

Consider these:

\(\dfrac{1}{1-x} = 1 + x + x^{2} + x^{3} + ...\)

\(\dfrac{1}{1+x} = 1 - x + x^{2} - x^{3} + ...\)

\(\dfrac{5}{1+x} = 5\cdot\left(1 - x + x^{2} - x^{3} + ...\right)\)

\(\dfrac{5}{1+9x} = 5\cdot\left(1 - (9x) + (9x)^{2} - (9x)^{3} + ...\right)\)

Then what?

\(\dfrac{5}{1+(9x)^{2}} = ??\)

Of course, things need to converge in order for this to work. The radius of convergence is quite important.

Answer 2

As far as the first, convince yourself that it converges for x < 0. It's no longer alternating for x < 0.

Answer 3

for your question it would be (5*(1-((9x)^2)+((9x)^3)....)

But yeah I'm not sure how to find the radius of convergence as I'm used to only used to find the interval of convergence which has the n-term - this question loses me

Answer 4

No, that's the same as \(\dfrac{5}{1 + (9x)}\). You should get something else for \(\dfrac{5}{1 + (9x)^{2}}\).

As far as radius of convergence, won't a ratio test do?

Answer 5

Surmising your own n-term can take a little thought. Give it a go.

Answer 6

Is it (5*(1-((9x)^2)+((9x)^4)+((9x^5)....) ? sorry my mind is really just a mess right now

And as for the convergence test I had

5(Σ 0 to inf) (-9x^2)^n
then had
abs(5 (-9x^2)^n) < 1
then abs(-9x^2) < 1/5
which gave an interval of conv of -sqrt(1/45) < x < sqrt(1/45)

then from there I thought my R was 0.14907 but that is wrong apparently

Answer 7

those questions marks is juat a sigma sing

Answer 8

sign*

Answer 9

i have solved c0 c1 c2 c3 and c4 but can't seem to get R right

Answer 10

update - have solved the question - I only need to know how to get the smallest nonzero coefficient question now

Answer 11

@Abhisar is it possible that you can help me with my problem?

Answer 12

Sorry, mathematics is not my subject :(

Answer 13

Ah darn sorryyyy

Answer 14

Still no. Those exponents should be 0, 2, 4, 6, 8, ... etc.

Answer 15

@tkhunny yeah I fixed that up already and solved that question - can you explain further on how to find the lowest nonzero coefficient ?

Answer 16

it was 5x^2(arctan(x^8))

Answer 17

and for the answer I cannot have an x in my term or it is considered wrong - so there has to be a value that must plug in for x - I got to that part but I guess I was off somewhere as I got 5x^10 for the answer I submitted

Answer 18

Got a hint and it said
The lowest term with non-zero coefficient would be when k=0. So, after finding the series for your given function, put k=0 and you will get the coefficient.

Not sure what K is referring to ? @tkhunny

Answer 19

k is just a sequence number.

Answer 20

Okay, you caught me sleeping.

\(\dfrac{d}{dx}\arctan(x) = \dfrac{1}{1+x^{2}} = 1 - x^{2} + x^{4} - x^{6} + ...\)

Integrate Termwise
\(\arctan(x) = x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7} + ...\)

Now!!!!
\(\arctan\left(x^{8}\right) = x^{8} - \dfrac{x^{11}}{3} + \dfrac{x^{13}}{5} - \dfrac{x^{15}}{7} + ...\)

Finally.
\(5x^{2}\cdot\arctan\left(x^{8}\right) = 5x^{10} - 5\dfrac{x^{13}}{3} + 5\dfrac{x^{15}}{5} - 5\dfrac{x^{17}}{7} + ...\)

Now, we can answer the question. Do we remember what it was?

Answer 21

What is the lowest nonzero coefficient - which should be 5x^10

Answer 22

Do you see a term with a lower degree? Let's go with the x^10, corresponding to k = 0, the first term.

Okay, you made me do all the work on these. Let's see you do one.

Answer 23

If we're going with the first term - I'm not sure what to do from here as this is where I was originally stuck on - you can see in the screenshot I posted I already had this answer but did not know what to do with it - sorry I'm just really lacking in knowledge