Question

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Answer 1

\[\int\limits e^{(sinx)}sinx dx\]

Answer 2

HEY LOKI

Answer 3

Evening :)

Answer 4

Make a substitution,\[u=\sin x\]

Answer 5

du=cos x

Answer 6

Yeah, hang on...

Answer 7

haha i tried..by parts..and substitution..

Answer 8

If you're looking for a solution in terms of standard integrals, I'm pretty sure it doesn't exist. You either have to write your answer in terms of the Bessel and Struve functions, or take a series solution, where you take the series expansion for sin(x), the expansion for e^(sin(x)) and then take the Cauchy product of the two series, and integrate term-by-term (which you can do since the series is uniformly convergent).

Where did this problem come from?

Answer 9

LOL umm someone asked this question yesterday, and i couldn't solve it.So i am asking you, the boss

Answer 10

Haha, well, boss says that.

Answer 11

oh dam be humble :P

Answer 12

lol, I am ;)

Answer 13

LOL sure boss :P

Answer 14

Well, I'm signing out...have a *lot* of work to do ><

Answer 15

I'll answer your ferris wheel question later :)

Answer 16

You should chillax, it's Friday.

Answer 17

question!!

Answer 18

What is the difference between power and geometric series!

Answer 19

I am chillaxing by chit chating with my roomate

Answer 20

Take your time on the wheel question :)

Answer 21

A geometric series is one where there's a constant ratio between each successive terms. A power series is different in that it is not always the case that the ratio between two successive terms is constant. In a geometric series, you'd have to successive terms,\[\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}=\frac{a_{n+1}}{a_n}(x-x_0)\]which is only constant if a_(n+1) and a_n give some number independent of n. In general, this isn't the case.

Answer 22

Thank you. I will try to do a little bit more research on my ow :)

Answer 23

This is research :)

Answer 24

LOL just go do your work!

Answer 25

Yeah I know...

Answer 26

I have Lammy below trying to figure something out. When he/she's done, I'm out... -.-

Answer 27

Alright i will ask questions later if have any :) Peace~

Answer 28

Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)

Answer 29

\[\sum_{1}^{\infty}\sin(9/x)\]

Answer 30

I don't think this converges. Let me check.

Answer 31

ok

Answer 32

I'm distracted because some moron in my building has been playing the pellettest music I've ever heard, all day, all loud and I'm about to go break down doors ><

Answer 33

Dude you are the boss, just order your followers to do it for you

Answer 34

lol, i wish :(

Answer 35

I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.

Answer 36

Sorry, had to go talk to the neighbor.

Answer 37

no problem :) i went through every single tests...but none of them worked

Answer 38

To use the limit comparison, we need a series that is divergent (since we're suspicious this thing won't converge). When I think 'non-convergence', the first series I look to to test against is the harmonic series.

Answer 39

Should I wait for you to finish typing?

Answer 40

lol what if i am suspicious this thing converge... then i will be in trouble

Answer 41

Lol, yes, you would be :D

Answer 42

Form the ratio\[\frac{1/n}{\sin \frac{9}{n}}\]and note that, in the limit, you get an indeterminate form, \[\frac{0}{0}\]

Answer 43

This is great, because you now have access to L'Hopital's rule.

Answer 44

So\[\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}-\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{-9\cos 9/n}{n^2}}=\frac{1}{9}\]

Answer 45

since cos(9/n) goes to 1 as n goes to infinity.

Answer 46

Now, since your ratio was that for a known divergent series with your own series, and because this ratio converged, you know that your series is divergent also.

Answer 47

So at the beginning, you have to guess whether this series diverges or converges?

Answer 48

Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.

Answer 49

A lot of mathematics is experimental.

Answer 50

But you yourself said you had a feeling that it was divergent...

Answer 51

On that integral before, you'd have to find more than just a couple of series expansions. I over-simplified it. If you ever see something like that in practice (i.e. engineering) you would use an approximation rule, like Simpson's or something.

Answer 52

I know it behaves like p-series as n->infinity, so if i had compared it to 1/n^2, it wouldn't follow any of the outcomes for the test?

Answer 53

1/n^2 is convergent, whereas your suspicion is that sin(9/n) is divergent. It wouldn't have helped.

Answer 54

i used the nth term test, and the limit goes to 0, so i thought it should be convergent.

Answer 55

Ahhhhh....no....

Answer 56

If you read the statement of the nth term test, it says:

Answer 57

If \[\lim_{n \rightarrow \infty}a_n \ne0 \] or if this limit does not exist as n tends to infinity, then\[\sum_{n}^{\infty}a_n \]does not converge.

Answer 58

The nth term test doesn't say that if your sequence tends to zero, your series converges.

Answer 59

It says if it DOESN'T tend to zero, your series DIVERGES.

Answer 60

Take 1/n for example. The limit is 0 as n goes to infinity, but the series itself does not converge.

Answer 61

yea, so , as a normal college kid, my suspicion would be convergent

Answer 62

SUSPICION

Answer 63

Now we come to 'instinct', which is a function of experience. It means you just need practice. You're looking at good questions. Trust your teachers - they'll show you all sorts of things you need to consider.

Answer 64

LOL what my math professors have taught me the most is HUMOR

Answer 65

Well, that *can* help..? :~S

Answer 66

i am stuck at number 3

Answer 67

you mean 4?

Answer 68

3..that answer is wrong

Answer 69

Does this have to be submitted now?

Answer 70

Nope, but soon~~->>>> tomorrow night

Answer 71

Oh, okay. I need to go out, that's all. I'll take a look later.

Answer 72

OH alright :) have fun!

Answer 73

Bye..

Answer 74

Did you figure it out?

Answer 75

Nope haha you told me to chillax

Answer 76

Slacker... ;p

Answer 77

I'll look now.

Answer 78

take your time. just look at it whenever u feel like it

Answer 79

The first one can be 2/361. You have 1/361.

Answer 80

Have you covered asymptotic equivalence?

Answer 81

what is that? might have..not sure

Answer 82

brb

Answer 83

kay kay

Answer 84

I know that as n-> infin, some terms become negligible

Answer 85

Yeah, basically. The definition is this:

Answer 86

But how you determine which term is going to infinity faster

Answer 87

Ah, there's an order to it. You can see these orders if you make plots of the corresponding functions.

Exponential functions will dominate polynomials at some point. You'll see, I'll write out how I worked out your question.

Answer 88

can you please do number 4 also :P

Answer 89

\[\frac{2^n+n^5+2}{19^{2n}+6^n+3} \iff \frac{2^n}{(19^2)^n}\]since in the numerator, the exponential dominates the quintic and 2, and in the denominator, (19^2)^n dominates 6^n+3 (19^2 > 6).

Answer 90

The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.

Answer 91

brb

Answer 92

i thought n^5 is going to infinity faster

Answer 93

No, if you plot n^5 and 2^n, you'll see 2^n overtake pretty quickly.

Answer 94

But listen, you can always check domination by using the definition. I'll scan in something.

Answer 95

You asked for tricks in math - I love asymptotic equivalence when it comes to limits - along with L'Hopital's rule. It saves your butt.