Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
MathSofiya
Group Title
Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty} a_n x^n\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]
\[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{2}\]
\[xy'=\sum_{n=1}^{\infty}a_nnx^n\]
\[\sum_{n=2}^{\infty} n(n1)a_nx^{2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 2 years ago
 2 years ago
MathSofiya Group Title
Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty} a_n x^n\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\] \[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{2}\] \[xy'=\sum_{n=1}^{\infty}a_nnx^n\] \[\sum_{n=2}^{\infty} n(n1)a_nx^{2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 2 years ago
 2 years ago

This Question is Closed

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yep, now what's first?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
changing the sums so that they all begin with n=2 changing the last sum from n=0 to n=1 would be \[a_1+\sum_{n=1}^\infty a_nx^n\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no we don't strip out the terms yet, we just do an index shift
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the first order of business is to get all the x's to be to the nth power through using index shifts
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the only series that has x to some power other than n is the first one how can we change the index of the first series to make the exponent on x be n?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
you mean on \[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{n2}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, we need to start from a different n to get x^n through an index shift
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=22} \[\sum_{n=0}^\infty n(n1)a_nx^n\] negative to make up for the negative?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty n(n+1)a_nx^n\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
still not right
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
because that would give us zero
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no but closer
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you didn't add 2 to every n
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
and I do mean \(every\) n (subscripts included)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
again better, but...^
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
can I add them?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
now that, I do believe, is correct :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
or multiply
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
or just leave as is?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
add what? oh that? I wouldn't yet
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah leave as is asd now rewrite what you have
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the whole expression please I'm lazy
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]okay next order of business is what you tried to do earlier; get all indices to start at the same n value the only problem child here is which term?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
the middle
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so what do we do to fix it?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you have to be a bit clever here...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
did you try to do an index shift?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it looks like you combined the idea of the index shift with stripping out the first term
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh my...let's try this again
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it is a bit tricky so let me give you a hint...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
just have the addition of an a_n up front like in the last example?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no, answer my question above and you will see why...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
times a_n
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no, be more careful
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
right, so...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so we can go ahead and just change the starting point of that series without further manipulation
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
always watch for that, it comes in handy
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand) next to get all the starting points for each series the same, we strip out terms/or do the trick we just did they are different tricks, and the order in which you do them is imortant
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
good, so now make it all into one summation...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
nice, so what will you set to zero to find the recurrence relation?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
(I am going to type minimally now as I am eating dinner at the same time :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
food comes first in my book... :P Ok let's see here...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I can eat and respond with "yes" or "no" pretty well, no worries :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=\frac{a_nna_n}{(n+2)(n+1)}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
si, pero si quieras podrias factor la \(a_n\)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
ok \[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
and now comes the tedious part that you can do while I eat plug in n=0,1,2,3 and check what results from that recurrence relation until you notice a pattern like last time so just cranks it out; plug in n=0,n=1,n=2 and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
go ahead and list out the first 5 or so and let's see if we see a relation
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=1:a_3=0\] \[n=2:a_4=a_2\frac{1}{12}\] \[n=3:a_5=a_3\frac{2}{20}\] \[n=4:a_6=a_4\frac{3}{30}\] \[n=5:a_7=a_5\frac{4}{42}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what about n=0 ? we need to start from n=0 since our series starts from n=0 this time (last time it started from n=1)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
whatcha eating?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
torta de pollo=chicken sandwich on a french sort of bread
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
con chipotle
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh yeah...the pattern
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah, the pattern write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
each one*
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
take you're time...i'm gonna watch a few more mins of my episode
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it's not different, it's the same as last time odd subscript gives zero even has a pattern reducible to a_0
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
okay I got a fairly simple answer, let me see if I can confirm it...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
stupid wolfram is of no help :/
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\] \[n=1:a_3=0\] \[n=2:a_4=a_2\frac{1}{12}=\frac{a_0}{2(3!)}\] \[n=3:a_5=0\] \[n=4:a_6=a_4\frac{3}{30}=\frac{a_0}{5(3!)}\] \[n=5:a_7=0\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
how did I do?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
almost, I actually realized that I made a mistake because I think I see a mistake from you so you did as well as me I'd say :D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
let me try on paper again...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
here I think I can demonstrate the benefit of not multiplying out terms and writing things overexplicitly to find patterns...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\] \[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so try to do the same thing as last time name all even numbers and make a new series with index k
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
then represent the coefficients and plug that in for \(a_n\)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
is it over k! ?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well that is the answer I get... so unless you have a way of confirming it apart from this I say that is correct btw you should put the (2k)! with parentheses to indicate that it is not 2(k!)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[a_n=a_{2k}=\frac{a_0}{(2k)!}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah, the sub it into the original guess for y
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
...and that's what I got :D I sure think it's right as it comes out quite nicely, as opposed to wolf's answer http://www.wolframalpha.com/input/?i=y''%2Bxy'%2By%3D0&t=crmtb01
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I forgot, what did you intend to study? Electrical Engineering or Mathematics?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it definitely has the e^(x^2) think in it though just by glancing, so I am even more confident now
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
why the e^?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh because the tayor expansion of e^(ax^2) always has x^(2k)/(2k)! in it (try the expansion yourself and see why, it's not that hard if you remember Taylor series)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
don't have the brain cells left to do so, but I trust your reasoning
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
haha, well thanks. I hope I'm right. Nos vemos!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
my spanish sucks as you might be able to tell
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Nos vemos ~ we'll see each other
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I was right!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sweet :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
we've made a mistake
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\] is 24 not 12...so the pattern is probably incorrect too
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
let's see...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
and why should 24 be 12 exactly?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
because we need 12 4*3 is 12
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you dropped a 2...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh I see
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you wrote\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]\[n=2:a_4=a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]but\[a_2=\frac{a_0}2\]and\[\frac1{12}=\frac1{2(3!)}\]so 1)\[a_4=a_2\frac1{2(3!)}=\frac{a_0}2\cdot\frac1{2(3!)}=\frac{a_0}{4!}\]and that minus sign you put that minus sign does not belong
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2 :/
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I'll finish y''=y tomorrow...I got it except for the parts stated above
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
once you find a base formula in terms of \(a_0\) or whatever the lowest you can express is, milk it to find the pattern
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
and like I say, sometime the writing out each term, even if it's just multiplying by one, helps make the pattern obvious g'night then!
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348252850986:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348252918564:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348253019736:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348253097421:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1348253165705:dw
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest would this statement about striping a term be true (generally speaking) \[0+a_1+\sum_{n=2}^\infty a_n nx^n\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yep, exactly
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh no, you dropped an x
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
at n=1 the exponent on x is 1
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n\]\[=a_1x+\sum_{n=2}^\infty a_nx^n\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
easier example, stripping out terms for the sum of the first n integers\[\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
much better
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
thought so, probably should have done that a while ago, hehe :P
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
cool thanks !
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Awesome, always a pleasure :D
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest shouldn't \[a_4 \] have a minus sign? I know you said it shouldn't but \[a_2=\frac{a_0}{2}\] and now we have \[a_4=a_2\frac{1}{12}=\frac{a_0}{24}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
hm... I guess you may be right
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yuuss!!!!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
^^^^my way of saying: yes!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
so it's an alternating series? for 2k+1?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I mean 2k
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sorry I was doublechecking this whole time yeah it alternates, so you know the only change we're going to make?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
let's see if I remember, can I cheat?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
cheating would be having me tell you, any way you wanna figure out how to make a series change sign every other term is up to you :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
good idea brb
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
argh! \[\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.\] \[\sum_{n=0}^{\infty}(1)x^n=0,x,x^2\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what can we do to 1 to make it even?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sorry, positive*
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[(1)^n\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
AWESOME!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
very awesome!!!!!!!!!!!!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
hey I got 240 for a_6
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you love multiplying the numbers out, I just call it 6!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
no that's 720
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
haha whatever I'm not even checking if you are having a problem with the constants again give me a min and I'll brb again to help
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
sounds good... sorry don't mean to be nitpicking :S
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh no, I;m glad you pointed it out, but.... must... eat
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
dude go for it LOL I'll see if I can figure this pattern out
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I think this is incorrect \[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\] because \[a_{n+2}=\frac{a_nn1}{(n+2)(n+1)}\neqa_n\frac{n1}{(n+2)(n+1)}\] It should be \[a_n\frac{(n+1)}{(n+2)(n+1)}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{a_n}{n+2}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[a_{n+2}=\frac{a_n}{n+2}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=\frac{a_nna_n}{(n+2)(n+1)}\]is what you wrote, let's see if that is correct...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=a_n\frac{n+1}{(n+2)(n+1)}\]aw crap you're right again
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yuuus!!!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well, being right that we were wrong is always a little bittersweet, haha gotta do the pattern over, but at least it's only\[a_{n+2}=\frac{a_n}{n+2}\]which should be easier I would think
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Yeah, let's see if I can figure the pattern out... (btw I did yoga today...feeling more alert, or it could be plain coincidence =)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well you're catching all my mistakes, so it seems to be working :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
the pattern seems harder to write concisely for me now... though it seems obvious
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[a_2=\frac{a_0}{2}=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3}\] \[a_4=\frac{a_2}{4}=\frac{a_0}{3!}\] \[a_5=\frac{a_3}{5}\] \[a_6=\frac{a_4}{6}=\frac{a_0}{36}\] what's withe the 36?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
it's like the odds get cut out of the factorials and I'm trying to figure out how to write that same for the evens
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yeah it's quite strange, but the formula is correct, right? could we have still made an error there?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
a mistake in the recursion relation? we could have... to be safe we ought to take it from the top, but I can't do that, I have to leave...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
sure I'll start from the beginning and post it in a new post...this one is getting quite long
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
good idea I am thinking\[a_{2k}=(1)^k\frac{a_0}{2^kk!}\]and do not yet have an expression for the odds
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh ok, I'll see ya trmw then
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
hasta luego
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
hasta luego mi amigo!
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.