Use power series to solve the differential equation.
y'=xy

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- anonymous

Use power series to solve the differential equation.
y'=xy

- schrodinger

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- anonymous

I guess the first step is this
\[y'-xy=0\]

- experimentX

hold on ... just let
\[ y = \sum_{n=0}^\infty x^n \]

- experimentX

\[ y = \sum_{n=0}^\infty a_n x^n \]

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## More answers

- anonymous

\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}\]
according to my book, same thing I guess

- experimentX

yes yes ...
\[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0\]
find the recurrence relation.

- anonymous

oh ok
\[y''=\sum_{n=2}^{\infty} n(n-1)c_nx^{n-2}\]
how do I apply these generalizations to this specific problem

- experimentX

hah?? why do you want more trouble? you don't need to find y'' for this.

- anonymous

really, I was just trying to follow this example in the book

##### 1 Attachment

- anonymous

I meant: really?

- experimentX

lol .. this example is of second order differential equation. you need it only for second order DE

- anonymous

because there is only a y' that makes it first order?

- experimentX

yep.

- anonymous

ok so I only consider this then, right?
\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}
\]

- experimentX

yes ... do you have a software called maple?

- anonymous

nope

- anonymous

but before solving for it, what do I sub?

- anonymous

or how do I know what do sub, I think I'm just afraid of sums :S

- anonymous

why do they have (n+2)(n+1) in the example?

- experimentX

from maple i got
y(x) = y(0)+(1/2)*y(0)*x^2+(1/8)*y(0)*x^4+O(x^6)

- experimentX

Honesty I don't like power series method.
here's how you do it. assume the solution has of the form
\[ y = \sum_{n=0}^\infty a_n x^n \]
so we are putting these vales in your DE and hunting a_n's

- anonymous

I'm back...

- anonymous

gtg again...

- experimentX

after putting the value of y in your DE, you get
\[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0 \\
\]
since on the left side you don't have negative power of x, put this \( \sum_{n=0}^\infty n \; c_n x^{n-1} \) = \( \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} \)
Also we can make \( \sum_{n=0}^\infty \; c_n x^{n+1} \) = \(\sum_{n=0}^\infty c_{n-1} x^{n} \)
now we have
\[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + \sum_{n=0}^\infty c_{n-1} x^{n} = 0
\\
\huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + c_{n-1} x^{n} = 0 \\
\huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} + c_{n-1} ) x^{n} = 0 \]

- experimentX

since you don't have power's of x on the right side, you have
\[ (n+1) c_{n+1} + c_{n-1} = 0 \\
\text{ or, } c_{n+1} = {- c_{n-1} \over n+1}\]
this is the recurrence relation you get
put n=1, you get
|dw:1348083846860:dw|
this way you find the coefficients. hence you have solution.

- experimentX

not a nice method ... purely brute. but this works for all non linear differential equations ... even if their solution is not closed, you (put it up) as special function like Airy function, Bessel function, Hermite ... etc.
and C_0 is your constant of integration. also you might wanna check Frobenius method.
Best of luck with these.

- TuringTest

I think you made a slight mistake @experimentX though I'm no fan of power series myself

- experimentX

hmm ... could you be more specific? power series is the method that is most prone to error ... I never get it correctly on first attempt (for second order)

- experimentX

oh ... i see that.
i put up + instead of -

- TuringTest

I'm just going to go ahead and give mine a shot, it's hard for me tow see how you ended up with c_{n+1} and c_{n-1} though

- TuringTest

yeah that

- experimentX

EDIT::
\[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - \sum_{n=0}^\infty c_{n-1} x^{n} = 0
\\
\huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} - c_{n-1} x^{n} = 0 \\
\huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} - c_{n-1} ) x^{n} = 0 \]

- experimentX

EDIT::
\[ (n+1) c_{n+1} - c_{n-1} = 0 \\
\text{ or, } c_{n+1} = {c_{n-1} \over n+1} \]

- experimentX

so the final solution is
\[ y = c_0\sum_{n=0}^\infty {1 \over 2^n n!}x^{2n}\]

- TuringTest

okay that was nice, let me see if doing it my way changes anything

- experimentX

this is same as
\[ \huge y = c_0 e^{x^2 \over 2}\]

- experimentX

2 am here. gotta sleep ... i'll be watching this thread tomorrow morning :)
gotta be careful with power series. this is quite annoying.

- TuringTest

yes your solution seems right @experimentX
sleep well, I still am gonna try to brush up myself :)

- TuringTest

@MathSofiya I have confirmed @experimentX 's answer in a slightly different way if you would like to go over it, let me know

- TuringTest

\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative)
Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series)
To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=1 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know.
So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\)
First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two.

- anonymous

I'm back

- TuringTest

at a very opportune time too :)

- anonymous

Sorry, women and hairstylist...
I'm reading everything now

- TuringTest

Take your time, it's not done... not sure if I should wait for you to try the next step.
I will prepare the answer in the meantime anyway :)

- anonymous

sounds good, It's gonna take me a while to understand it. I'm gonna type questions as I come upon them.

- anonymous

\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]
because
\[\sum_{n=0}^\infty a_nx^n x \] since the exponents are added that's why it's \[x^{n+1}\] correct?

- TuringTest

yep

- anonymous

I understand what you did by changing, we start one lower at n=1,
\[na_nx^{n-1} \implies (n+1)a_nx^n\]
but why does that change where the sum starts from n=1 instead now from n=0
\[\sum_{n=1}^{\infty} \implies \sum_{n=0}^{\infty}\]
sorry I guess I'm still struggling with the basics of summations

- TuringTest

try it both ways, write out the first two or three terms and convince yourself how starting earlier means you need to add to the argument

- TuringTest

for\[\sum_{n=1}^\infty na_nx^{n-1}\]what is the first term?

- anonymous

\[1\]

- TuringTest

not quite, it's \(a_n\)

- anonymous

right

- TuringTest

what about the first term of\[\sum_{n=0}^\infty(n+1)a_nx^n\]?

- anonymous

\[a_n\]

- anonymous

?

- anonymous

weird

- TuringTest

see how by starting our index at n=0 we had to add 1 everywhere to keep the terms the same
try it for more terms until you get the feel if you still doubt

- anonymous

does this mean that we're trying to manipulate the way we write the sum without changing the value of it?

- TuringTest

yes
if we started from zero the first term would be zero, because it has n as a coefficient
the next term would have a coefficient of 1, the one after of 2 etc.
if we add 1 then we get the correct coefficients the whole way through because we made up for changing the starting point of the count

- anonymous

ok so we change the starting point and made up for it.
Why are we changing the starting points, what is our eventual goal....ahhh, we want to be able to write the two sums as one sum?

- TuringTest

exactly :)

- TuringTest

we can only combine the sums once we have their indices at the same starting point
and as you will see, we also need the powers on x to be the same, customarily x^n....

- TuringTest

these are not short problems :P

- anonymous

indices...

- anonymous

{n=..... } are those the indices? I'm kinda rusty on the terminology

- TuringTest

yes exactly, n is the index of a sum with n=a under it, and "a" would be the starting point of the index
changing the starting point of an index is called an index shift, and requires you to alter the summand (the thing under the summation sign) as I have just tried to describe.

- TuringTest

that should say "we explicitly plugged in x=0..." *

- TuringTest

I mean n=0
:/

- anonymous

by taking an \[a_1\] out of the summation how did that change the indices? nothing else seems to have been changed.
\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]

- TuringTest

because we wrote out the first term explitily

- TuringTest

explicitly*

- anonymous

ohh...so either way the first answer would be a_1 is the point again I guess

- TuringTest

we don't need to start from n=0 anymore because we already wrote it out and took it out of the series, so repeating the n=0 term would add an extra term

- anonymous

on either side of that equal sign I mean.

- TuringTest

what is important right now is that we get the index to start at n=1 like the other term does

- TuringTest

if you have a series that starts at n=0, and you want to write it starting at n=1 you can either do an index shift (which we already did to get x^n on both series) or write out the n=0 term explicitly.

- anonymous

got it. I am down to this line now
\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
and I do understand how you came to this

- TuringTest

cool :)

- anonymous

I see a pattern
something like this
\[\frac{a_0}{2^n(n!)}\]

- anonymous

well I probably shouldn't use "n"

- TuringTest

yes! that is exactly right for half the n

- anonymous

oh so I can use "n"

- TuringTest

well no, we need to differentiate between the two cases for n
n only equals what you wrote under what condition for n ?

- TuringTest

a_n only equals what you wrote for which n ?

- anonymous

when n=0

- anonymous

no no no

- TuringTest

actually the n in your formula is outright wrong sorry
but look at it another way so you can figure how to fix it...
for which values of n is a_n zero?

- anonymous

durrr...all the even numbers

- anonymous

haha it took me while to see that :P

- TuringTest

yes, but I screwed up the question again, sorry I must phrase more carefully

- TuringTest

for which *subscripts on a* is a_n zero ?

- anonymous

all the odd subscripts

- TuringTest

right :) so and we can write any odd number as n=2k+1 with k=0,1,2..., right?
so any term with a coefficient we can write as \(a_n=a_{2k+1}=0\)
therefore our formula is only concerned with the even number subscripts since the odd ones will drop out

- TuringTest

so we only need to think about the terms with n=2k
by the way we skipped a set; plugging in n=0 to find what a_1 is (we can't do it with our formula because our series does not have a_1 defined)
to back up just for a moment, what does a_1=

- anonymous

0?

- TuringTest

yes, how did you know?

- anonymous

part of the pattern

- TuringTest

that does not illustrate the proper reasoning, and I think you know that ;)

- TuringTest

that reasoning will certainly not always work and will in fact often fail.
the real reason is as follows:\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]plug in n=0 and notice something: the series is not defined at n=0, only from n=1 on, so we just wind up with the whole series not in the equation\[a_1=0\]

- TuringTest

hence the problem we are now dealing with is really just\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]

- anonymous

interesting, yes

- TuringTest

so where were we....?

- anonymous

one more thing

- anonymous

so you're saying that I could drop the \[a_1\] because it is not defined at n=0
yes...ok

- TuringTest

a_1 is defined at n=0 (a_1=0), but the series is not

- anonymous

oh ok the series is not defined at n=0 and this is left
\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]

- anonymous

my goodness this is exhausting!

- TuringTest

\[a_1+\cancel{\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n}^{\text{not defined}}=0\]\[a_1=0\]leaving\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since \(x^n\neq0\) then the coefficients must be zero\[(n+1)a_{n+1}-a_{n-1}=0\]and we get the recurrence relation we were doing

- anonymous

agreed

- TuringTest

and yes, it is extremely exhausting I think most people who do DE's whould agree, but they handle non-constnat coefficient problems well

- anonymous

so would this finally be the solution
\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]
I doubt it...seems like it's just the beginning...now it's on to solving the DE

- TuringTest

I'm afraid we have ways to go dear, I would understand if you want a break or something...
I am mainly practicing, and I have a final solution from beginning to end prepared on a word doc, but it would be better to guide you of course.

- TuringTest

that solution above is not an explicit solution for\[ y=\sum_{n=0}^\infty a_nx^n\]which is what we want for an answer

- anonymous

Wow an example they did in my book took up 2 pages. yes.
I don't think I need a break yet...but when we're done with this problem I'm watching an episode of the Big Bang Theory and just laugh at all of their stupid jokes.
continue.....

- TuringTest

we need a formula for a_n in terms of the index, that's how we get our answer
we are mostly there actually, I just want to make sure you understand each step

- anonymous

"index"?

- TuringTest

you will see what I mean: that the index n (which don't get attached to cuz we're gonna change the name of it in a second) is going to give an explicit formula for \(a_n\) in terms of the only constant we can't determine \(a_0\)
so you need a formula for all even \(a_n\) in terms of n

- TuringTest

you almost had it, but you wrote\[a_n=\frac{a_0}{2^nn!}\]but check the numbers and you can see that's not quite right

- anonymous

what I screwed up on earlier by calling it \[\frac{a_0}{2^n(n!)}\]

- anonymous

ok I'll fix it...let's see here...

- TuringTest

hint: We are only talking about even numbers, the odd subscripts are all zero.
All even numbers can be written as...?

- anonymous

\[a_n= \frac{a_0}{2^{2k}(2nk!)}\]
If this is right.....

- TuringTest

almost

- anonymous

woops hold your horses

- anonymous

\[a_n= \frac{a_0}{2^{2k}(2k!)}\]

- TuringTest

but 2k=n....
the power on 2 is not the same as the subscript on a, check again the relation

- anonymous

\[a_{2k}=\frac{a_0}{2^{2k}(2k!)}\]
That just looks odd

- TuringTest

plug in k=2 and check if that's true based on the table I made way back up there

- anonymous

that gives me a_1 instead of a_0

- anonymous

am I loosing it?

- TuringTest

?
k=2
n=2k=4
check a_4

- anonymous

oh I see what you mean

- anonymous

\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]
uhm? what do you want me to check? the fact that it is not zero?

- anonymous

which was our goal

- TuringTest

no, I want you to check if that fits your formula\[a_{2k}=\frac{a_0}{2^{2k}(2k)!}\]or if not, how to fix it :)

- anonymous

divide it by half because it seems like double the answer
so \[a_{2k}=\frac{a_0}{2^{k}(k)!}\]

- TuringTest

yay!

- anonymous

no way!

- anonymous

now is that my y_p or something?

- TuringTest

and since these are the only a_n we care about we can now write\[a_n=a_{2k}=\frac{a_0}{2^kk!}\]and no, it's much better than our y_p...

- anonymous

it's my "y" final answer

- TuringTest

basically, yes
it is the key to the final answer

- TuringTest

first notice that the above is true for \(k=0,1,2,...\) and we already made the tacit assumtion that our answer is\[y=\sum_{n=0}^\infty a_nx^n\]now since all odd will give zero terms for \(a_n\), there will be only even n in our formula
so we can change all n in the series to 2k, start our series from k=0, and substitute the above formula for \(a_n\) and that will be our final answer

- anonymous

\[y=\sum_{k=0}^\infty \frac{a_0}{2^kk!}x^{2k}\]

- TuringTest

you got it :)

- TuringTest

and just to make things crystal clear that we are in fact right...

- TuringTest

separation of variables gives
y'=yx
dy/y=xdx
ln y=x^2/2+C
y=Ce^(x^2/2)
the Tayore xpansion for e^y is\[e^y=1+\frac y{1!}+\frac{y^2}{2!}+...=\sum_{n=0}^\infty\frac{y^n}{n!}\]let \[y=\frac12x^2\]and plug that into the taylor series and you get the same answer

- TuringTest

with \(C=a_0\) of course

- TuringTest

experimentX was right too, but I thought I would try to practice through explanation myself :)

- anonymous

Amazing!
I owe you half of my paychecks when I begin working as civil engineer. Dude, you are awesome. Thanks for helping me.

- anonymous

patience is a virtue...you have an abundance of that...oh yeah, sorry I forgot you don't like compliments :P

- TuringTest

Haha, don't tempt me I could use them. And thanks as always :)
For posterity's sake I wanna post my full explanation without interruption, just because I already had it prepared and it may be easier to read should you decide to review this.
It will slow down the page though...

- anonymous

Thank you. I'm off to watch an episode of the Big Bang Theory. Is that silly? LOL

- TuringTest

Makes more sense than doing more series solution DE's to me!
I'm off to do something unproductive too, see ya!

- anonymous

see ya!

- anonymous

medal?

- TuringTest

gotta take it away from experimentX...
which I'll do, he has enough
sorry @experimentX ;)

- TuringTest

\[y'=yx\implies y'-xy=0\]\[y=\sum_{n=0}^\infty a_nx^n\implies xy=\sum_{n=0}^\infty a_nx^{n+1}\]\[y'=\sum_{n=1}^\infty a_nnx^{n-1}\](notice the index starts at n=1 for the series for y' because the first term in the series for y is a constant \(a_n\) and disappears upon taking the derivative)
Our problem then becomes\[\sum_{n=1}^\infty a_nnx^{n-1}-\sum_{n=0}^\infty a_nx^{n+1}=0\]The first thing is to get all the x's to the \(n^{th}\) power with index shifts, so we will start the first series one lower at n=1 (so we add 1 to n in the summand) and the second series we will shift up by 1 (so that n shifts down by 1 in the summand). Note these shifts also change the indices of the constants \(a_n\)\[\sum_{n=0}^\infty( n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]Next we need to get the indices the same again (make n start at the same number for each series)
To avoid doing another index shift that will take us back where we started, we will "strip" the n=0 term out of the first series\[\sum_{n=0}^\infty(n+1)a_{n+1}x^n=a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n\]notice all we did was plug in n=0 and explicitly write that term out in front of the rest of the series. They call that "stripping out a term" in case you did not know.
So we now have\[a_1+\sum_{n=1}^\infty(n+1)a_{n+1}x^n-\sum_{n=1}^\infty a_{n-1}x^n=0\]\[a_1+\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]Here we now need to figure out a formula for any pattern(s) in the constants \(a_n\)
First, plugging in \(n=0\) since the series is not defined we get\[n=0:a_1=0\] so now we just have\[\sum_{n=1}^\infty[(n+1)a_{n+1}-a_{n-1}]x^n=0\]and since exponentials like \(x^n\) can never be zero, and the summation sign will not change the fact that what is within the brackets must be zero we have the "recurrence relation"\[(n+1)a_{n+1}-a_{n-1}=0\implies a_{n+1}={a_{n-1}\over n+1}\]so again we start plugging in values for n starting from n=1\[n=1:a_2=\frac{a_0}2\]\[n=2:a_3=\frac{a_1}3=\frac03=0\]\[n=3:a_4=\frac{a_2}4=\frac{a_0}{2\cdot4}=\frac{a_0}{2^2(1\cdot2)}=\frac{a_0}{2^2(2!)}\]\[n=4:a_5=\frac{a_3}5=\frac05=0\]\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]\[~~~~~~~\vdots\]so what's the pattern for the constants? There seem to be two; one for even subscripts and one for odd. Even numbers can be written as \(n=2k\) and odd as \(n=2k+1\) for \(k=0,1,2,...\)
That means that all the \(a_n\) for even n can be written as \(a_{2k}\) and have the form\[a_{2k}=\frac{a_0}{2^kk!}\]and for all odd n we have \[a_{2k+1}=0\]therefor all those terms will disappear from the series. That lets us replace all the n's in the summand with 2k and make a new representation of the series based on that. This way we don't include the odd n terms, which are zero anyway.
We can now rewrite \[a_n=a_{2k}=\frac{a_0}{2^kk!}~~~~\text{where}~~~~k=0,1,2,3...\]and our guess for y becomes\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{2^kk!}\]

- experimentX

you guys have patience of a saint ... usually i don't write more than lines.

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