Question

Maclaurin expansion series = x cos(x^2)

Answer 1

if you don't already, you should half the maclaurin series for \(\cos(x)\) memorized

Answer 2

easy to do since cosine is even
it has all even exponents and it alternates

Answer 3

then replace \(x\) by \(x^2\) in the previous maclaurin series
finally, multiply your result by \(x\) i.e. kick up the exponent by 1 in each term

Answer 4

if this is not clear, let me know, it is only 3 steps the first being a memory step

Answer 5

so it would be x^2 cos x^3?

Answer 6

oh no
you are looking for the maclaurin series right? a power series

Answer 7

do you know the maclaurin series for \(\cos(x)\) ?

Answer 8

yeah and no i dont know

Answer 9

ok
you want to
a) look it up
b) derive it
c) me to tell you what it is?
either way
which ever way, once you have it you should memorize it

Answer 10

can you tell me what it is?

Answer 11

sure
and i will tell you how i memorize it too
first, \(\cos(0)=1\) so it starts with 1
second, it alternates
thirds, cosine is an even function, so all exponents are even
we can write it out as
\[1-\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}...\]

Answer 12

damn typo, it alternates!\[1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}...\] that's better

Answer 13

more succinctly written in sigma notation as
\[\large \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}\]

Answer 14

memorize that one!
also the one for \(\sin(x)\) and also \(e^x\)

Answer 15

now for \(\cos(x^2)\) replace each \(x\) by \(x^2\)
term by term it looks like
\[1-\frac{x^4}{2!}+\frac{x^6}{4!}-\frac{x^8}{6!}...\]

Answer 16

damn another typo
square means double each exponent, let me try again

Answer 17

\[1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}...\] that's better

Answer 18

more succinctly written in sigma notation as
\[\cos(x^2)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n}}{(2n)!}\]

Answer 19

how we doing so far? one more step...

Answer 20

So far so good. i understand everything you're saying. You break the problem down and everthing sound so clear.

Answer 21

good
then finally for \(x\cos(x^2)\) multiply all that mess by \(x\) term by term, which means really just add 1 to each exponent

Answer 22

\[x(1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}...\]
\[x-\frac{x^5}{2!}+\frac{x^9}{4!}-\frac{x^{13}}{6!}...\]

Answer 23

Thank you so much. I appreciate of what you did.

Answer 24

in sigma notation i guess you can write this as \[\cos(x^2)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+1}}{(2n)!}\]

Answer 25

oops i meant \(x\cos(x^2)\) but you know what i mean i hope
yw