find the maclaurin series for f(x) using known maclaurin series and evaluate $$f^{(4)}(0)$$
$$f(x)=e^{\cos(2x)}$$

Question

find the maclaurin series for f(x) using known maclaurin series and evaluate $$f^{(4)}(0)$$
$$f(x)=e^{\cos(2x)}$$

xapproachesinfinity · Answer

i tried to use maclaurin series for cos2x and take exp
but seems bad
tried to derived seems really tedious work with all those derivatives

anonymous · Answer

thank god for the internet, it is so smart http://www.sosmath.com/CBB/viewtopic.php?t=51552 scroll down to third answer

xapproachesinfinity · Answer

oh nice lol
i didn't put on google to see.

xapproachesinfinity · Answer

that looks neat:)

anonymous · Answer

yours will be slightly different because of the $2x$ but just as before the fact that it is even means there will only be even powers
i have no idea what is means "use know maclaurin"

xapproachesinfinity · Answer

yeah i know sub will do :)

amistre64 · Answer

ln(f(x)) = cos(2x)

ln(f(x)) = sum (-1)^n (2x)^(2n)/(2n)!

$$f(x)=exp\left(\sum_0 \frac{(-1)^n}{(2n)!} (2x)^{2n}\right)$$

$$f'(x)= \left(\sum_1 \frac{(-1)^n4n}{(2n)!} (2x)^{2n-1}\right)~exp\left(\sum_0 \frac{(-1)^n}{(2n)!} (2x)^{2n}\right)$$

$$f''(x)= \left(\sum_2 \frac{(-1)^n8n(2n-1)}{(2n)!} (2x)^{2n-2}+\sum_1 \frac{(-1)^n4n}{(2n)!} (2x)^{2n-1}\right)\~~~~~~~~~~~~~~~~~~~~~~exp\left(\sum_0 \frac{(-1)^n}{(2n)!} (2x)^{2n}\right)$$

pretty sure theres a pattern going here that can be simplified in a generality

amistre64 · Answer

4 derivatives of the summations, and a multiple of the exp(...)

xapproachesinfinity · Answer

it crossed my mind to use logs

amistre64 · Answer

with any luck
$$f^{(4)}(x)= \
\left(

\sum_4 \frac{(-1)^n32n(2n-1)(2n-2)(2n-3)}{(2n)!} (2x)^{2n-4}\
~~~+\sum_3 \frac{(-1)^n16n(2n-1)(2n-2)}{(2n)!} (2x)^{2n-3}\
~~~+\sum_2 \frac{(-1)^n8n(2n-1)}{(2n)!} (2x)^{2n-2}\
~~~+\sum_1 \frac{(-1)^n4n}{(2n)!} (2x)^{2n-1}\right)\

~~~~~~~~~~~~~~~~~~~~~~exp\left(\sum_0 \frac{(-1)^n}{(2n)!} (2x)^{2n}\right)$$

amistre64 · Answer

i dont think i see any constant in the summations

x, x^2, x^3, x^4 ... at x=0 is 0

f^(4)(0) = 0

xapproachesinfinity · Answer

that's not the answer should be 64e

amistre64 · Answer

yeah, 64e is the wolf .... do you see any error?

amistre64 · Answer

cos(2x) power series is good

xapproachesinfinity · Answer

hmm seems good to me
i think we need to align the indices
there is a technique that we can use

amistre64 · Answer

0:e^a

1:a' e^a

2:(a''+ a') e^a

3:(a'''+ a'' + a') e^a

4:(a''''+ a''' + a'' + a') e^a

amistre64 · Answer

pulling out indexes just pulls out an x factor

xapproachesinfinity · Answer

i was doing a similar thing with u=a in my case
could not keep track of the derivatives hahahA

amistre64 · Answer

power series derivativs tend to be mechanical power rules ....

xapproachesinfinity · Answer

ok so a'=-2sin2x
a''=-4cos2x
a'''=8sin2x
a''''=16cos2x
f''''(x)=(16cos2x+8sin2x-4cos2x-2sin2x)e^cos2x
f''''(0)=(16-4)e=12e

xapproachesinfinity · Answer

something is wrong with that hehe

amistre64 · Answer

$$a=\sum_0 \frac{(-4)^n}{(2n)!} ~x^{2n}\implies 1-x^2+...$$
$$a'=\sum_1 \frac{(-4)^n~2n}{(2n)!} ~x^{2n-1}\implies x+...$$
$$a''=\sum_2 \frac{(-4)^n~2n(2n-1)}{(2n)!} ~x^{2n-2}\implies x^2+...$$
$$a'''=\sum_3 \frac{(-4)^n~2n(2n-1)(2n-2)}{(2n)!} ~x^{2n-3}\implies x^3+...$$
$$a''''=\sum_4 \frac{(-4)^n~2n(2n-1)(2n-2)(2n-3)}{(2n)!} ~x^{2n-4}\implies x^4+...$$

xapproachesinfinity · Answer

i think taking derivatives and ignoring taylor is good for this

amistre64 · Answer

e^1 = e,  but im getting 0s for the rest ... 

1 + x^2 +x^4 + x^6 + x^8
2x +4x^3 + 6x^5 + 8x^7
2 +12x^2 + 30x^4 + 56x^6
24x +...

amistre64 · Answer

i spose we should expect at 2 of the derivatives to have constants

amistre64 · Answer

f'' and f''''

xapproachesinfinity · Answer

hmm I spent a lot of time on this
this was given on the test can't believe it hahah

xapproachesinfinity · Answer

the prof is nuts, this needs at 2 hours by itself and there is another that is also challenging

amistre64 · Answer

f^0 = 1 -2 x^2 + 2x^4/3 -4x^6/45 + ...
f^1 = -4 x + 8x^3/3 -24x^5/45 + ...
f^2 = -4 + 24x^2/3 -120x^4/45 + ...
f^3 = 48x/3 -480x^3/45 + ...
f^4 = 48/3 -480x^3/45 + ...

(48/3 - 4) e

(36/3) e

12e

xapproachesinfinity · Answer

should be 12e i keep getting the same thing too

amistre64 · Answer

we have the power series right, we have the derivatives right by expansion ....

we are centered at 0, which doesnt seem to give us a bad domain ....

amistre64 · Answer

we know how to take the derivative of e^a,  or e^u

amistre64 · Answer

dont we?

e^u

u' e^u

u'' e^u + u'u' e^u    well, maybe we dont lol

xapproachesinfinity · Answer

i think we made a mistake with derivatives there lol

amistre64 · Answer

yeah, looks that way

xapproachesinfinity · Answer

eh i give up i have others to finish

xapproachesinfinity · Answer

i will return to it and give another round

amistre64 · Answer

we are almost done ... if we define the 4th derivative in terms of u ... we have the functon already derived and know some of its zero


2nd:  u'' e^u + u'u' e^u 

3rd:  (u''' + u''u' + u'u'') e^u  + (u'' + u'u')u' e^u 

3rd:  (u''' + 2u''u') e^u  + (u'' u' + u'u'u') e^u 

4th:  (u'''' + 2u''' u'+ 2u'' u'') e^u  + (u''' + 2u''u')u'  e^u
            + (u''' u' + u'' u'' + 3u''(u')^2) e^u + (u'' u' + u'u'u')u' e^u 

u' = u''' = 0

(u'''' + 0+ 2u'' u'') e^u  + (u''' + 2u''u')0  e^u
            + (0 + u'' u'' + 0) e^u + (u'' u' + u'u'u')0 e^u 


(u''''+ 3(u'')^2) e^u 

(48/3+ 3(4)^2) e

(16+3(16)) e

4(16) e

64 e

amistre64 · Answer

define u as a power series, so that you can define values of u,u',u'',u''',u''''

generalize the derivative process, and sub in afterwards ... thats the only way i would approach this

xapproachesinfinity · Answer

thanks a lot i will retry it myself:)