# Taylor Series of exp(cos(x))

Watch
Announcements

OmegaKaos

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1

Find the Taylor Series for exp(cos(x)) about the point x=0 up to x^4

Really no clue where to even begin.

I know the Taylor series for e^x and cos(x)

My only idea was to take the series of cos(x) and substitute that for x in the series for e^x but that seems very daunting and i would be prone to make mistakes. So any better methods out there??

Really no clue where to even begin.

I know the Taylor series for e^x and cos(x)

My only idea was to take the series of cos(x) and substitute that for x in the series for e^x but that seems very daunting and i would be prone to make mistakes. So any better methods out there??

1

reply

Username1818

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2

Report

#2

I might be wrong... but wouldn't you just let f(x) = e^(cosx) and then differentiate it 4 times to get f'(x), f''(x), f'''(x), f''''(x), then substitute x=0 in each of the f(x)s and then plug the results into the formula ?

2

reply

user2020user

Badges:
16

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3

Report

#3

(Original post by

Find the Taylor Series for exp(cos(x)) about the point x=0 up to x^4

Really no clue where to even begin.

I know the Taylor series for e^x and cos(x)

My only idea was to take the series of cos(x) and substitute that for x in the series for e^x but that seems very daunting and i would be prone to make mistakes. So any better methods out there??

**OmegaKaos**)Find the Taylor Series for exp(cos(x)) about the point x=0 up to x^4

Really no clue where to even begin.

I know the Taylor series for e^x and cos(x)

My only idea was to take the series of cos(x) and substitute that for x in the series for e^x but that seems very daunting and i would be prone to make mistakes. So any better methods out there??

(Original post by

I might be wrong... but wouldn't you just let f(x) = e^(cosx) and then differentiate it 4 times to get f'(x), f''(x), f'''(x), f''''(x), then substitute x=0 in each of the f(x)s and then plug the results into the formula ?

**Username1818**)I might be wrong... but wouldn't you just let f(x) = e^(cosx) and then differentiate it 4 times to get f'(x), f''(x), f'''(x), f''''(x), then substitute x=0 in each of the f(x)s and then plug the results into the formula ?

As a side note, the Taylor expansion of a function about the point is called a Maclaurin expansion:

1

reply

OmegaKaos

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4

(Original post by

I think this method would be simpler as opposed to substituting the expansion of into that of .

As a side note, the Taylor expansion of a function about the point is called a Maclaurin expansion:

**Khallil**)I think this method would be simpler as opposed to substituting the expansion of into that of .

As a side note, the Taylor expansion of a function about the point is called a Maclaurin expansion:

when r=0 the value of f(r)=e^(cosx) is 1

when r=1 the value of d/dr(f(r))= -(sinx) * e^(cosx) is 0

when r=2 the value of d/dr(f(r))= -(sinx)^2 * e^(cosx) is 0

when r=3 the value of d/dr(f(r))= -(sinx)^3 * e^(cosx) is 0

when r=4 the value of d/dr(f(r))= -(sinx)^4 * e^(cosx) is 0

0

reply

Username1818

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5

Report

#5

(Original post by

Is the following at all right?? All those 0s worry me!!

when r=0 the value of f(r)=e^(cosx) is 1

when r=1 the value of d/dr(f(r))= -(sinx) * e^(cosx) is 0

when r=2 the value of d/dr(f(r))= -(sinx)^2 * e^(cosx) is 0

when r=3 the value of d/dr(f(r))= -(sinx)^3 * e^(cosx) is 0

when r=4 the value of d/dr(f(r))= -(sinx)^4 * e^(cosx) is 0

**OmegaKaos**)Is the following at all right?? All those 0s worry me!!

when r=0 the value of f(r)=e^(cosx) is 1

when r=1 the value of d/dr(f(r))= -(sinx) * e^(cosx) is 0

when r=2 the value of d/dr(f(r))= -(sinx)^2 * e^(cosx) is 0

when r=3 the value of d/dr(f(r))= -(sinx)^3 * e^(cosx) is 0

when r=4 the value of d/dr(f(r))= -(sinx)^4 * e^(cosx) is 0

if y=uv then dy/dx= u*dv/dx + v*du/dx

0

reply

user2020user

Badges:
16

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6

Report

#6

**OmegaKaos**)

Is the following at all right?? All those 0s worry me!!

when r=0 the value of f(r)=e^(cosx) is 1

when r=1 the value of d/dr(f(r))= -(sinx) * e^(cosx) is 0

when r=2 the value of d/dr(f(r))= -(sinx)^2 * e^(cosx) is 0

when r=3 the value of d/dr(f(r))= -(sinx)^3 * e^(cosx) is 0

when r=4 the value of d/dr(f(r))= -(sinx)^4 * e^(cosx) is 0

Also, remember that

0

reply

OmegaKaos

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7

f(x)=e^cosx

f '(x)= -sinx *f(x)

f ''(x)=sin^2(x)*f(x) - cosx*f(x)

f '''(x)=2sin(x)cos(x)f(x)-sin^3(x)f(x)+cos(x)sin(x)f(x)+sin(x)f(x)

f''''(x)=-2sin^2(x)cos(x)f(x) - 2[cos^2(x)-sin^2(x)]f(x)+sin^4(x)f(x)-3sin^2(x)cos(x)f(x)-sin^2(x)cos(x)f(x) + [cos^2(x)-sin^2(x)]f(x) - sin^2(x)f(x) + cos(x)f(x)

Therefore

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = -2e +e+e= 0

Does that look anything like it should??

0

reply

user2020user

Badges:
16

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8

Report

#8

I haven't looked over the derivatives but:

I got the same values for and which are affected by the terms in your answer. However, for I got .

Try re-calculating your first and fourth derivatives at when

Another bit of helpful advice when approaching questions like these is to leave and it's derivatives in the successive derivatives, i.e.

This way you can calculate , then

using , followed by

using and

and so on and so forth.

I got the same values for and which are affected by the terms in your answer. However, for I got .

Try re-calculating your first and fourth derivatives at when

(Original post by

...

**OmegaKaos**)...

This way you can calculate , then

using , followed by

using and

and so on and so forth.

0

reply

Username1818

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9

Report

#9

(Original post by

Ok, rectifying my earlier brainlapse

f(x)=e^cosx

f '(x)= -sinx *f(x)

f ''(x)=sin^2(x)*f(x) - cosx*f(x)

f '''(x)=2sin(x)cos(x)f(x)-sin^3(x)f(x)+cos(x)sin(x)f(x)+sin(x)f(x)

f''''(x)=-2sin^2(x)cos(x)f(x) - 2[cos^2(x)-sin^2(x)]f(x)+sin^4(x)f(x)-3sin^2(x)cos(x)f(x)-sin^2(x)cos(x)f(x) + [cos^2(x)-sin^2(x)]f(x) - sin^2(x)f(x) + cos(x)f(x)

Therefore

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = -2e +e+e= 0

Does that look anything like it should??

**OmegaKaos**)Ok, rectifying my earlier brainlapse

f(x)=e^cosx

f '(x)= -sinx *f(x)

f ''(x)=sin^2(x)*f(x) - cosx*f(x)

f '''(x)=2sin(x)cos(x)f(x)-sin^3(x)f(x)+cos(x)sin(x)f(x)+sin(x)f(x)

f''''(x)=-2sin^2(x)cos(x)f(x) - 2[cos^2(x)-sin^2(x)]f(x)+sin^4(x)f(x)-3sin^2(x)cos(x)f(x)-sin^2(x)cos(x)f(x) + [cos^2(x)-sin^2(x)]f(x) - sin^2(x)f(x) + cos(x)f(x)

Therefore

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = -2e +e+e= 0

Does that look anything like it should??

You are supposed to get...

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = e

so you are really close

0

reply

Badges:
16

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10

Report

#10

(Original post by

I think your derivative of f''''(x) might be wrong

You are supposed to get...

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = e

so you are really close

**Username1818**)I think your derivative of f''''(x) might be wrong

You are supposed to get...

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = e

so you are really close

(I'll reply to your PM soon! I was a bit busy )

0

reply

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#11

**Username1818**)

I think your derivative of f''''(x) might be wrong

You are supposed to get...

f(0)= e

f ' (0)= 0

f '' (0)= -e

f ''' (0)= 0

f ''''(0) = e

so you are really close

for f''''(0), i now think it is 4e

0

reply

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#12

Report

#12

(Original post by

Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e

**OmegaKaos**)Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e

0

reply

ghostwalker

Badges:
17

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#13

Report

#13

**OmegaKaos**)

Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e

0

reply

ghostwalker

Badges:
17

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#14

Report

#14

(Original post by

...

**Khallil**)...

0

reply

Badges:
16

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#15

Report

#15

**OmegaKaos**)

Damn, so i am 100% positive of the first four coefficients.

**for f''''(0), i now think it is 4e**

(Original post by

PRSOM for some of the best LaTex I've ever seen on here.

**ghostwalker**)PRSOM for some of the best LaTex I've ever seen on here.

To be honest, I've picked it up by hovering over your and other members' posts

0

reply

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#16

Report

#16

**OmegaKaos**)

Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e

0

reply

interstitial

Badges:
18

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#17

Report

#17

That would probably take me like half an hour to type

Posted from TSR Mobile

0

reply

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#18

(Original post by

It is!

Thanks ever so much!

To be honest, I've picked it up by hovering over your and other members' posts

**Khallil**)It is!

Thanks ever so much!

To be honest, I've picked it up by hovering over your and other members' posts

How do i transfer it to the student room??

0

reply

Badges:
0

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#19

(Original post by

sorry... you were right It's 4e, I was just being stupid Happens to everyone

**Username1818**)sorry... you were right It's 4e, I was just being stupid Happens to everyone

=[e*x^0]/0! + [0x]/1! + [-ex^2]/2!] + [0x^3]/3! + [4e*x^4]/4!

= e - [ex^2]/2! + [ex^4]/3!

=e [ 1 - (x^2)/2! + (x^4)/3!]

0

reply

Badges:
2

Rep:

?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#20

Report

#20

(Original post by

So now i know the coefficeitns, does the following look right??

=[e*x^0]/0! + [0x]/1! + [-ex^2]/2!] + [0x^3]/3! + [4e*x^4]/4!

= e - [ex^2]/2! + [ex^4]/3!

=e [ 1 - (x^2)/2! + (x^4)/3!]

**OmegaKaos**)So now i know the coefficeitns, does the following look right??

=[e*x^0]/0! + [0x]/1! + [-ex^2]/2!] + [0x^3]/3! + [4e*x^4]/4!

= e - [ex^2]/2! + [ex^4]/3!

=e [ 1 - (x^2)/2! + (x^4)/3!]

0

reply

X

### Quick Reply

Back

to top

to top