OmegaKaos
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#1
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#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??
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Username1818
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#2
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#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 ?
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user2020user
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#3
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#3
(Original post by 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 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 ?
I think this method would be simpler as opposed to substituting the expansion of \cos x into that of e^x.

As a side note, the Taylor expansion of a function about the point x=a=0 is called a Maclaurin expansion:

\displaystyle \begin{aligned} f(x) = \underbrace{ \sum_{r=0}^{\infty} \dfrac{f^{(r)}(a)}{r!} \ (x-a)^r }_{\text{Taylor Expansion}} \ \overset{a=0}= \underbrace{ \sum_{r=0}^{\infty} \dfrac{f^{(r)}(0)}{r!} \ x^r }_{\text{Maclaurin Expansion}} \end{aligned}
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OmegaKaos
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#4
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#4
(Original post by Khallil)
I think this method would be simpler as opposed to substituting the expansion of \cos x into that of e^x.

As a side note, the Taylor expansion of a function about the point x=a=0 is called a Maclaurin expansion:

\displaystyle \begin{aligned} f(x) = \underbrace{ \sum_{r=0}^{\infty} \dfrac{f^{(r)}(a)}{r!} \ (x-a)^r }_{\text{Taylor Expansion}} \ \overset{a=0}= \underbrace{ \sum_{r=0}^{\infty} \dfrac{f^{(r)}(0)}{r!} \ x^r }_{\text{Maclaurin Expansion}} \end{aligned}
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
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Username1818
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#5
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#5
(Original post by 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
Yyy... nope, you need to use product rule when differentiating

if y=uv then dy/dx= u*dv/dx + v*du/dx
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user2020user
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#6
Report 7 years ago
#6
(Original post by 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
Not quite.

\begin{aligned} f(x) = e^{\cos x}

\implies f'(x) = -\sin x \ e^{\cos x}

\begin{aligned} \implies f''(x) & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right) + \left( -\sin x \cdot \dfrac{d}{dx} \left( e^{\cos x} \right) \right) \\ & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right)  -\sin x f'(x) \\ & = \ ?

\implies f'''(x) = \cdots \end{aligned}

Also, remember that \cos 0 = 1

\therefore \ f(0) = e^{\cos 0} = e^1 = e
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OmegaKaos
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#7
Report Thread starter 7 years ago
#7
(Original post by Khallil)
Not quite.

\begin{aligned} f(x) = e^{\cos x}

\implies f'(x) = -\sin x \ e^{\cos x}

\begin{aligned} \implies f''(x) & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right) + \left( -\sin x \cdot \dfrac{d}{dx} \left( e^{\cos x} \right) \right) \\ & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right)  -\sin x f'(x) \\ & = \ ?

\implies f'''(x) = \cdots \end{aligned}
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??
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user2020user
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#8
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#8
I haven't looked over the derivatives but:

f(0) = e^{\cos 0} = e^1 = e

I got the same values for f'(0) and f'''(0) which are affected by the \sin(0) terms in your answer. However, for f''''(0) I got 4e.

Try re-calculating your first and fourth derivatives at x=0 when f(0) = e

(Original post by OmegaKaos)
...
Another bit of helpful advice when approaching questions like these is to leave f(x) and it's derivatives in the successive derivatives, i.e.

f(x) & = e^{\cos x}

f'(x) = -\sin x \cdot f(x)

f''(x) = -\cos x \cdot f(x) - \sin x \cdot f'(x)

f'''(x) = -\sin x \cdot f''(x) - f'(x) \cdot (2\cos x + 1)

This way you can calculate f(0), then

f'(0) using f(0), followed by

f''(0) using f'(0) and f(0)

and so on and so forth.
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Username1818
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#9
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#9
(Original post by 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??
I think your derivative of f''''(x) might be wrong, but I got same results for f'(x), f''(x) and f'''(x)

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
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user2020user
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#10
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#10
(Original post by 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 got f''''(0) = 4e instead and Wolfram agrees with my final answer too. You might've made an arithmetical error when calculating (3cosx + 1) in your fourth derivative

(I'll reply to your PM soon! I was a bit busy )
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OmegaKaos
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#11
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#11
(Original post by 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
Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e
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Username1818
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#12
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#12
(Original post by OmegaKaos)
Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e
Let me double check
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ghostwalker
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#13
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#13
(Original post by OmegaKaos)
Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e
Agreed.
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ghostwalker
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#14
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#14
(Original post by Khallil)
...
PRSOM for some of the best LaTex I've ever seen on here.
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user2020user
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#15
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#15
(Original post by OmegaKaos)
Damn, so i am 100% positive of the first four coefficients.

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

(Original post by ghostwalker)
PRSOM for some of the best LaTex I've ever seen on here.
Thanks ever so much!

To be honest, I've picked it up by hovering over your and other members' posts
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Username1818
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#16
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#16
(Original post by OmegaKaos)
Damn, so i am 100% positive of the first four coefficients.

for f''''(0), i now think it is 4e
sorry... you were right It's 4e, I was just being stupid Happens to everyone
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interstitial
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#17
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#17
(Original post by Khallil)
Not quite.

\begin{aligned} f(x) = e^{\cos x}

\implies f'(x) = -\sin x \ e^{\cos x}

\begin{aligned} \implies f''(x) & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right) + \left( -\sin x \cdot \dfrac{d}{dx} \left( e^{\cos x} \right) \right) \\ & = \left( e^{\cos x} \cdot \dfrac{d}{dx} \left( -\sin x \right) \right)  -\sin x f'(x) \\ & = \ ?

\implies f'''(x) = \cdots \end{aligned}

Also, remember that \cos 0 = 1

\therefore \ f(0) = e^{\cos 0} = e^1 = e
:eek: How long did all that Latex take, and how did you get so good

That would probably take me like half an hour to type

Posted from TSR Mobile
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OmegaKaos
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#18
Report Thread starter 7 years ago
#18
(Original post by Khallil)
It is!



Thanks ever so much!

To be honest, I've picked it up by hovering over your and other members' posts
I know how to write in latex code.

How do i transfer it to the student room??
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OmegaKaos
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#19
Report Thread starter 7 years ago
#19
(Original post by Username1818)
sorry... you were right It's 4e, I was just being stupid Happens to everyone
So now i know the coefficients, 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!]
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Username1818
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#20
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#20
(Original post by 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!]
Yep ^.^ well done x
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