Mathematicus65
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Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give
Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...
Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:
1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?
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TeeEm
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(Original post by Mathematicus65)
Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give
Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...
Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:
1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?
I do not understand why sin(cosx) this is different from cos(sinx)
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the bear
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maybe the 1s are alternately + and - ?
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InOrbit
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http://mathworld.wolfram.com/MaclaurinSeries.html

I think trying to manipulate the sinx and cosx expansion is incorrect and unwieldy. Try differentiating a few times and then substituting zero for x to find the coefficients.
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DFranklin
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(Original post by Mathematicus65)
Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give
Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...
Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:
1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?
Are you certain you are NOT supposed to do it the "normal" Maclaurin way? (That is: define f(x) = \sin(\cos(x)) then f(x) = \sum_0^\infty f^{(n)}(0) x^n / n!).

Trying to do it any other way, I think you're in for a fairly significant world of pain. The coefficients involve terms like sin(1) and if you want to find those by summing the series you're going to need to consider an infinite number of terms and then spot things like

1 - 1/3! + 1/5! - 1/7! + ... = \sin(1)
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DFranklin
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(Original post by TeeEm)
I do not understand why sin(cosx) this is different from cos(sinx)
You can find cos(sin(x)) to O(x^4) by substituting x-x^3/6 for sin(x) and then using the known expansion of cos x. Because there's no constant term, you don't need to worry about (x-x^3/6)^5 and higher terms.

But for sin(cos(x)) the same approach has you considering terms of the form (1-x^2/2 + x^4/24)^n and because of the constant term you can't discard any of them if you want to get the right answer.
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TeeEm
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(Original post by DFranklin)
You can find cos(sin(x)) to O(x^4) by substituting x-x^3/6 for sin(x) and then using the known expansion of cos x. Because there's no constant term, you don't need to worry about (x-x^3/6)^5 and higher terms.

But for sin(cos(x)) the same approach has you considering terms of the form (1-x^2/2 + x^4/24)^n and because of the constant term you can't discard any of them if you want to get the right answer.
I would never substitute in these type of expansions ...
maybe I had to do them a few times and I remember the pain
Standard differentiation (with a few tricks such as bringing y and y' back in) is the way forward for those
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DFranklin
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(Original post by TeeEm)
I would never substitute in these type of expansions ...
maybe I had to do them a few times and I remember the pain
I think it's reasonable for the cos(sin) case, but it fails horribly for the other (as we have been telling the OP).

For context, he's made a couple of other posts where he's said he needs to do it by substitution/manipulation rather than differentiation, so I was pretty sure that was the direction his brain was running in, but I can't believe you were supposed to not use differentiation here.
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Mathematicus65
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(Original post by DFranklin)
I think it's reasonable for the cos(sin) case, but it fails horribly for the other (as we have been telling the OP).

For context, he's made a couple of other posts where he's said he needs to do it by substitution/manipulation rather than differentiation, so I was pretty sure that was the direction his brain was running in, but I can't believe you were supposed to not use differentiation here.
the question specifically asks for me to substitute
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InOrbit
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(Original post by Mathematicus65)
the question specifically asks for me to substitute
I don't think it's wise or possible.

http://www.wolframalpha.com/input/?i...8cos%28x%29%29

You're getting sin(1) and cos(1), both of which aren't rational numbers.
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Mathematicus65
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Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?
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InOrbit
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(Original post by Mathematicus65)
Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?
Okay, you've got to be using differentiation now. Leave manipulation and using standard series behind unless it is trivially easy. Differentiating ln(1+e^x) isn't too difficult.
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DFranklin
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(Original post by Alex:)
I don't think it's wise or possible.

You're getting sin(1) and cos(1), both of which aren't rational numbers.
For sure it's possible - but it requires some fairly inspired recognition to convert various infinite series into sin 1 and cos 1.

Wise? Definitely not...

EdIt: Thought of a better approach (if we absolutely insist on doing this by substitution). It's the fact that cos(0) = 1 that kills us, so we need to finesse that.

So rewrite sin(cos(x)) as sin(1+(cos(x) - 1) = sin(1)cos(cos(x)-1) + cos(1) sin(cos(x) - 1).

Then cos(x) - 1 = (x^4/24 - x^2/2) + O(x^6). So our problem reduces to finding sin(1) cos(x^4/24 - x^2/2) + cos(1) sin(x^4/24-x^2/2) up to terms in O(x^4), which is not TOO bad and doesn't involve infinite sums.
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Mathematicus65
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(Original post by Alex:)
Okay, you've got to be using differentiation now. Leave manipulation and using standard series behind unless it is trivially easy. Differentiating ln(1+e^x) isn't too difficult.
I did differentiate it at first and got the correct solution, was just wondering if I could manipulate it in another way. Finding f(0), f'(0) f''(0) and f'''(0) was easy but, I had to construct the Maclaurin polynomial of degree 4 and finding f''''(0) took a long time, which I wouldn't have in an exam, but having said that they would either ask me to expand something similar or expand that to a lower degree which would be fine.

Thank you for your help!
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DFranklin
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(Original post by Mathematicus65)
Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?
Again, if you insist on doing this by substitution, you're going to need a "trick" to replace e^x by something without a leading term.

So if you write ln(1+e^x) as ln(2 + (e^x - 1)) = (ln 2) + ln(1 + (e^x-1)/2) then you can replace (e^x-1)/2 by a power series starting (x/2 + x^2/4 + x^3/12 + ...) and sub in).

I would still be surprised if this was quicker than differentiating.
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