anonymous
  • anonymous
Find the first few terms of the Maclaurin series for the following function: secx=1/(cosx) I know the series for cosx: 1-(x^2/2!)+(x^4/4!)-(x^6/6!)...+... would i just take that series and 1 by that?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
divide by 1*
anonymous
  • anonymous
Yes pretty much! That's the way it goes. If you see "cos(x)" anywhere in a expression, you can replace it with its series
anonymous
  • anonymous
ok thanks!

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dape
  • dape
That would work perfectly fine, but sadly it isn't the Maclaurin series of sec(x). To get that you need to put 0 into sec(x) and it's derivatives, adding the terms together.
anonymous
  • anonymous
hmmm..well for the series 1/cosx i got:
anonymous
  • anonymous
|dw:1377558059394:dw|
anonymous
  • anonymous
you must be writing alot lol
anonymous
  • anonymous
?
dape
  • dape
That's the wrong series, let's try it from the beginning. The first term of the maclaurin series of sec(x) is simply sec(0)=1/cos(0)=1. So you got the first term right. Then for the second term we need the derivative of sec(x), which is \[\frac{d}{dx}(\sec x)=\frac{d}{dx}(\frac{1}{\cos x})=\frac{1}{\cos x}\frac{\sin x}{\cos x}\] Now put in x=0, since sin(0)=0 this is 0, so our second term is 0. For the third term you will get \[ \frac{d^2}{d^2x}(\sec x)=\frac{d}{dx}(\sec x\tan x) \] When you have this put in x=0 and don't forget to multiply by \(x^2/2\).
dape
  • dape
In general, the nth term of a maclaurin series for some function f(x) is: \[\frac{x}{n!}\times\left.\frac{d^nf}{dx^n}\right|_{x=0}\] The bar with the x=0 at the bottom means "calculate this at x=0".
dape
  • dape
But you could also use the cos-series cut off after some point and do some nasty algebra to get rid of all the x's in the denominator.
anonymous
  • anonymous
hmm ok... im working on it now
dape
  • dape
Crap, it should be \(x^n/n!\), not \(x/n!\)
anonymous
  • anonymous
Honestly I think i was orginally right...my book lists the series that i typed as an important Maclaurin series for cosx...
anonymous
  • anonymous
so if i need 1/cosx then i should just divide the series by 1
dape
  • dape
But then you don't have a polynomial, a maclaurin series is a polynomial. So you have to do some hairy algebra to get it into the right form.
dape
  • dape
If you just do one over you get a rational function, which is not a polynomial.
anonymous
  • anonymous
there has to be an easier way than just taking 4 or 5 derivatives
dape
  • dape
The first two terms of the sec(x) maclaurin series are \[ \sec x=1+\frac{x^2}{2}+...\] and not \[ \sec x=\frac{1}{1-\frac{x^2}{2}+...} \] These two things are equal to each other but one is a series and polynomial, the other is not.
anonymous
  • anonymous
i see where you are coming from, but im trying to learn a way without taking successive derivatives esp when they start becoming super long ya know?
dape
  • dape
That's how you calculate Taylor/Maclaurin series, it's a long process. But you will only need one more term, for that you need to do the 4th derivative of sec(x). Usually one gets the derivatives one by one along the way of getting all the terms.
anonymous
  • anonymous
ok ill try it out
anonymous
  • anonymous
what did you get for the 3rd term in the maclaurin series?
dape
  • dape
I got 0 for the derivative when i put in x=0, so the 3rd term should be 0.
anonymous
  • anonymous
f(0)=1 f'(0)=0 f""(0)= x^2/2!
anonymous
  • anonymous
i got f"""(0) = 0
dape
  • dape
Yeah, the third derivative is zero for x=0, \(f'''(0)=0\).
dape
  • dape
Your thing looks like the 6th derivative on my screen.

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