Maclaurin series

  • #1
Work out the first five derivatives of the function [itex]f(x)=sec(x)[/itex], and hence deduce the Maclaurin series of [itex]g(x)=sec(x)(1+tan(x))[/itex] up to and including the term of order [itex]x^4[/itex].

(Hint: why have you been asked for five derivatives of [itex]f(x)[/itex]?)




The Maclaurin series for function [itex]g(x)[/itex] is given by [tex]g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k[/tex]



I know how to differentiate [itex]g(x)[/itex], and although it would take a long time, I could differentiate [itex]g(x)[/itex] 4 times, evaluate the derivatives at [itex]x=0[/itex] and substitute the values in the series equation above to deduce the Maclaurin series up to the [itex]x^4[/itex] term. But I'm not sure how differentiating [itex]f(x)=sec(x)[/itex] 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.
 
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  • #2
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Work out the first five derivatives of the function [itex]f(x)=sec(x)[/itex], and hence deduce the Maclaurin series of [itex]g(x)=sec(x)(1+tan(x))[/itex] up to and including the term of order [itex]x^4[/itex].

(Hint: why have you been asked for five derivatives of [itex]f(x)[/itex]?)




The Maclaurin series for function [itex]g(x)[/itex] is given by [tex]g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k[/tex]



I know how to differentiate [itex]g(x)[/itex], and although it would take a long time, I could differentiate [itex]g(x)[/itex] 4 times, evaluate the derivatives at [itex]x=0[/itex] and substitute the values in the series equation above to deduce the Maclaurin series up to the [itex]x^4[/itex] term. But I'm not sure how differentiating [itex]f(x)=sec(x)[/itex] 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.

This is a Maclaurin series, so 1) each derivative is evaluated at x = 0, and 2) it's a sum of powers of x. Once you get to the 5th derivative of sec(x) I think you'll see what the hint is about.
 
  • #3
This is a Maclaurin series, so 1) each derivative is evaluated at x = 0, and 2) it's a sum of powers of x. Once you get to the 5th derivative of sec(x) I think you'll see what the hint is about.

The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.
 
  • #4
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The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.
That looks fine. So with f(x) = sec(x), you have f(5)(x) = tan(x)sec(x)(120sec^4(x)-60sec^(x)+1).
What is f(5)(0)?
 
  • #5
Dick
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The 5th derivative of sec(x) is tan(x)sec(x)(120sec^4(x)-60sec^(x)+1). Sorry, but I still don't get the hint.

I think the sense of the hint is that you should compute the Maclaurin series of sec(x) up to power x^4 and then multiply by the Maclaurin series of 1+tan(x), rather than computing the series of sec(x)(1+tan(x)) directly. The differentiation is easier.
 
  • #6
That looks fine. So with f(x) = sec(x), you have f(5)(x) = tan(x)sec(x)(120sec^4(x)-60sec^(x)+1).
What is f(5)(0)?

f(5)(0) = 0

I think the sense of the hint is that you should compute the Maclaurin series of sec(x) up to power x^4 and then multiply by the Maclaurin series of 1+tan(x), rather than computing the series of sec(x)(1+tan(x)) directly. The differentiation is easier.

Oh, I see. But to compute the Maclaurin series for sec(x) up to the power x^4, I only need to differentiate it 4 times. Why did the question ask for the 5th derivative of sec(x)?
 
  • #7
vela
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I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.
 
  • #8
I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.

Ohhhh! How did I miss that... :surprised

I now understand why they asked for 5 derivatives of sec(x). It's so that I can obtain the x4 order term for the Maclaurin series of ##f'(x)##.

Thanks for the replies, everyone :)
 
  • #9
Dick
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I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.

Ohhh. Tricky!
 

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