Find Maclaurin Series for g(x) with 5 Derivs of f(x)=sec(x)

[subzero0137]

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

(Hint: why have you been asked for five derivatives of $f(x)$?)
The Maclaurin series for function $g(x)$ is given by $$g(x)=\sum\limits_{k=0}^\infty \frac{g^{k}(0)}{k!}x^k$$
I know how to differentiate $g(x)$, and although it would take a long time, I could differentiate $g(x)$ 4 times, evaluate the derivatives at $x=0$ and substitute the values in the series equation above to deduce the Maclaurin series up to the $x^4$ term. But I'm not sure how differentiating $f(x)=sec(x)$ 5 times would make this problem less time consuming. In short, I do not understand the hint. Could someone please explain the hint? Thanks.

 

[Mark44]

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

(Hint: why have you been asked for five derivatives of $f(x)$?)



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



I know how to differentiate $g(x)$, and although it would take a long time, I could differentiate $g(x)$ 4 times, evaluate the derivatives at $x=0$ and substitute the values in the series equation above to deduce the Maclaurin series up to the $x^4$ term. But I'm not sure how differentiating $f(x)=sec(x)$ 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.

 

[subzero0137]

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.

 

[Mark44]

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)?

 

[Dick]

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.

 

[subzero0137]

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)?

 

[vela]

I figured the hint had to do with the fact that $g(x) = f(x) + f'(x)$.

 

[subzero0137]

I figured the hint had to do with the fact that $g(x) = f(x) + f'(x)$.

Ohhhh! How did I miss that...

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 :)

 

[Dick]

I figured the hint had to do with the fact that $g(x) = f(x) + f'(x)$.

Ohhh. Tricky!