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

• subzero0137
In summary: Thank you for the explanation. I never would have thought of that. In summary, the hint was to compute the Maclaurin series of sec(x) up to the power x^4, then multiply by the Maclaurin series of 1+tan(x) to find the Maclaurin series of g(x). The reason for computing the 5th derivative of sec(x) was to obtain the x^4 term for the Maclaurin series of f'(x). This method is easier than directly computing the series of g(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.

Last edited by a moderator:
subzero0137 said:
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.

Mark44 said:
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.

subzero0137 said:
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)?

subzero0137 said:
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.

Mark44 said:
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

Dick said:
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)?

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

vela said:
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 :)

vela said:
I figured the hint had to do with the fact that ##g(x) = f(x) + f'(x)##.

Ohhh. Tricky!

## 1. What is the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x)?

The Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x) is 1 + x^2/2 + 5x^4/24 + 61x^6/720 + 277x^8/8064.

## 2. How do you find the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x)?

To find the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x), you can use the formula for the Maclaurin series of a function, which is f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + f''''(0)x^4/4! + ..., where f^(n)(x) represents the nth derivative of f(x). For this specific case, you will need to calculate the first 5 derivatives of f(x) = sec(x) and evaluate them at x = 0.

## 3. What is the significance of using the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x)?

The Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x) is important because it allows us to approximate the value of g(x) for any value of x by using a finite number of terms from the series. This is useful in many areas of science and engineering, where it may be difficult to compute the exact value of a function at a specific point.

## 4. Can the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x) be used to find the value of the function at x = 1?

Yes, the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x) can be used to find the value of the function at x = 1. However, the more terms you use from the series, the more accurate your approximation will be. In this case, using all 5 terms will give you a more precise estimate than using only 3 terms.

## 5. Are there any limitations to using the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x)?

Yes, there are some limitations to using the Maclaurin series for g(x) with 5 derivatives of f(x) = sec(x). One limitation is that the series only converges for certain values of x, specifically for |x| < π/2. Additionally, using a finite number of terms from the series will always result in an approximation, so it may not give you the exact value of g(x) for a specific value of x.

• Calculus and Beyond Homework Help
Replies
3
Views
377
• Calculus and Beyond Homework Help
Replies
26
Views
949
• Calculus and Beyond Homework Help
Replies
1
Views
384
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
298
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
16
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
5K
• Calculus and Beyond Homework Help
Replies
1
Views
917
• Calculus and Beyond Homework Help
Replies
7
Views
2K