Expansion of sec²x - Stuck :(

Talvon · 2006-11-15 04:45:43

I have to show sec²x is approximately 1 + x^2 + 2x^4/3 + 17x^6/45

Anything x^8 and above is negligible.

I have looked at the standard expansions on the site, however, I have not yet studied the En number, so I won't be able to get away with it like that

I tried using the identity that sec²x=1/cos²x, and using the standard Maclaurin expansion of cos (1-x²/2+x^4/4!-x^6/6!+...) and just squared it out but I got nowhere

Any help would be appreciated

Last edited by Talvon (2006-11-15 04:50:20)

Dross · 2006-11-15 05:20:48

Have a look at your textbook - there will be an example somewhere (certainly an introductory one, and probably others elsewhere) on how a McLaurin series is constructed from first principles. Do this with sec[sup]2[/sup]x.

As a very, very rough outline, you want to say that:

1) Let sec[sup]2[/sup]x = a[sub]0[/sub] + a[sub]1[/sub]x[sup]1[/sup] + a[sub]2[/sub]x[sup]2[/sup] + ...

2) Set x = 0 in (1), above

3) Find a[sub]0[/sub]

4) Differentiate (1) with respect to x

5) Repeat (3) and (4) to get the required values of a[sub]1[/sub], a[sub]2[/sub], et cetera. Obviously, the series expansion you get will hopefully be the same as that which is given by the question!

Is this all clear? Feel free to post if you get stuck at any point.

luca-deltodesco · 2006-11-15 05:52:33

dross, according to wolfram, a mclauring series, is constructed a bit differently

for a function f:
it can be expanded into a polynomial of

for example:

etc:

so cosx can be expanded to a mclaurin series by

Talvon · 2006-11-15 07:36:13

Cheers but I think I'm meant to derive it from a list of standard expansions that I already have, and I figured the best one to use would be the cos expansion and invert it for the sec=1/cos thing, but I tried that and ended up getting it wrong I have others, like the expansion of sin x and 1/1+-x, etc.

Don't think I'm meant to do the differentiating and stuff myself.

Would sec²x= (1/cosx)(1/cosx)?

Last edited by Talvon (2006-11-15 07:36:29)

Dross · 2006-11-15 10:00:14

luca-deltodesco wrote:

dross, according to wolfram, a mclauring series, is constructed a bit differently...

What you've given is like a "ready made formula" for the process I've given (or, alternatively, it's what you get when you use the process i've given for a generalised function f(x)). Try it on cos(x) and a general function f(x) and see - you'll get the same result either way. My method just takes more work as I can hardly ever remember formulas like you've posted!

Talvon wrote:

Cheers but I think I'm meant to derive it from a list of standard expansions that I already have, and I figured the best one to use would be the cos expansion and invert it for the sec=1/cos thing, but I tried that and ended up getting it wrong sad I have others, like the expansion of sin x and 1/1+-x, etc.
Don't think I'm meant to do the differentiating and stuff myself.
Would sec²x= (1/cosx)(1/cosx)?

Oh - sorry, I kind of assumed it was an exercise in deriving a McLaurin series from scratch.

In that case, sec[sup]2[/sup]x = tan[sup]2[/sup]x + 1 will do the trick if you're given the expansion for tan(x).

luca-deltodesco · 2006-11-15 10:20:57

What you've given is like a "ready made formula" for the process I've given (or, alternatively, it's what you get when you use the process i've given for a generalised function f(x)). Try it on cos(x) and a general function f(x) and see - you'll get the same result either way. My method just takes more work as I can hardly ever remember formulas like you've posted!

yours was mine, minus the division by factorials in the series, which gives a very different result no?

Dross · 2006-11-15 11:27:43

No - the factorials appear when you differentiate the series. You start off with:

Then you can set x = 0 and get the value of a[sub]0[/sub] = f(0). Differentiating term-by-term, you get:

Set x = 0 and you get your a[sub]1[/sub] = f'(0). Now you should start seeing some factorials, just like what you posted. Differentiating again, term-by-term, you get:

Setting x = 0 shows us that:

You may think that last step is fudged because of the "pure coincidence" that the number happens to be a factorial, but if you look for the term a[sub]n[/sub], you'll see that you have to differentiate the series n times. When you do that, the coefficient in front of the a[sub]n[/sub] term is the product of all the powers it's been through for each differentiation: n(n-1)(n-2)...(3)(2) = n!

So we always have that

No I'm wondering why I can't always think this clearly and come up with a formula when I do these series!

If you're still not convinced, continue expanding the series for f(x) above to see that it fits in with the one you've given. Also, try it for cos(x), and any function that takes your fancy - do it both ways and you'll see they are exactly the same.

Last edited by Dross (2006-11-15 11:29:54)

luca-deltodesco · 2006-11-15 11:38:01

ah right, i get what you meant in youre method now, i thought you just mean to differentiate f, and then take f(0), not to differentiate the series aswell

Talvon · 2006-11-17 01:42:21

A bunch of us got it in the end, we differentiated the tan x expansion to get the answer, if it's right or not we'll see on monday. So just so you know lol

Math Is Fun Forum

#1 2006-11-15 04:45:43

Expansion of sec²x - Stuck :(

#2 2006-11-15 05:20:48

Re: Expansion of sec²x - Stuck :(

#3 2006-11-15 05:52:33

Re: Expansion of sec²x - Stuck :(

#4 2006-11-15 07:36:13

Re: Expansion of sec²x - Stuck :(

#5 2006-11-15 10:00:14

Re: Expansion of sec²x - Stuck :(

#6 2006-11-15 10:20:57

Re: Expansion of sec²x - Stuck :(

#7 2006-11-15 11:27:43

Re: Expansion of sec²x - Stuck :(

#8 2006-11-15 11:38:01

Re: Expansion of sec²x - Stuck :(

#9 2006-11-17 01:42:21

Re: Expansion of sec²x - Stuck :(

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