3 questions (Sqrt, Maclaurin Series, 0!)

**Timbuk2**

Hi, I have some small trivial questions (they are unrelated to each other)

1. Does [latex]\sqrt{9}[/latex] = 3 or is it = [latex]\pm3[/latex]. I know that solving [latex]x^2 = 9[/latex] yields 3 and -3 as solutions, but I'm wondering about just [latex]\sqrt{9}[/latex] on its own


2. Why does 0! (zero factorial) = 1? I'm finding it hard to get my head around this one, as in my head it should be undefined, but the calculator gives me 1 when I type in 0!


3. For finding the general term in the Maclaurin series, for f(x) = cosx, the series works out as
[latex]\displaystyle f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots[/latex]

The way I see it, taking [latex]T_1[/latex] as the first term, [latex]T_2[/latex] as the second etc, that [latex]\displaystyle T_n = \frac{(-1)^{n-1}x^{2n-2}}{(2n-2)!}[/latex]
but my friend says that [latex]T_0[/latex] is the first term, and thus calculates the general term as [latex]\displaystyle T_n = \frac{(-1)^{n}x^{2n}}{(2n)!}[/latex]

Which of us is correct? Does it matter whether you take n=0 or n=1 as the first term? Is there ever a situation that warrants taking n=0 over n=1, or vice versa?

Thanks for reading

Eliot Rosewater

**Timbuk2** wrote: »

1. Does [latex]\sqrt{9}[/latex] = 3 or is it = [latex]\pm3[/latex]. I know that solving [latex]x^2 = 9[/latex] yields 3 and -3 as solutions, but I'm wondering about just [latex]\sqrt{9}[/latex] on its own

Depends on the context. In general it's just the positive square root, so disregard -3.

**Timbuk2** wrote: »

2. Why does 0! (zero factorial) = 1?

It's just defined that way. It's like an axiom with no proof. In the same way that [LATEX]n^0=1[/LATEX]. The reason for these arbitrary definitions is that they seem to make maths simpler and more consistent, and seem to work out nicely.

**Timbuk2** wrote: »

3. For finding the general term in the Maclaurin series...

Asfaik I know they're both correct, however I would tend to go for your way. For example, if you want to sum the first n terms of the series, you have to add up all the elements to n, whereas your friend has to add up all the elements to n-1, which is a little counter-intuitive.

Setting the first element of a series to be element 0 is more a computer programming thing, really.


Just going back to question 2, note your general formula:

[latex]\displaystyle T_n = \frac{(-1)^{n-1}x^{2n-2}}{(2n-2)!}[/latex]

If you sub in n=1, you get hit with a 0!. If 0! were equal to 0, as might seem intuitive, your formula would be broken with a division by zero. And I think this is how mathematicians decided that 0!=1: from personal experience they concluded that it would be the most normal and consistent thing.


In the same vein,

[LATEX]0^n=0[/LATEX]

[LATEX]n^0=1[/LATEX]

But to avoid a contradiction,

[LATEX]0^0[/LATEX] is undefined.

sponsoredwalk

1: By virtue of the fact that (-3)² = 3² = 9 there is no reason to assume that √9 = 3 and not √9 = ±3.

What I think is the most shocking thing about this mathematical "oddity" is that physicist Paul Dirac predicted Anti-Matter based on a square root result in his final answer a few years before it's discovery!

[FONT=Trebuchet MS, arial, helvetica, sans-serif]The square root was a problem, as every square root has both a positive and negative solution[FONT=Trebuchet MS, arial, helvetica, sans-serif]1[/FONT]. This suggested that a particle could have negative energy. As no physicist had ever heard of negative energy, most would simply sweep the negative energy result under the carpet. However, when combined with quantum mechanics, it suggested the existence of negative energy states.

http://www.bbc.co.uk/dna/h2g2/A599853 [/FONT]

You'll never forget ± again

2: I actually heard in a lecture that it was by definition that 0! = 1 was chosen in order to simplify calculations 'and it actually simplifies calculations'.

I could find the lecture online if you're adamant for a source to this info.

3: I'm pretty sure that by definition in all series calculations the series goes from 1 to n, i.e. the standard thing to do is to use the Natural numbers.

Fringe

I've seen this before:

n! = n.(n-1)!

Subbing in n=1,

1! = 1.0!
0! = 1

Iwasfrozen

Eliot Rosewater wrote: »

It's just defined that way. It's like an axiom with no proof. In the same way that [LATEX]n^0=1[/LATEX]. The reason for these arbitrary definitions is that they seem to make maths simpler and more consistent, and seem to work out nicely.

There is proof, [latex]n^0 = 1[/latex] because [latex] n^-1 = 1/n[/latex] and [latex] n^1 = n[/latex] so [latex] (n^1)(n^-1) = n^0 = n/n = 1[/latex].

Eliot Rosewater

Iwasfrozen wrote: »

There is proof, [latex]n^0 = 1[/latex] because [latex] n^-1 = 1/n[/latex] and [latex] n^1 = n[/latex] so [latex] (n^1)(n^-1) = n^0 = n/n = 1[/latex].

Yes, but why should [latex]\displaystyle n^{-1} = \frac{1}{n}[/latex]?

You see, the original definition of a power [latex]\displaystyle a^{b}[/latex] is a.a.a.a.a ... b times. So [latex]\displaystyle a^{3}=a.a.a[/latex] How does one do that with -1? With 1/3? With 1.5? With pi?



EDIT: I think I'm lining myself up for a fall

Fremen

Quick answers, because I'm goin' drinkin'.

**Timbuk2** wrote: »

1. Does [latex]\sqrt{9}[/latex] = 3 or is it = [latex]\pm3[/latex]. I know that solving [latex]x^2 = 9[/latex] yields 3 and -3 as solutions, but I'm wondering about just [latex]\sqrt{9}[/latex] on its own

As far as I am aware, that particular symbol refers to the positive square root.

2. Why does 0! (zero factorial) = 1? I'm finding it hard to get my head around this one, as in my head it should be undefined, but the calculator gives me 1 when I type in 0!

Read the wikipedia articles on the Empty sum and Empty product.

3. For finding the general term in the Maclaurin series, for f(x) = cosx, the series works out as
[latex]\displaystyle f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots[/latex]

The way I see it, taking [latex]T_1[/latex] as the first term, [latex]T_2[/latex] as the second etc, that [latex]\displaystyle T_n = \frac{(-1)^{n-1}x^{2n-2}}{(2n-2)!}[/latex]
but my friend says that [latex]T_0[/latex] is the first term, and thus calculates the general term as [latex]\displaystyle T_n = \frac{(-1)^{n}x^{2n}}{(2n)!}[/latex]

Which of us is correct?

Easy way to test this: sub in known values for cos x (eg. x=0). If a formula disagrees with cos 0 = 1, then it's wrong.

Edit: actually, the two forms look to be equivalent. Test this by subbing in, say, x=1 in both formulas and summing over the first few terms.

Iwasfrozen

Eliot Rosewater wrote: »

Yes, but why should [latex]\displaystyle n^{-1} = \frac{1}{n}[/latex]?

You see, the original definition of a power [latex]\displaystyle a^{b}[/latex] is a.a.a.a.a ... b times. So [latex]\displaystyle a^{3}=a.a.a[/latex] How does one do that with -1? With 1/3? With 1.5? With pi?



EDIT: I think I'm lining myself up for a fall

To be honest, I have no idea. I'm only a Leaving Cert student.

But anything to the power of a minus is always that number to the power of a plus divided by one.
or [latex] n^-1 = 1/n [/latex] or [latex] n^-2 = 1/(n^2) [/latex]

As for your question, I don't know why but [latex]a^-b = 1/(a^b)[/latex], [latex]a^(1/3) = a^3^(-1)[/latex].

I'm sure however that some of the many more talented people in this forum can give you the proof that I can't.

Eliot Rosewater

I've been trawling Wiki.

[LATEX]\displaystyle x^{n}=x^{n}[/LATEX]

[LATEX]\displaystyle x.x^{n-1}=x^{n}[/LATEX]

[LATEX]\displaystyle x^{n-1}=\frac{x^{n}}{x}[/LATEX]

Let n=1.

[LATEX]\displaystyle x^{0}=\frac{x^{1}}{x}[/LATEX]

[LATEX]\displaystyle x^{0}=1[/LATEX]

Let n=0.

[LATEX]\displaystyle x^{-1}=\frac{x^{0}}{x}[/LATEX]

[LATEX]\displaystyle x^{-1}=\frac{1}{x}[/LATEX]


However there are technical issues. For example, if we had subbed in n=1 at the start of the process we would have gotten

[LATEX]\displaystyle x^{1}=x^{1}[/LATEX]

[LATEX]\displaystyle x^{1}x^{0}=x^{1}[/LATEX]

Which doesn't make much sense, unless you know [LATEX]\displaystyle x^{0}=1[/LATEX], but we don't because we are trying to prove it! Wikipedia (I know, crap source) it says that the above things I have supposedly proved are only defined.

Iwasfrozen

Eliot Rosewater wrote: »

However there are technical issues. For example, if we had subbed in n=1 at the start of the process we would have gotten

Which doesn't make much sense, unless you know , but we don't because we are trying to prove it! Wikipedia (I know, crap source) it says that the above things I have supposedly proved are only defined.

I think that is still alright though as you are proving it for all numbers, sure when you put in n = 1 it doesn't work but if you use any other number it will.

For example put in n = 2.
[latex] x^1 = x^2/x = x [/latex]

and x = 3
[latex] x^2 = x^3/x = x^2 [/latex]

So this will basically work with any number I try as long as you can prove it with letters, which we have.

As for (x^1)(x^0) = x^(1+0) = x^1 which still works out.

EDIT: I hate latex.

Fringe

All those properties you've mentioned can be derived from the exponential/log function. For example, x^0 = exp(log(x^0)) = exp(0logx) = exp(0) = 1 by the definition of the exponential function. I'd say if you play around with it a bit more, you'll get everything you need.

Michael Collins

**Timbuk2** wrote: »

Hi, I have some small trivial questions (they are unrelated to each other)

1. Does [latex]\sqrt{9}[/latex] = 3 or is it = [latex]\pm3[/latex]. I know that solving [latex]x^2 = 9[/latex] yields 3 and -3 as solutions, but I'm wondering about just [latex]\sqrt{9}[/latex] on its own

Hey Timbuk2. Good questions! The radical sign [latex] \sqrt(x) [/latex] refers to what is known as the principle square root, which means only the positive root, for x a positive, real number. One of the main reasons for doing this is so that we have [latex] f(x) = \sqrt(x) [/latex] a function - if both answers were allowed it wouldn't be a standard function (since functions can only be single valued, i.e. each point [latex] x [/latex] in the domain maps onto exactly one point [latex] f(x) [/latex] in the codomain).

2. Why does 0! (zero factorial) = 1? I'm finding it hard to get my head around this one, as in my head it should be undefined, but the calculator gives me 1 when I type in 0!

One way to explain this is to say it's "just defined" that way! Which I always hate as an answer, especially when you know there's usually a better reason, which is true in this case.
Two other reaons:

The factorial [latex] n! [/latex] - as I'm sure you know - is often used to count how many ways we can permute or arrange [latex] n [/latex] distinct items in different orders. If we have no objects, does it stand to reason that we do have 1 arrangement, i.e. the arrangement of nothing? Hmm, maybe that's a bit too philosophical...

[LIST=2]
A more mathematical reason for [latex] 0! =1 [/latex] can be derived from the Gamma Function [latex] \Gamma(z) [/latex]. I don't know what your level of maths so I'll just give the idea here. Bascially the Gamma function is equivalent to the factorial function at all the positive integers, in fact:

[latex] \displaystyle \Gamma(n + 1) = n!, \forall n>0 \in \mathbb{N} [/latex]

but the Gamma function [latex] \Gamma(z) [/latex] is described at all points z in the complex plane so is more general than the factorial, and you can show that [latex] \Gamma(1) = 0! = 1[/latex]. In other words the factorial can be defined as the Gamma function at the positive integers, in which case you get [latex]0! = 1[/latex] from that.
I hope that hasn't confused you more and you see some idea that's [latex]0! = 1[/latex] is not just an arbritarty choice. [/LIST]
Feel free to ask any more questions about this if you like, I'll try my best to answer them.

3. For finding the general term in the Maclaurin series, for f(x) = cosx, the series works out as
[latex]\displaystyle f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots[/latex]

The way I see it, taking [latex]T_1[/latex] as the first term, [latex]T_2[/latex] as the second etc, that [latex]\displaystyle T_n = \frac{(-1)^{n-1}x^{2n-2}}{(2n-2)!}[/latex]
but my friend says that [latex]T_0[/latex] is the first term, and thus calculates the general term as [latex]\displaystyle T_n = \frac{(-1)^{n}x^{2n}}{(2n)!}[/latex]

Which of us is correct? Does it matter whether you take n=0 or n=1 as the first term? Is there ever a situation that warrants taking n=0 over n=1, or vice versa?

Thanks for reading

Yeh, you are both right is the answer. You've just labelled the terms different. If you put [latex] n= n+1 [/latex] in yours, i.e. shift the starting term to zero, you'll get your mate's answer. In my experience I've come across the notation of labelling the terms starting with zero more often...I think. Mainly due to the fact that series often start with a constant term, which works out well if you have [latex]x^n, n=0[/latex] means the first term is a contant, 1. But it really doesn't matter, once you make it clear.