Help a kid with Calculus!

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Dason
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Re: Calculus (more specifically Integral Calculus)

Postby Dason » Fri Oct 22, 2010 11:58 am UTC

keeperofdakeys wrote:Now try this:
1. d/dx(x*y2 + tan(32/x)) =
(hint: d/dx(tan(x)) = sec2(x) = 1/cos2(x) )

2. d/dx( y*ln(2*y3*x) ) =

Are you assuming y is a function of x or that y is a constant?
double epsilon = -.0000001;

tckthomas
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 12:05 pm UTC

? I got equations with x and y in them when I learnt them from the net. So what am I supposed to do?

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Dason
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Re: Calculus (more specifically Integral Calculus)

Postby Dason » Fri Oct 22, 2010 12:25 pm UTC

It's implied that y is a function of x but I was just asking because it makes a difference. In those cases you would use implicit differentiation (like they said in their post).
double epsilon = -.0000001;

tckthomas
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 12:37 pm UTC

do I assume it's equal to 0?
EDIT: oh well I'll just go and try it out

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 1:21 pm UTC

keeperofdakeys wrote:1. d/dx(x*y2 + tan(32/x)) =
(hint: d/dx(tan(x)) = sec2(x) = 1/cos2(x) )

d/dx(xy2) + d/dx(tan(32/x))
2y*dy/dx + d/dx(tan(32 x-1))
2y*dy/dx + sec2(32 x-1)(-32 x-2)
2y*dy/dx + (-32 x-2)/cos2(32 x-1)
2y*dy/dx + (-32/x2)/cos2(32/x)
2y*dy/dx - 32/(cos2(32/x)*x2)

Wolfram Alpha prefers secant, 2y*dy/dx - (32 sec2(32/x))/x2


keeperofdakeys wrote:2. d/dx( y*ln(2*y3*x) ) =

y * d/dx(ln(2xy3)) + dy/dx * ln(2xy3) //did I use the product rule right? am I supposed to use it? I haven't learned it yet. :|
y * 1/(2xy3) * d/dx(2xy3) + dy/dx * ln(2xy3)
y * 1/(2xy3) * (d/dx(2x) * y3 + d/dx(y3) * 2x) + dy/dx * ln(2xy3)
y * 1/(2xy3) * (2y3 + 6xy2 * dy/dx) + dy/dx * ln(2xy3)
(2y3 + 6xy2 * dy/dx)/(2xy2) + dy/dx * ln(2xy3)

AHHH! I really don't know!

vilidice
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Re: Calculus (more specifically Integral Calculus)

Postby vilidice » Fri Oct 22, 2010 8:47 pm UTC

I'm going to try to take a stab at the sin(x) -> cos(x) thing that you're having problems with (I was always dissatisfied with the way calculus was taught so I've been trying to figure out an elementary way of explaining stuff like this in an elementary way).

The first big thing I need to explain is summations and functions.

The general idea is that some functions and limits can be described using an infinite order polynomyal (there are other kinds too but we'll stick to simple stuff now). The most common introduction to this is the exponential function:
[math]e^x[/math] which is approximated by the series (infinite sum): [math]\sum_{n=0}^{\infty} \frac{1}{n!} X^n = 1+\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6}+...[/math] which is clearly a polynomial (which can be differentiated as such)

to show that this is equal to [imath}e^x[/imath] we can determine properties of [imath]e^x[/imath]:
1. The derivative is equal to the function itself.
2. The function is equal to 1 at 0.

to see that #2 is true is trivial, so we'll focus on #1.

[math]\frac{dy}{dx} \sum_{n=0}^{\infty} \frac{1}{n!} X^n = \sum_{n=0}^{\infty} \frac{1}{n!} \frac{dy}{dx} X^n[/math] (by linearity of derivatives (aka the sum rule for differentiation and the constant multiple rule))

[math]= \sum_{n=1}^{\infty} \frac{n}{n!} X^{n-1}[/math] (power rule, note that the index starts a 1 now, because the constant term ([imath]\frac{X^0}{0!}=1[/imath]) goes to 0, so now we can write the sum out (simplifying the factorial):

[math]\sum_{n=1}^{\infty} \frac{1}{n-1!} X^{n-1} = \frac{1}{1} +\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6}...[/math]

which is equal to the original function (so this is equal to [imath]e^x[/imath]) (this is also why factorials are important)

while I think you know this [imath](n-1)!= \frac{n}{n!}[/imath] and [imath]n!=n* ((n-1)!)[/imath].

so, now we can determine that [math]e^{ax}=\sum_{n=0}^{\infty} \frac{1}{n!} a^n X^n[/math]
so if we let a = [imath]\sqrt{-1} = i[/imath] then we have: [math]e^{ix}=\sum_{n=0}^{\infty} \frac{1}{n!} i^n X^n =[/math]

now, this gets interesting because for all even n [imath]i^n = -1[/imath] so we can write the series as the some of two series:

[math]e^{ix}=\sum_{n=0}^{\infty} \frac{1}{n!} i^n X^n = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1!} X^{2n+1}[/math] (for the second series I factored out the i using the fact [imath]\frac{1}{i} = -1[/imath]

now the first series can be written as:[math]\frac{\sum_{n=0}^{\infty} \frac{1}{n!} i^n X^n +\sum_{n=0}^{\infty} \frac{1}{n!} (-i)^n X^n}{2}[/math] to show this:
[math]=\frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1!} X^{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n(-1)^n}{2n+1!} X^{2n+1}}{2}[/math][math]= \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1!} X^{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^{2n}}{2n+1!} X^{2n+1}}{2}[/math][math]= \frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1!} X^{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{-(-1)^n}{2n+1!} X^{2n+1}}{2}[/math][math]=\frac{\sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} +i\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1!} X^{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n} -i\sum_{n=0}^{\infty} \frac{(-1)^n(-1)^n}{2n+1!} X^{2n+1}}{2}[/math][math]= \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} X^{2n}[/math]
[math]=\frac{e^{ix}+e^{-ix}}{2}[/math]

the same can be shown for the fact that the second series is equal to \frac{e^{ix}-e^{-ix}}{2i}

Now we're going to take a bit of a diversion down a different path, but we'll wind up back here in a moment.

There were two mathematicians named Taylor and MacLauren who worked extensively with series and realized that you could compute a function's series expansion based on the derivative, specifically the derivative about the point 0, and came up with this formula to go about it:

[math]f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n[/math] called a MacLauren series (where n is the th derivative of f). To use this we then only need to know the slope of a line tangent to the function at a single point (x=0), so let's look at the cos(x) function:

cos(0) = 1, and we know that the curve is a perfectly "flat" bell at the top, with equally increasing and decreasing rates on either side, so the slope is 0.

sin(0) = 0 and has the same slope as the function f(t)=t, so we know that the slope is equal to 1.

We know that the values of the functions also oscillate, and so their derivatives should too, so we're going to make a crude assumption now (because to go about this another way would take a bit longer), that since cos(0)=1 for the 0 derivative, for every 3rd derivative it will be equal to -1, and sin will do the same oscillation for the even derivatives, with all of the odd ones being 0.


so now we have a taylor series for cos(x) of the form:
[math]cos(x) = \sum_{n=0}^{\infty} \frac{cos^{(n)}(0)}{n!}x^n[/math]
[math]=\sum_{n=0}^{\infty} \frac{0}{(2n+1)!}x^{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}x^{2n}[/math]
[math]=\sum_{n=0}^{\infty} \frac{(-1)^{2n}}{n!}x^{2n} = \frac{e^{ix}+e^{-ix}}{2}[/math]

and the fact that the taylor series for sin(x) looks like:
[math]cos(x) = \sum_{n=0}^{\infty} \frac{sin^{(n)}(0)}{n!}x^n[/math]
[math]=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} + \sum_{n=0}^{\infty} \frac{(0)^{n}}{n!}x^{2n}[/math]
[math]\sum_{n=0}^{\infty} \frac{(-1)^{2n+1}}{n!}x^{2n+1} = \frac{e^{ix}-e^{-ix}}{2i}[/math]

so now we know sin and cos have exponential approximations, and we know that the derivatives of those approximations work in a particular way, so let's take the derivative of cos(x):

[math]cos(x)=\frac{e^{ix}+e^{-ix}}{2}[/math]
[math]\frac{d}{dx}cos(x)=\frac{1}{2}(\frac{d}{dx}e^{ix}+\frac{d}{dx}e^{-ix})[/math]
[math]\frac{1}{2}ie^{ix}-ie^{-ix}=-\frac{1}{2i}e^{ix}-e^{-ix} = -sin(x)[/math]

and for sin(x):
[math]sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/math]
[math]\frac{d}{dx}sin(x)=\frac{1}{2i}(\frac{d}{dx}e^{ix}-\frac{d}{dx}e^{-ix})[/math]
[math]\frac{1}{2i}ie^{ix}+ie^{-ix}=\frac{1}{2}e^{ix}-e^{-ix} = cos(x)[/math]

so all that long stuff was just to get the sin and cos into exponentials, and also to let you know how to use series to demonstrate something, and to show why the derivative of [imath]e^x[/imath] is [imath]e^x[/imath], it's also a really useful formula to have that [imath]e^{ix}=cos(x)-isin(x)[\imath]. For your own enrichment you might want to use this technique to show the derivatives of tan, sec, cot, csc, cosh, and sinh.

Still not totally elementary, but it's better than "because that's the way it is," which is what I found most newer (post 1978-1986) calculus texts seem to say about all of the rules, try ordering some old (like 60s and early 70s) calculus books off amazon, you can find them really cheap and they cover the material a lot better, tending to prove most of the rules and explain things rigorously, which is good experience.

Also at 13, you're a very impressive student to have gotten this far, and if you work to most of my post you're ahead of most non-senior-level students and my school when it comes to trig functions, complex numbers, and series. Also once you understand the basics of calculus in many variables and vector calculus, look at set theory as a really good introduction subject following caclulus, once you have that you can do real analysis, which explains calculus a lot better, and once you have that you're beyond set for most university mathematics.

voidPtr
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Re: Calculus (more specifically Integral Calculus)

Postby voidPtr » Fri Oct 22, 2010 11:55 pm UTC

vilidice wrote:I'm going to try to take a stab at the sin(x) -> cos(x) thing that you're having problems with (I was always dissatisfied with the way calculus was taught so I've been trying to figure out an elementary way of explaining stuff like this in an elementary way).

The first big thing I need to explain is summations and functions.

The general idea is that some functions and limits can be described using an infinite order polynomyal (there are other kinds too but we'll stick to simple stuff now). The most common introduction to this is the exponential function:
[math]e^x[/math] which is approximated by the series (infinite sum): [math]\sum_{n=0}^{\infty} \frac{1}{n!} X^n = 1+\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6}+...[/math] which is clearly a polynomial (which can be differentiated as such)




Vilidice, you lost me right here. That e^x is equal to that series is a huge assumption, that involves calculus to prove it. It may be correct ( I didn't go through your entire post), but it's way overkill, and I know the Taylor Series is way beyond anyone trying to learn the derivative of sinx.

When trying to prove that something foo logically follows from something bar, why not start with the definition of bar? In this case, the definition of the derivative is [imath]\lim_{h\to0}f(x+h)-f(x)/h[/imath]. Straight from this definition of the deriviative, the derivatives of sinx and cosx are not that bad, not beyond the reach of honours high school students, and certainly less work and assumptions than what you proposed.

It does involve proving the limits [imath]\lim_{x \to 0} sinx/x = 1[/imath] and [imath]\lim_{x\to 0}(cosx-1)/x = 0[/imath], but they're not hard to prove.The first one does depend on the limit squeeze theorem and a little geometry, but it's intuitive to show it is true as you zoom in closer to zero. The second limit directly follows from the first.

Why use a powertool to kill a fly when a swatter will suffice?

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314man
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Re: Calculus (more specifically Integral Calculus)

Postby 314man » Sat Oct 23, 2010 2:28 am UTC

tckthomas wrote:
First, thanks!
I should give myself a test, then tell me what I have left to learn :D

d/dx ln(x) = 1/x
and you can convert any base log into natural log: base a log x = base e log x / base e log a

there seems to be too much trig identities here! i can only remember the reciprocal ones, sine(x) = 1/cosecant(x), cosine(x) = 1/secant(x), tangent(x) = 1/cotangent(x)

this i need more experience, this is the reason why Calculus in 20 minutes fails.

factorials? i think they are just too weird. is it impossible to derive a factorial if it isn't over another factorial?

implicit differentiation i think i get it. derive both sides with respect to x, then dy/dx magically appears, then put that on one side and done.

chain rule/product rule/quotient rule, i need to find out when to use the chain rule. I just don't know why to use the chain rule when sin(3x^2+1)... can you tell me more?

squeeze theorem, heh this is quite cool! squeezing functions into a point... its true when there is upper bound and lower bound functions and when they are equal to something, the middle is squeezed to also equal that.

spins on calculus, making it to a series, AAAHHHHH! freaky stuff!

thanks! this will help me the next 2 years when i do additional math. i would learn calculus secondary 3 or 4...

EDIT: yeah, either 0, -infinity, or infinity, 3 values...


There've been a few posts since this so I might be repeating some things but I'll try to be as simple and helpful as I can.

So first off, I like to think that the chain rule is used in two different ways. It's really one, which is dy/dt = dy/dx * dx/dt. But when dealing with functions that pretty much has the chain rule apart of it (like the sin function), I just like to think of it as "and derive what's inside", or in the case of e^x, "derive what's in the exponent".
So for sin(3x^2 + 1), you start by deriving the sin function. The derivative of sinx is cosx, so sin(3x^2 + 1) = cos(3x^2 + 1). Then you take what's inside and derive that and put it outside. The derivative of the inside is 6x. So the answer comes to cos(3x^2 + 1) * 6x. It's essentially working the outside going in.

You can't derive factorials, and it doesn't come up too often with integral and derivative calculus (at least that I found so far... I'm in 2nd year actuarial science so I'm on multivariable calculus right now). But it comes up quite often when you get into series. Pretty much what happens is factorials pop up, it's a big pain in the ass, so you want to get rid of it. So essentially you want to know how to cancel out factorials. There are the easier ways like n! / (n-1)! = n. But there are some that kind of go against logic. For example, when n goes to infinite, n! / n^n is 1. You can break it down by ((n)(n-1)(n-2)(n-3)...etc.) / (n * n * n * n * etc.). Essentially all the terms have a limit of 1 so the whole thing goes to one. The reason I think it goes againt logic because personally, I think it should go to 0. Because the first term is 1/n, (the 1 being the first number in the factorial). And that limit is 0. And the rest of the terms is less than 1, so it isn't growing so I think it should go to 0. Does anyone want to help me out explaining this to me? (wooh asking help when I'm trying to help :) )

There are quite a bit of trig identities, but honestly I forgot most of them, but there are a few ones to know that are easy to see and it'll make your life a whole lot easier.
cos^2(x) + sin^2(x) = 1
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2cos^2(x) = 2sin^2(x) + 1 ** (I haven't used this one in a while and I might've made a mistake. It might be the signs are reverse so double check on me before using this one. Also it is an identity to both because to switch from one to the other, you just use the cos^2(x) + sin^2(x) = 1 indentity)
sin(x + y) = ?? (I forgot... I think it is (sin(x/2) + sin(y/2)) / 2 but my memory is a little iffy

Yep that's pretty much how the squeeze theorem works. It's also why you want to know your general function knowledge because you are choosing the function with the upper bounds/lower bounds. So you always want to choose a similar function. A REALLY simple example is the limit of x^3/x when x goes to 0 (so essentially x^2 but not defined at 0). The functions to choose would be y = 2x^2 and y=0. So you have 0 < x^3/x < 2x^2 (should less than or equal to signs... I don't know how to type them). The limits of the functions you chose when x goes to 0 is 0. So x^3/x has a limit of 0. (Of course you can just tell that by looking at x^3 / x but there will be functions that you can't really do that and L'Hopital won't be useful)

Oh when I mean spins of calculus, I mean other ways you can use it that doesn't generally teach anything new but it has an application. For example integrating by rotation. For example, if you have a function y = 1 and you want to find the area of its rotation about the x-axis. Imagine the function rotates going out of the paper and leaves a permanent trail. The area under that trail is the area you are looking for. So for y = 1 about the x-axis, you are making a horizontal cylinder with 1 radius and x height.
This kind of knowledge is neat to know but it doesn't lead to more complicated calculus as far as I know.

Series on the other hand you will have to learn eventually. More specifically, infinite series. Then it's learning to tell if a series converges or diverges (the sum approaches a number or goes to infinte). And then you'll get into the Taylor series which you'll be seeing a lot.

Anyways the stuff I covered comes from high school calc and first year university calc, so you're way ahead of your time. Again I have to give you a congratulations, you have some great motivation. I didn't have this kind of motivation when I was that young (well not for calculus) so keep it up! You'll do great in life

Ps: At the beginning I said I'd keep it simple and whatnot but I feel like I rambled a lot, so sorry if I got confusing or anything

keeperofdakeys
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Re: Calculus (more specifically Integral Calculus)

Postby keeperofdakeys » Sat Oct 23, 2010 3:27 am UTC

tckthomas wrote:AHHH! I really don't know!

Sorry, I may have gone a little bit overboard :(
You managed to use the product rule perfectly, and in-fact the second one is right :D

Now wolfram alpha was lying to you for the first one, as it treated y as a constant for the first part (this is called a partial derivative, and is used in multi-variable calculus). You need to enter d/dx(x*y2 + tan(32/x) = 0) to make wolfram alpha work correctly (even I didn't know this). It is actually easier then the second, so give it a go yourself!

tckthomas
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Sat Oct 23, 2010 5:00 am UTC

Here I go again! :D
keeperofdakeys wrote:1. d/dx(x*y2 + tan(32/x)) =
(hint: d/dx(tan(x)) = sec2(x) = 1/cos2(x) )

d/dx(x*y2)+d/dx(tan(32/x))
y2+2xy(dy/dx)+sec2(32/x) * d/dx(32/x)
y2+2xy(dy/dx)+sec2(32/x) * -32/x^2
y2+2xy(dy/dx)-(32sec2(32/x))/x^2

I think i got it right! :D

Edit: no, wolfram alpha didn't lie to me, I didn't use it. I just didn't use the product rule.

keeperofdakeys
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Re: Calculus (more specifically Integral Calculus)

Postby keeperofdakeys » Sat Oct 23, 2010 5:09 am UTC

tckthomas wrote:Here I go again! :D
keeperofdakeys wrote:1. d/dx(x*y2 + tan(32/x)) =
(hint: d/dx(tan(x)) = sec2(x) = 1/cos2(x) )

d/dx(x*y2)+d/dx(tan(32/x))
y2+2xy(dy/dx)+sec2(32/x) * d/dx(32/x)
y2+2xy(dy/dx)+sec2(32/x) * -32/x^2
y2+2xy(dy/dx)-(32sec2(32/x))/x^2

I think i got it right! :D

Edit: no, wolfram alpha didn't lie to me, I didn't use it. I just didn't use the product rule.

Well done, you have go it right.
I would suggest you next have a look into indefinite integrals, they are harder than derivatives in that you can't integrate every funtion; and when you can, you have to know which rules to apply to get a simpler integral.

tckthomas
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Re: Help a kid with Calculus!

Postby tckthomas » Sat Oct 23, 2010 9:06 am UTC

Ok, I'm prepared.

I'm ready for a range of questions starting from easy to overkill. :D Then maybe you would know what I have yet to search.

mike-l
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Re: Calculus (more specifically Integral Calculus)

Postby mike-l » Sat Oct 23, 2010 10:56 am UTC

314man wrote:You can't derive factorials, and it doesn't come up too often with integral and derivative calculus (at least that I found so far... I'm in 2nd year actuarial science so I'm on multivariable calculus right now). But it comes up quite often when you get into series. Pretty much what happens is factorials pop up, it's a big pain in the ass, so you want to get rid of it. So essentially you want to know how to cancel out factorials. There are the easier ways like n! / (n-1)! = n. But there are some that kind of go against logic. For example, when n goes to infinite, n! / n^n is 1. You can break it down by ((n)(n-1)(n-2)(n-3)...etc.) / (n * n * n * n * etc.). Essentially all the terms have a limit of 1 so the whole thing goes to one. The reason I think it goes againt logic because personally, I think it should go to 0. Because the first term is 1/n, (the 1 being the first number in the factorial). And that limit is 0. And the rest of the terms is less than 1, so it isn't growing so I think it should go to 0. Does anyone want to help me out explaining this to me? (wooh asking help when I'm trying to help :) )

Generally the factorials that appear in calculus are constants, so of course you can derive them, just like any other constant. Their derivative is 0. In some more advanced stuff you'll get to where the factorial is not a constant, but a function of x, but then we need to replace the factorial (which is only defined on the natural numbers) with something defined on the real numbers, and the definition makes the derivative not so bad.

Now, as for n!/n^n... that's definitely 0 and is a simple application of Stirling's Approximation, which shows that as n gets large, the terms inside the limit are approximately
[math]\frac{\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n}{n^n} =\frac {\sqrt{2 \pi n}}{e^n}[/math]
and [imath]e^n[/imath] grows waaaay faster than [imath]\sqrt n[/imath] so it's 0.

Your argument only shows that nPr/n^r goes to 1. It doesn't work for n! because while each quotient goes to 1, you keep adding more and more smaller quotients. In fact, the argument you gave for why it should be 0 is exactly right. You showed that the limit grows is smaller than 1/n, and it's obviously greater than 0, so by the squeeze theorem it must be 0.

There are quite a bit of trig identities, but honestly I forgot most of them, but there are a few ones to know that are easy to see and it'll make your life a whole lot easier.
cos^2(x) + sin^2(x) = 1
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2cos^2(x) = 2sin^2(x) + 1 ** (I haven't used this one in a while and I might've made a mistake. It might be the signs are reverse so double check on me before using this one. Also it is an identity to both because to switch from one to the other, you just use the cos^2(x) + sin^2(x) = 1 indentity)
sin(x + y) = ?? (I forgot... I think it is (sin(x/2) + sin(y/2)) / 2 but my memory is a little iffy

You can get all the addition formulae quickly from expanding out [imath]e^{i(a+b)}=e^{ia}e^{ib}[/imath] using [imath]e^{ix}=\cos x + i \sin x[/imath].

Anyway, cos(2x) = 2cos^2 x - 1=1-2sin^2 x. The missing ones are
sin (x+y) = sin x cos y + sin y cos x
cos (x+y) = cos x cos y - sin x sin y

For cases like this you can do some simple error checking based on other facts you know. For example, cos(2x) couldn't possibly be 2sin^2 x + 1, since the first summand is nonnegative, this value is always at least than 1, and cos(2x) is at most 1. If you have a guess for what sin (x+y) is, you can plug in x=y to see if it agrees with what you know sin (2x) to be.
Last edited by mike-l on Sat Oct 23, 2010 7:13 pm UTC, edited 1 time in total.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

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Dopefish
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Re: Help a kid with Calculus!

Postby Dopefish » Sat Oct 23, 2010 5:29 pm UTC

Dear various people, the verb corresponding to taking a derivative is differentiate, not derive.

Thank you.

Nitrodon
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Re: Help a kid with Calculus!

Postby Nitrodon » Sat Oct 23, 2010 6:42 pm UTC

Also, cos(2x) = cos2(x) - sin2(x) = 1 - 2sin2(x) = 2cos2x - 1. Just thought I'd clear this up.

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314man
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Re: Calculus (more specifically Integral Calculus)

Postby 314man » Sat Oct 23, 2010 9:32 pm UTC

mike-l wrote:
314man wrote:You can't derive factorials, and it doesn't come up too often with integral and derivative calculus (at least that I found so far... I'm in 2nd year actuarial science so I'm on multivariable calculus right now). But it comes up quite often when you get into series. Pretty much what happens is factorials pop up, it's a big pain in the ass, so you want to get rid of it. So essentially you want to know how to cancel out factorials. There are the easier ways like n! / (n-1)! = n. But there are some that kind of go against logic. For example, when n goes to infinite, n! / n^n is 1. You can break it down by ((n)(n-1)(n-2)(n-3)...etc.) / (n * n * n * n * etc.). Essentially all the terms have a limit of 1 so the whole thing goes to one. The reason I think it goes againt logic because personally, I think it should go to 0. Because the first term is 1/n, (the 1 being the first number in the factorial). And that limit is 0. And the rest of the terms is less than 1, so it isn't growing so I think it should go to 0. Does anyone want to help me out explaining this to me? (wooh asking help when I'm trying to help :) )

Generally the factorials that appear in calculus are constants, so of course you can derive them, just like any other constant. Their derivative is 0. In some more advanced stuff you'll get to where the factorial is not a constant, but a function of x, but then we need to replace the factorial (which is only defined on the natural numbers) with something defined on the real numbers, and the definition makes the derivative not so bad.

Now, as for n!/n^n... that's definitely 0 and is a simple application of Stirling's Approximation, which shows that as n gets large, the terms inside the limit are approximately
[math]\frac{\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n}{n^n} =\frac {\sqrt{2 \pi n}}{e^n}[/math]
and [imath]e^n[/imath] grows waaaay faster than [imath]\sqrt n[/imath] so it's 0.

Your argument only shows that nPr/n^r goes to 1. It doesn't work for n! because while each quotient goes to 1, you keep adding more and more smaller quotients. In fact, the argument you gave for why it should be 0 is exactly right. You showed that the limit grows is smaller than 1/n, and it's obviously greater than 0, so by the squeeze theorem it must be 0.

There are quite a bit of trig identities, but honestly I forgot most of them, but there are a few ones to know that are easy to see and it'll make your life a whole lot easier.
cos^2(x) + sin^2(x) = 1
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2cos^2(x) = 2sin^2(x) + 1 ** (I haven't used this one in a while and I might've made a mistake. It might be the signs are reverse so double check on me before using this one. Also it is an identity to both because to switch from one to the other, you just use the cos^2(x) + sin^2(x) = 1 indentity)
sin(x + y) = ?? (I forgot... I think it is (sin(x/2) + sin(y/2)) / 2 but my memory is a little iffy

You can get all the addition formulae quickly from expanding out [imath]e^{i(a+b)}=e^{ia}e^{ib}[/imath] using [imath]e^{ix}=\cos x + i \sin x[/imath].

Anyway, cos(2x) = 2cos^2 x - 1=1-2sin^2 x. The missing ones are
sin (x+y) = sin x cos y + sin y cos x
cos (x+y) = cos x cos y - sin x sin y

For cases like this you can do some simple error checking based on other facts you know. For example, cos(2x) couldn't possibly be 2sin^2 x + 1, since the first summand is nonnegative, this value is always at least than 1, and cos(2x) is at most 1. If you have a guess for what sin (x+y) is, you can plug in x=y to see if it agrees with what you know sin (2x) to be.



Ah thanks a lot! Can't believe how many mistakes I've made, good thing I made them on a forum where I'm anonymous instead of on a midterm haha.


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