Thread: Equations 2

Yeah radius 1 center 0,0. It's easy to deduce using Pythagoras' theorem.

Code:

4(∫(0, 1) (x^2 + y^2 = 1))
x^2 + y^2 = 1 = { y = sqrt(1-x^2)
                { y = -sqrt(1-x^2)
4(∫(0, 1) (sqrt(1-x^2)-(-sqrt(1-x^2)) dx)
4(∫(0, 1) 2sqrt(1-x^2) dx)
4(xsqrt(1-x^2)+arcsin(x)](0,1))
4((sqrt(1-1)+arcsin(1))-(arcsin(0)))
4(arcsin(1))
4(π/2)
2π

I don't really get what constant it derives.. π? Wouldn't you need that to find arcsin 1 anyway?

Your'e supposed to find the area under the curve and the multiply by 4. It's how to derive pi without using trig.

Yeah, I'll get to yours in a sec.

Originally Posted by ninja9578

Your'e supposed to find the area under the curve and the multiply by 4. It's how to derive pi without using trig.

Yeah, I'll get to yours in a sec.

Oh, so 0 to 1 on just sqrt(1-x^2)... but wouldn't you have to use trig (arcsin) to get the antiderivative of that function in order to evaluate the integral? Is there a way to do it without? (I'm a senior in high school taking AP Calc atm, so I've learned this all pretty recently, not an expert, but I'm interested )

Originally Posted by ninja9578

Your'e supposed to find the area under the curve and the multiply by 4. It's how to derive pi without using trig.

Yeah, I'll get to yours in a sec.

aha, so you DO intergrate it then

You can't integrate it. You can only integrate it if you extrapolate what the question was supposed to be, which was 4S(1,0)(1-x2)1/2dx.

And yes, you would use a trig substitution to do this, which gets you nowhere. Well, it gets you pi (which makes sense, pi*1^2), but not an expression for pi.

I think ninja wants a binomial expansion or something.

what was wrong with that integration? I could have sworn Ive done that kind of thing before. seperating the xs and ys and integrating both sides.

No, you take one of the variables and turn it into a function of the other one, that's how to do multiple variables in simple thing like this. I think... I know I've done this equation before I thikn that's how I did it.

FYI, Alt + B will give you the ∫ symbol.

Yup

Code:

x^2 + y^2 = 1
1 - x^2 = y^2

x^2 + (1 - x^2) = 1

Then solve under the curve

hmm...Im thinking of something different then. dang it my old papers are at home grrr

Originally Posted by ninja9578

Yup

Code:

x^2 + y^2 = 1
1 - x^2 = y^2

x^2 + (1 - x^2) = 1

Then solve under the curve

1 = 1. How useful.

No, you can't integrate an equation, you integrate functions. x2 + y2 = 1 is not something you can 'integrate'. You have to write it as a function, so: y = (1 -x2)1/2. We have had to restrict the range to a semicircle (by only taking the positive root), to make it one-one. Therefore we'll have to double the integral from -1 to 1. Alternatively we could write it as 4*the integral from 0 to 1.

You then integrate it using x = sinu, and a bit of messing about will get you to pi, which is clearly correct.

Please explain how this gives us an expression for pi?

Also please answer my one.

Solve for x

(3x2 +1)1/2 -(2x)1/2 +x -1 = 0

oh right i see what your talking about now xei, you see thats just so obvious i assumed he did it, so i didnt really pay much attention to what he wrote.

Originally Posted by Xei

Simple quotient rule, I believe. vdu/dx is 0 at x=5. The denominator tends to 0 as x tends to 5, and the numerator, x(9x3 +x2 -10x +25) has no (x-5) factor so we have an asymptote?

Here's something a bit less formulaic:

(3x2 +1)1/2 -2x1/2 +x -1 = 0

Indicies are on the right, so (3x2 +1)1/2 is root(3xsquared +1), etcetera.

I promised I'd get to it... I'm confused as to why you put this one in here. I didn't require anything more than high school algebra to do it, perhaps I read it wrong


I added all the proper parens to make it easier to see

Code:

((3x^2) + 1) / 2 - (2x/2) + x - 1 = 0
(3x^2 -2x + 1) / 2 + x - 1 = 0  //consolidate fractions
(3x^2 -2x + 1) / 2 + 2x/2 - 2/2 = 0  //multiple x and -1 by 2/2
(3x^2 - 1) / 2 = 0   //cancel out 2x and -2x
(3x^2) / 2 = 1/2  //move constant to other side of equation
3x^2 = 1  //remove duplicate division
x^2 = 1/3
x = √(1/3)

Indicies are on the right, so (3x2 +1)1/2 is root(3xsquared +1), etcetera.

Also how do you deduce pi with that integral?

I have no clue, just found the sight today and havent done math since high school but I want to learn if you can help?

Originally Posted by Xei

Solve for x

(3x2 +1)1/2 -(2x)1/2 +x -1 = 0

(3x2 +1)1/2 -(2x)1/2 = 1-x

3x^2 +1 + 4x - 2*sqrt(3x^2 + 2x +1) = 1 -2x + x^2

-2*sqrt(3x^2 + 2x +1) = -2x^2 -6x

sqrt(3x^2 + 2x+1) = x^2 + 3x

3x^2 + 2x +1 = x^4 + 9x^2 + 6x^3

x^4 + 6x^3 + 6x^2 -2x -1 = 0

*phew* Now all it needs is a little long division.

I won't do this, I will just say that by the Fundamental Theorem of Algebra, it has 4 complex roots. Actually, they're all real from the looks of it. Real, but ugly.

Pretty sure you fluffed the algebra (the roots are nice and it should be easy to factorise) but obviously that kind of question's pretty error prone...

There's a necessary step you need to take at the end, as well.

But apart from that you did the difficult work which was seeing that the binomial step would simplify things.

Originally Posted by Xei

Indicies are on the right, so (3x2 +1)1/2 is root(3xsquared +1), etcetera.

Also how do you deduce pi with that integral?

Oh. Please use the √ symbol (Alt + V, Alt 251 if you're on a PC) I'll figure it out later then, I got other stuff to do right now.


Deducing pi is easy. Take the formula for a circle at the origin with radius 1, find the area under the curve from 0 to 1 (which equals exactly one quarter of the area of the circle.) Multiply by four for the exact area of the circle. A = pi r^2. r^2 = 1 so the area that you got after the multiplication equals pi if you drop the units.

I think that's how I did it

Code:

(3x2 +1)1/2 -(2x)1/2 +x -1 = 0

(3x2 +1)/ 2 - 1 = 0

3x2 + 1 = 2

3x2 = 1

x2 = 1/3

x = √(3)/3

Hm.

EDIT: FUCK, I left out the +1.

Damnit.

Originally Posted by partly Exoteric

I have no clue, just found the sight today and havent done math since high school but I want to learn if you can help?

lol this is a site where you learn to lucid dream, yet you come to learn maths

Indicies are on the right argh.

I don't really like the root symbol as it's just a special case of any real exponent. Might as well just stick to normal notation...

Deducing pi is easy. Take the formula for a circle at the origin with radius 1, find the area under the curve from 0 to 1 (which equals exactly one quarter of the area of the circle.) Multiply by four for the exact area of the circle. A = pi r^2. r^2 = 1 so the area that you got after the multiplication equals pi if you drop the units.

I think that's how I did it

You can't integrate a formula.

Just look at the graph; you can't find the area under the curve if there are actually two curves above and below each other.

You have to express it as a function.

And that's really very circular (...lol.). How exactly did you integrate it? Because I used a trig substitution, which ended up with me arcsining 1 to get pi, and arcsin 1 is only pi because of the definition of pi in the first place.

It doesn't help to 'determine' pi at all; you just get the symbol pi at the end. To deduce pi you've got to use a Maclaurin expansion. Alternatively you could use the binomial theorem to give an infinite polynomial, I think, and then integrate it.

Indicies are on the right

What does that mean?

(I'm only up to Analytical Geometry, and I'm taking that right now...)

Oh... Root index... Duh...

Well.

Uhm, Xei, maybe you can enlighten me i where I went wrong, I got an answer of (3+√17) / 2 and (3-√17) / 2. I graphed your original equation and I can tell that the answer is supposed to be 1 and i.

Code:

√(3x^2 + 1) - √2x + x - 1 = 0
√(3x^2 + 1) - √2x = 1 - x  //pull apart equation
3x^2 + 2x + 1 - 2√(3x^2 + 2x + 1) = x^2 - 2x + 1  //square both sides to get one radical
2x^2 = 2√(3x^2 + 2x + 1)  //simplify
x^4 = 3x^2 + 2x + 1  //square both sides to remove radical
x = (3 +√17) / 2  //use quadratics to solve for x

Where'd I fuck up?

Pi can be defined using the taylor (macclaurin) expansion of arcsin, arccos, or arctan I think.

Like, arctan 1 == pi/4, so pi = 4* [1 - 1/3 + 1/5 -1/7 + ...], which is the nice for approximations since it alternates, so error is less than the size of the first dropped term.

So, with 5 terms, we get pi~= 1052/315

Ok, maybe there are faster ways to approximate it. This way isn't even linear, but you get the point.