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[Therion]

thanks for the help, but i think im just gonna guess on a problem like this on my final, way too much freakin work, they even want the answer to the nearest hundredth...

Are there no calculators allowed? I wouldn't think you'd need to take square roots on paper, would you? If you can use a calculator, the problem is pretty much solved from the step you're at.

 

[Jaladinozozo]

they are allowed but no matter what i put it, my webpage webwork says the answer is wrong

 

[Therion]

they are allowed but no matter what i put it, my webpage webwork says the answer is wrong

Ah, sorry to hear that. Those systems are evil for math, where you can have all kinds of equivalent answers. It's not like it would take that much longer to grade these things by hand, anyway.

 

[Jaladinozozo]

my final answer comes out to -81/140 + or - sqrt9641/140 btw

 

[Jaladinozozo]

omg i finaly got it right, stupid fuckin ti-89...i keep forgetting you cant put the 9641/140 into the entire square root, you have to seperate the 9641 in ( ) then divide over 140

jeez this problem took me like 2 hours just because of calculator and online functions

 

[Therion]

my final answer comes out to -81/140 + or - sqrt9641/140 btw

Those answers are correct. Rounded to hundredths, they are 0.12 and -1.28. Is that what you are trying? Or do you still need to enter it as one number +/- another number?

Edit: Frustrating, but at least you got it right and can move on.

 

[Feep]

So, quick question.

In a program I'm making, I wanted to draw a parabolic arc between any two points on the screen. Originally, what I did was the following:

1) Create a third point as a midpoint of the other two, and boost its y-coordinate up by a bit,
2) Generate three matrices to represent three equations to solve for a, b, and c in the standard form y = ax^2 + bx + c,
3) Use Cramer's Rule to solve for a, b, and c. Use a simple shader to shade the arc.

While this works beautifully in most situations, things start getting weird when the two points get very close to each other horizontally (which will happen regularly). Values for A, B, and C get sky-rocketingly large, and the vertex of the parabola zooms off the screen.

There is one of two solutions, here, I think:

1) Is there any way to generate the third midpoint precisely as the vertex, so I keep reasonable values for A, B, C?

or...

2) Can I generate the parabola parametrically? The x-coordinate is easy, but how do I find the parametric equation of the y-coordinate to start at one value and end up at another, arbitrarily? I was never good with parametrics. = P

Thanks in advance, math-GAF!

Edit: Option 2 is looking like the only viable one at the moment; I can't bring the parabola's vertex down even by manually moving the midpoint by miniscule amounts. I think it has something to do with an inherent inaccuracy and rounding error in using Cramer's Rule computationally, which is warned of in many texts.

 

[cpp_is_king]

So, quick question.

In a program I'm making, I wanted to draw a parabolic arc between any two points on the screen. Originally, what I did was the following:

1) Create a third point as a midpoint of the other two, and boost its y-coordinate up by a bit,
2) Generate three matrices to represent three equations to solve for a, b, and c in the standard form y = ax^2 + bx + c,
3) Use Cramer's Rule to solve for a, b, and c. Use a simple shader to shade the arc.

While this works beautifully in most situations, things start getting weird when the two points get very close to each other horizontally (which will happen regularly). Values for A, B, and C get sky-rocketingly large, and the vertex of the parabola zooms off the screen.

There is one of two solutions, here, I think:

1) Is there any way to generate the third midpoint precisely as the vertex, so I keep reasonable values for A, B, C?

or...

2) Can I generate the parabola parametrically? The x-coordinate is easy, but how do I find the parametric equation of the y-coordinate to start at one value and end up at another, arbitrarily? I was never good with parametrics. = P

Thanks in advance, math-GAF!

Edit: Option 2 is looking like the only viable one at the moment; I can't bring the parabola's vertex down even by manually moving the midpoint by miniscule amounts. I think it has something to do with an inherent inaccuracy and rounding error in using Cramer's Rule computationally, which is warned of in many texts.

Let's pretend the two points formed a horizontal line. Problem would be easy then right? You've got:

(x1,y) and (x2,y)

You could decide up front how "high" you wanted the parabola to be. Let's say you want the height to be Length/3 of the line.

Height = |x2 - x1| / 3

Thus, the maximum point on this parabola is going to be horizontally at the midpoint between the two points, and vertically a distance of |x2 - x1| / 3 above the line. This point is easily seen to be ((x1+x2)/2, y + |x2-x1|/3)

Now, construct a parabola of the form f(x) = A(x+B)^2 + C that matches this new standard parabola between these two points on the horizontal line.

When you're done, simply apply a simple rotation matrix to each point by computing arctan(y/x) to figure out the angle the *original* line made with the x-axis.

 

[Feep]

Let's pretend the two points formed a horizontal line. Problem would be easy then right? You've got:

(x1,y) and (x2,y)

You could decide up front how "high" you wanted the parabola to be. Let's say you want the height to be Length/3 of the line.

Height = |x2 - x1| / 3

Thus, the maximum point on this parabola is going to be horizontally at the midpoint between the two points, and vertically a distance of |x2 - x1| / 3 above the line. This point is easily seen to be ((x1+x2)/2, y + |x2-x1|/3)

Now, construct a parabola of the form f(x) = A(x+B)^2 + C that matches this new standard parabola between these two points on the horizontal line.

When you're done, simply apply a simple rotation matrix to each point by computing arctan(y/x) to figure out the angle the *original* line made with the x-axis.

This was a good idea, and I thank you for it...the rotation was clever.

However, I just spend 40 minutes implementing it, and while it works just as well as my first method under normal circumstances, it STILL screws up when the x-coordinates get close together. I'm absolutely convinced it has to do with round-off errors from Cramer's Rule. I might try another method to solve for A, B, C, but row-reduction seems very difficult to implement programmatically...

Anyone with an idea for the parametrics?

 

[ScrabbleDude]

I need some math help and wasn't aware this thread existed. Glad it hit the first page when I was on.

I'm taking a data structures and algorithm analysis course and I'm currently covering a unit on Big-O notation. The main problem is that out of nowhere they've started using what seems to me to be complex summation to calculate O and T. I haven't done a math course yet that covers summation and I feel like I'm over my head.

Does anyone know of a page that will introduce it to me on a good curve so I can understand this and be successful at it? I've already checked Google out and checked a couple tutorials, but only feel a little better off than I did in the beginning.

 

[cpp_is_king]

This was a good idea, and I thank you for it...the rotation was clever.

However, I just spend 40 minutes implementing it, and while it works just as well as my first method under normal circumstances, it STILL screws up when the x-coordinates get close together. I'm absolutely convinced it has to do with round-off errors from Cramer's Rule. I might try another method to solve for A, B, C, but row-reduction seems very difficult to implement programmatically...

Anyone with an idea for the parametrics?

that's interesting. I'm not sure of a way off the top of my head of doing a parameterization without starting with a parameterization of the horizontal line and then applying a rotation to that parameterization.

But that doesn't really help you, because a parameterization of a parabloa is trivial.

When the x coordinates are close together, could you simply just use a straight line? How close are we talking about? Is a parabola really going to look all that much different? Could you upload a screenshot on imgur of what the incorrect results look like?

 

[cpp_is_king]

I need some math help and wasn't aware this thread existed. Glad it hit the first page when I was on.

I'm taking a data structures and algorithm analysis course and I'm currently covering a unit on Big-O notation. The main problem is that out of nowhere they've started using what seems to me to be complex summation to calculate O and T. I haven't done a math course yet that covers summation and I feel like I'm over my head.

Does anyone know of a page that will introduce it to me on a good curve so I can understand this and be successful at it? I've already checked Google out and checked a couple tutorials, but only feel a little better off than I did in the beginning.

Can you give a concrete example of a situation you don't understand?

 

[Genesis Knight]

My mind is melted. How the heck do I do this:

Let f : A -> B. Let X and Y be subsets of A. Prove that f(X ∪ Y) = f(X) ∪ f(Y)

It seems so simple, but I just can't think of a way to do it.

 

[Lonely1]

My mind is melted. How the heck do I do this:

Let f : A -> B. Let X and Y be subsets of A. Prove that f(X ∪ Y) = f(X) ∪ f(Y)

It seems so simple, but I just can't think of a way to do it.

If x is an element of f(X ∪ Y) then exist an x' in X ∪ Y so that f(x')=x, so it exist an element x' in either X or Y so that f(x')=x, hence x is an element of f(X) ∪ f(Y) and f(X ∪ Y) is a subset of f(X) ∪ f(Y).

The converse is about the same.

 

[Feep]

that's interesting. I'm not sure of a way off the top of my head of doing a parameterization without starting with a parameterization of the horizontal line and then applying a rotation to that parameterization.

But that doesn't really help you, because a parameterization of a parabloa is trivial.

When the x coordinates are close together, could you simply just use a straight line? How close are we talking about? Is a parabola really going to look all that much different? Could you upload a screenshot on imgur of what the incorrect results look like?

I'm way late on this one, but this is literally what I did. I switched to a linear path if the x-coordinates were within 4-5 pixels of each other.

 

[Genesis Knight]

If x is an element of f(X ∪ Y) then exist an x' in X ∪ Y so that f(x')=x, so it exist an element x' in either X or Y so that f(x')=x, hence x is an element of f(X) ∪ f(Y) and f(X ∪ Y) is a subset of f(X) ∪ f(Y).

The converse is about the same.

Thanks. I think that makes sense.

 

[ThanksVision]

Just starting AP Calc again in my sr year of HS and I've forgotten some of the basics over the summer. Doing a few logarithms for review and I don't remember how to solve this one:

(1+X)^15=2

Anybody know? Thanks!!

 

[hemtae]

Just starting AP Calc again in my sr year of HS and I've forgotten some of the basics over the summer. Doing a few logarithms for review and I don't remember how to solve this one:

(1+X)^15=2

Anybody know? Thanks!!

There shouldn't be any logarithms in that. You raise both sides to the (1/15)th power so you have

1+x = 2^(1/15)

so x = 2^(1/15)-1

 

[Bryan1321]

Just starting AP Calc again in my sr year of HS and I've forgotten some of the basics over the summer. Doing a few logarithms for review and I don't remember how to solve this one:

(1+X)^15=2

Anybody know? Thanks!!

Solving X ?

Because it may be something totally diferent to what ive thinked,

Anyway, if you meant solving X = .047294122820648


Edit: Solved ^^^^^^

 

[ThanksVision]

Awhhh yesh that's right. I forgot about making it to the 1/15. Thanks

 

[Unconditional]

Anyone have knowledge about hypothesis testing? The problem is set up as
H1 : R = S + N
H0 : R = N, where S and N are independent random variables distributed exponentially therefore the pdf of S and N are:

fS(s) = a*e^-(a*s), if s>= 0,
0 , otherwise

fN = b*e^-(b*n), if n>= 0,
0 , otherwise

where a>0, b>0, and b>a.

I'm trying to prove via Likelihood Ratio Test, Choose H1 if r > eta; Choose H0 if r < eta.

I determine fR(r|H1) via convolution of S and N, but when I determine L(r) = fR(r|H1)/fR(r|H0) >< eta, likelihood ratio, I don't get Choose H1 if r > eta; Choose H0 if r < eta.

 

[ThanksVision]

So there is some extra credit on this take home test. Stuff on it that we haven't gone over yet. I don't even know what it really means hah. Any help is appreciated, I just want to learn...

 

[Takao]

Okay, I'm having a mental freak with this Data Management stuff:

An interior deisnger is creating a pattern of tiles for a border. If the pattern is to contain an equal number of tiles of each of four colours, how many different patterns can be made if:

a) only four tiles are used in the pattern? Answer key: 362 880
b) eight tiles are used in the pattern? Answer key: 27. 5 times more
c) eight tiles are used and a green tile must begin and end the pattern? Answer key: 45 times more

 

[louis89]

First person to explain how to answer this question wins a prize

 

[CrudeDiatribe]

First person to explain how to answer this question wins a prize

The important information is the last row and the notes. There's nothing to suggest that the distribution of products amongst client types is anything other than uniform.

So, we can ignore everything than the relative size of the 'home' and 'international' client types.

54/16 = 3.375, so ~ 3:1

 

[f0rk]

Was that for a job application?

 

[mcrae]

im doing this report for my materials lab in college, civil engineering.

we've established the blend of the aggregates needed for concrete, for a total of 1210grams per specimen

next we need to figure out how much asphalt cement we need to add, in grams. for 4.3% AC by mass you figure it out by doing 1210*(4.3%/(100-4.3%)) = 4.49%

i get the math, im just having a lot of trouble wrapping my head around the concept, if we only want 4.3% then why are we adding 4.49% ?

edit: nvm, its just late lol.

 

[mcrae]

Okay, I'm having a mental freak with this Data Management stuff:

An interior deisnger is creating a pattern of tiles for a border. If the pattern is to contain an equal number of tiles of each of four colours, how many different patterns can be made if:

a) only four tiles are used in the pattern? Answer key: 362 880
b) eight tiles are used in the pattern? Answer key: 27. 5 times more
c) eight tiles are used and a green tile must begin and end the pattern? Answer key: 45 times more

yeah i dont understand that at all. b should have more answers than c, not the other way around. A should just be 4cPr4=24

 

[louis89]

The important information is the last row and the notes. There's nothing to suggest that the distribution of products amongst client types is anything other than uniform.

So, we can ignore everything than the relative size of the 'home' and 'international' client types.

54/16 = 3.375, so ~ 3:1

But how do you know that it is uniform? I can't see how it's justified to make that assumption.

 

[CrudeDiatribe]

But how do you know that it is uniform? I can't see how it's justified to make that assumption.

There is absolutely not enough information to figure it out if it wasn't— so it's either uniform or it's unanswerable. Note #3 might be an attempt to suggest this, but it's poorly written if so.

 

[louis89]

Was that for a job application?

Yeah. Normally questions like this have a "Cannot say" option when it's impossible to know the answer, but this one doesn't and I can't help but think that it's a mistake.

 

[hawkshockey11]

Anyone know an easy way to remember Reduction Formulas?

Like those ^
but ALL of them....

 

[ant_]

Anyone know an easy way to remember Reduction Formulas?

Like those ^
but ALL of them....

In all honesty there's no way to completely remember all of them unless you are crazy. The best way to remember them is to know how to do each integral without a formula.

 

[cpp_is_king]

Anyone know an easy way to remember Reduction Formulas?

Like those ^
but ALL of them....

Honestly there's no reason to remember these when you can just remember how to derive them. Just apply integration by parts once and that's exactly what you end up with.

Basically you're asking something like "How do I remember that that the derivative of x^2 is 2x and the derivative of x^3 is 3x^2, and the derivative of x^4 is ...". You don't remember every single one, you just apply a single rule and you have the answer in like one step.

 

[ChocolateCupcakes]

Anyone know an easy way to remember Reduction Formulas?

Like those ^
but ALL of them....

I'm taking Calc II now and in short, no. But I wish there was.

 

[hawkshockey11]

Honestly there's no reason to remember these when you can just remember how to derive them. Just apply integration by parts once and that's exactly what you end up with.

Basically you're asking something like "How do I remember that that the derivative of x^2 is 2x and the derivative of x^3 is 3x^2, and the derivative of x^4 is ...". You don't remember every single one, you just apply a single rule and you have the answer in like one step.

Except many times you will need to do integration by parts like 5 times to get a function you can integrate....

I was never taught the shortcuts for these like with regular derivatives so I don't know the single rule. In fact I haven't been taught anything. My teacher is from China and whispers in broken English to the board and writes incredibly small. I am fucked for Calc 2...

 

[ant_]

Except many times you will need to do integration by parts like 5 times to get a function you can integrate....

I was never taught the shortcuts for these like with regular derivatives so I don't know the single rule. In fact I haven't been taught anything. My teacher is from China and whispers in broken English to the board and writes incredibly small. I am fucked for Calc 2...

PatrickJMT is one of the best math youtube teachers available. His videos are short and a great complement to learning Calculus. I have high praise for him because his videos saved me quite a few times when I didn't fundamentally understand what was going on.

http://patrickjmt.com/#calculus

^ Calculus videos in general

Integration by Parts

http://patrickjmt.com/integration-by-parts-a-loopy-example/

^ one video of many

 

[hawkshockey11]

PatrickJMT is one of the best math youtube teachers available. His videos are short and a great complement to learning Calculus. I have high praise for him because his videos saved me quite a few times when I didn't fundamentally understand what was going on.

http://patrickjmt.com/#calculus

^ Calculus videos in general

Integration by Parts

http://patrickjmt.com/integration-by-parts-a-loopy-example/

^ one video of many

I love you. Khan Academy didn't have much for what I needed and I didn't have much luck finding a good series of videos on youtube.

 

[Sealda]

I am having problem with this one

I think the limits are

EDIT:
But probably wiser write them as this maybe?

and start with dx?


EDIT:
gives us

then we take dy

guess, i figured it out eventually with the help of wolfram with the last integration

 

[Genesis Knight]

So, if I'm trying to prove that the sequence

{x} : n / (n^2 + 2)

is null, I need to show that given any epsilon, there is some value for n such that the "nth" term of x is less than that epsilon.

To do that, I first need to do some preliminary analysis and assume that

[n / (n^2 + 2)] < e

Then if I can get n and e on opposite sides of the equation I can have a 'guess' for n that I can then prove.

My problem is that I can't for the life of me get n an e isolated on opposite sides without damaging the strict inequality! Has anyone done something like this before?

 

[cpp_is_king]

So, if I'm trying to prove that the sequence

{x} : n / (n^2 + 2)

is null, I need to show that given any epsilon, there is some value for n such that the "nth" term of x is less than that epsilon.

To do that, I first need to do some preliminary analysis and assume that

[n / (n^2 + 2)] < e

Then if I can get n and e on opposite sides of the equation I can have a 'guess' for n that I can then prove.

My problem is that I can't for the life of me get n an e isolated on opposite sides without damaging the strict inequality! Has anyone done something like this before?

I assume by "is null" you mean "the limit is 0 as n goes to infinity"

Can you do it a different way?

0 < n / (n^2 + 2) < n / n^2 = 1/n

Thus, n/(n^2+2) < 1/n

Since 1/n -> 0 as n->infinity, so does n/(n^2+2)

If that last sentence is something you need to prove instead of being able to say "it's trivial to see that ...", then just repeat your original idea, but using 1/n instead of n/(n^2+2). Once you prove that 1/n goes to 0, then it follows that n/(n^2+2) does, since it's always less than 1/n and greater than 0.


The general technique being used here, which is applicable to many limit problems, is this: Find an expression which is always greater than or equal to the original expression and is very easy to prove that it tends to 0.

 

[Genesis Knight]

That's the Squeeze Rule, right?

The question specifies that it should be done 'from the definition (that is, find the X and show it works for any given epsilon)', so I don't think I can get away with that.

 

[TimesEunuch]

To do that, I first need to do some preliminary analysis and assume that

[n / (n^2 + 2)] < e

Then if I can get n and e on opposite sides of the equation I can have a 'guess' for n that I can then prove.

You're not going to be write an explicit inequality for n in terms of e this way. Instead you need to be a bit cleverer. Remember, you just want to find *any* N such that when n>=N, [n/(n^2+2)]<e; it doesn't have to be the `best' (i.e. smallest) possible such N. So use the observation cpp_is_king made, namely that n/(n^2+2)<1/n. Now pick an N which makes 1/n less than e for n>=N (this should be easy), and n/(n^2+2) will also be less than e.

 

[cpp_is_king]

That's the Squeeze Rule, right?

The question specifies that it should be done 'from the definition (that is, find the X and show it works for any given epsilon)', so I don't think I can get away with that.

Just because it says "from the definition" doesn't mean you can't make observations. There's nothing wrong with using the observation that n / (n^2 + 2) < 1/n and then using the definition against 1/n.

 

[fanboi]

I suck at math so bare with me.

Geometry:

A triangle has the lenghts 6, 7, 8 cm. The height to the shortest side is 6,7 cm.
What is the area in cm2?

Also, how do I calclulate above and how do I draw it...

Halp

 

[hemtae]

I suck at math so bare with me.

Geometry:

A triangle has the lenghts 6, 7, 8 cm. The height to the shortest side is 6,7 cm.
What is the area in cm2?

Also, how do I calclulate above and how do I draw it...

Halp

I think you just apply the formula Area = (1/2)*(base)*(height)

it gives you the height and the shortest side is 6 cm so I get 20,1 cm^2

 

[Feep]

Europeans *do* understand that using commas to denote decimals is confusing as fuck, right?

 

[fanboi]

I think you just apply the formula Area = (1/2)*(base)*(height)

it gives you the height and the shortest side is 6 cm so I get 20,1 cm^2

Thank you!

I think the sentence is what made me confused... (the swedish one)

 

[survivor]

So I'm doing some questions practicing for a linear algebra test, but this question is kinda stumping me. I guess I'm not exactly sure if I'm doing the right thing.

Let X = {a, b, c, d} with f: X -> X

The subset { (x, f(x)) in X x X } defines a relation on X. For which f below it defines an equivalence relation ? If f defines an equivalence relation, find the respective partition of X.

Ok for example one of the f is defined such as this

f(c) = d
f(d) = a
f(a) = b
f(b) = c

Ok so would my relation end up being like { (c,d), (d,a), (a,b), (b,c) }?

I'm just being unsure since for all the 5 different examples of f I keep not getting an equivalence relation so I just want to make sure I'm at least getting the right relations.

 

[TimesEunuch]

So I'm doing some questions practicing for a linear algebra test, but this question is kinda stumping me. I guess I'm not exactly sure if I'm doing the right thing.

Let X = {a, b, c, d} with f: X -> X

The subset { (x, f(x)) in X x X } defines a relation on X. For which f below it defines an equivalence relation ? If f defines an equivalence relation, find the respective partition of X.

Ok for example one of the f is defined such as this

f(c) = d
f(d) = a
f(a) = b
f(b) = c

Ok so would my relation end up being like { (c,d), (d,a), (a,b), (b,c) }?

I'm just being unsure since for all the 5 different examples of f I keep not getting an equivalence relation so I just want to make sure I'm at least getting the right relations.

Yeah, that's correct. Recall though that an equivalence relation has to contain (x, x) for any x in X (i.e. we must have x~x for all x). So for { (x, f(x)) | x in X } to be an equivalence relation, f(x)=x always. In other words, the only f that gives an equivalence relation is the identity function (and the corresponding partition consists of singleton sets).