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[boviscopophobic]

I'm a little hazy on subtracting functions. Let's see.

If (f-g)x=(h)x then g(x)=(h+f)x? That's the only conclusion I can draw, unless I'm missing something. But I doubt that's what you're suggesting.

Well, that's part of it -- that tells us that if you have h(x), then you can get g(x), which is the function you actually wanted.

What can you say about the value of h(x) at the given x-values, namely h(2), h(3), and h(4)? Can you write down, for example, a polynomial for h(x) that does the right thing?

 

[EightBitNate]

Well, that's part of it -- that tells us that if you have h(x), then you can get g(x), which is the function you actually wanted.

What can you say about the value of h(x) at the given x-values, namely h(2), h(3), and h(4)? Can you write down, for example, a polynomial for h(x) that does the right thing?

lol thanks for the help but I'm completely lost at what you're getting at.

At h(2), h(3), and h(4), what I do know? That if you subtract f(2), f(3), and f(4) you'll get a g-function? But what good to me is the h-function if I don't know what it looks like?

Not sure what time it is by you, but if it's late thanks for staying up.

 

[runningjoke]

lol thanks for the help but I'm completely lost at what you're getting at.

At h(2), h(3), and h(4), what I do know? That if you subtract f(2), f(3), and f(4) you'll get a g-function? But what good to me is the h-function if I don't know what it looks like?

Not sure what time it is by you, but if it's late thanks for staying up.

I think he means that you know h(x) has to be 0 for x=2, 3 and 4?

 

[EightBitNate]

I think he means that you know h(x) has to be 0 for x=2, 3 and 4?

Ok that makes sense. So if we're subtracting 2-h and the resulting answer is 0, then h=0. I gotcha.

So in polynomial form:

H(x)=(x-2)(x-3)(x-4)

Yea?

4x-5-(x-2)(x-3)(x-4)
4x-5-(x^2-5x+6)(x-4)
4x-5-(x^3-5x^2+6x-4x^2+20x-24)
4x-5-(x^3-9x^2+26x-24)
4x-5-x^3+9x^2-26x+24
-x^3+9x^2-22x+19

That works out right?

 

[boviscopophobic]

Ok that makes sense. So if we're subtracting 2-h and the resulting answer is 0, then h=0. I gotcha.

So in polynomial form:

H(x)=(x-2)(x-3)(x-4)

Yea?

That's one possible choice of h(x), yes. That yields

g(x) = 4x-5 + (x-2)(x-3)(x-4) = x^3 - 9x^2 + 30x - 29

EDIT: Your computation above works too, it's just differing sign conventions.

 

[EightBitNate]

That's one possible choice of h(x), yes. That yields

g(x) = 4x-5 + (x-2)(x-3)(x-4) = x^3 - 9x^2 + 30x - 29

Oh gee man thanks and you too, running joke

 

[justjohn]

Never mind. Got it. Thanks.

 

[hitsugi]

What do you guys think of the Schaum Outline Books? I've been out of school for a (long) while, and I need to get myself up to College Algebra level asap for upcoming classes, which (I assume) means I need to have Elementary Algebra down at the very least.

My understanding is that the only way to really learn math is to grind it out, so I thought an actual book may serve me better than Khan Academy would.

 

[Irobot82]

IS 1+or-1sqrt(2) as a solution set

1+1sqrt(2),1-1sqrt(2)

or 2(sqrt)2,sqrt(2)

 

[Anustart]

When solving for p, why is 1/f = (1/p) + 1/q not equal to P = (FQ)/Q-F?

Nevermind, it is. Dumb ass webassign didn't take it because I capitalized the variables.

 

[Dropped_Off]

IS 1+or-1sqrt(2) as a solution set

1+1sqrt(2),1-1sqrt(2)

or 2(sqrt)2,sqrt(2)

It's the first one: (1 + sqrt(2), 1 - sqrt(2) )

 

[Vigilant Walrus]

I am slowly progressing into Algebra on Khan Academy. I am not getting much out of my math at my community college, but Khan is helping me. I just watched 4-5 videos about what abstraction is, and this is amazing.

This is what I have been looking for in all my years of math. All my questions of WHY? HOW? YOU JUST SAID? BUT BEFORE YOU SAID - ?

All these questions that made my mind shut off. Math teachers always be lying when they say math is logical and then they start talking equations like it's wizard speak in ancient hebrew.
These guys videos seems to tell me everything. It's really difficult but I hate math a lot less in the past 2 weeks than Ihave done in my entire life. It's embarrassing that I have to sit and do 8th grade level stuff, but I guess this is what I need to re-learn Math right.

To me, all you math nerds are my heroes. Like I discovered in my philosophy class, here is a couse were you are not supposed to progress by not being able to think. Must other courses have me just follow the instructions like I did at my office job. Read the manual, do the shit. Monkey see, monkey do. No! Here I need to think. I really am using my brain in areas I don't do otherwise.
Why could all my math teachers, most of whom were great people, not help me? What is so wrong with me that made me except for all these years? I've quitted college-level math exams 3 times over the last 4 years. It's hard to explain to you how much math has held my life back.
Though I am still only on step one. I have so many things needs I need to re-learn and learn. Just being able to effectively using a calculator, sin, cos, tan.. equations, formulas. Formulas most of all. geometry. measurements.


I am just happy that Khan Academy is free. They haven't tried to fork anything out of me. When I almost gave up, the sent me this mail about if I could figure out the secret of Pythagoras. It really intrigued me and I got re-invested again.

Yesterday I learned about the origin of why x is the unknown. And I learned about Galileo and Decadent-es. Pretty cool guys. I heard about them before in math classes, but was never able to concentrate.

 

[RELAYER]

Speaking of Algebra, how does

g * sqrt(2h/g)

=

sqrt(2hg)

?

 

[Kazerei]

g * sqrt(2h/g)
= sqrt(g^2) * sqrt(2h/g)
= sqrt (g^2 * 2h/g)
= sqrt(2hg)

 

[RELAYER]

Thanks never learned that trick but it seems obvious now.


I missed a lecture last week and I'm trying to do my calculus homework but I'm not sure of the method. It's about finding the sum of infinite series.

The first one is
[series from n=1 to infinity] of n/(n+12)

I just evaluated the limit of the sequence like you would normally do, multiply the fraction by 1/n.

If you do that you get 1/(1+12/n) and 12/n goes to zero as n approaches infinity.

So then for the sum of the series I put in my answer as 1 but it's wrong.

What am I doing wrong?

 

[Calvero]

Slightly OT, but did any of you ever run a Math Club in High School?

 

[fireside]

Thanks never learned that trick but it seems obvious now.


I missed a lecture last week and I'm trying to do my calculus homework but I'm not sure of the method. It's about finding the sum of infinite series.

The first one is
[series from n=1 to infinity] of n/(n+12)

I just evaluated the limit of the sequence like you would normally do, multiply the fraction by 1/n.

If you do that you get 1/(1+12/n) and 12/n goes to zero as n approaches infinity.

So then for the sum of the series I put in my answer as 1 but it's wrong.

What am I doing wrong?

that series diverges by the limit/term test. since it diverges it doesn't have a sum

you guys probably went over convergence/divergence tests in the class you missed

 

[boviscopophobic]

Thanks never learned that trick but it seems obvious now.


I missed a lecture last week and I'm trying to do my calculus homework but I'm not sure of the method. It's about finding the sum of infinite series.

The first one is
[series from n=1 to infinity] of n/(n+12)

I just evaluated the limit of the sequence like you would normally do, multiply the fraction by 1/n.

If you do that you get 1/(1+12/n) and 12/n goes to zero as n approaches infinity.

So then for the sum of the series I put in my answer as 1 but it's wrong.

What am I doing wrong?

An infinite series is an infinite summation -- roughly speaking, it's what you would get if you added all the terms up. For instance, babby's first geometric series might look like

1 + (1/2) + (1/4) + (1/8) + ... + (1/2^n) + ... = 2

Notice that the series converges to 2, but the limit of the sequence of individual summands 1/2^n is 0. So those are different concepts, although they are in fact related.

How are they related? Well, not every infinite series converges to a real number. For example,

1 + 1 + 1 + ...

runs off to infinity, so it doesn't converge to any finite number. Even worse, something like

1 - 1 + 1 - 1 + ... + (-1)^(n-1) + ...

never settles down at all. So when looking at an infinite series, we have to figure out if it converges or not. There are a number of standard tests that can be applied to help determine convergence. One of the easiest such tests is that if the *sequence* of terms doesn't converge to 0, then the *series* can't converge. (Can you figure out why this is the case?) So in your example, you found that the sequence of terms converges to 1, therefore it must be the case that the series is divergent.

The true relationship between sequences and series is that the technical definition of the sum of a (convergent) series is the limit of the sequence of partial sums. That is, if we have a series

a_1 + a_2 + ... + a_n + ...

and we define the partial sums

s_j = a_1 + ... + a_j (for j = 1, 2, 3, ...)

then the limit of the sequence {s_j} is *defined* to be the sum of the series.

 

[octopiggy]

Is there a simple formula for creating an s-curve output on a graph?

I've just started playing around making my own game and I'm working on acceleration where I want speed to build up slowly, then quicker and then slowdown to a stop as top speed is approached.

 

[Water]

Is there a simple formula for creating an s-curve output on a graph?

I've just started playing around making my own game and I'm working on acceleration where I want speed to build up slowly, then quicker and then slowdown to a stop as top speed is approached.

Smoothstep and cubed smoothstep are the usual go-to options for easy S-curves. I guess you could also take a weighted average of one of them and linear curve to get finer adjustment of the "strength" of the curve.
http://en.wikipedia.org/wiki/Smoothstep
http://sol.gfxile.net/interpolation/index.html

 

[octopiggy]

Smoothstep and cubed smoothstep are the usual go-to options for easy S-curves. I guess you could also take a weighted average of one of them and linear curve to get finer adjustment of the "strength" of the curve.
http://en.wikipedia.org/wiki/Smoothstep
http://sol.gfxile.net/interpolation/index.html

This seems to be what I'm looking for. Thanks for the help!

 

[CO_Andy]

Anybody here know anything about discrete math?

Let p represent the statement "Students are males" and let q represent the statement "Teachers are males."

~(p v ~q)

 

[Therion]

Anybody here know anything about discrete math?

~(p v ~q) = ~p ^ q

So that would read: "Students are not males and teachers are males."

 

[Two Words]

Discrete math is fun. It has helped me understand how to not make crappy loops and if statements.

 

[Granadier]

Discrete math is fun. It has helped me understand how to not make crappy loops and if statements.

You're crazy

 

[MisterLuffy]

Hello. I need help with this problem for Discrete Math homework. I have to prove that:
(A-B)U(C-B) = (AUC)-B

Showing Left side subsets to Right side, and vice versa. Can someone help me?

 

[TangerineCloud]

Help with derivatives, pls.

I need to find the equation of the tangent line to this curve:

Fourth root of X (or X to the one-fourth power) at position (1,1)

The answer is at that back of the book, but I'm not getting it right. :/

 

[Incandenza]

Help with derivatives, pls.

I need to find the equation of the tangent line to this curve:

Fourth root of X (or X to the one-fourth power) at position (1,1)

The answer is at that back of the book, but I'm not getting it right. :/

Your equation is f(x)=x^(1/4).

We take the derivative by sticking the exponent up in front as a coefficient and subtracting 1 from the exponent. So f'(x)=(1/4)x^(-3/4).

f'(1)=(1/4). So the slope of the tangent line is (1/4).

Our point is (1,1) and our slope is (1/4). Plug numbers into point-slope form, (y-y1)=m(x-x1).
(y-1)=(1/4)(x-1)
y = (1/4)x + (3/4).

 

[Therion]

Hello. I need help with this problem for Discrete Math homework. I have to prove that:
(A-B)U(C-B) = (AUC)-B

Showing Left side subsets to Right side, and vice versa. Can someone help me?

Here's one direction. If you get this you should be able to do the other way yourself:

x in (A-B)U(C-B) => x in (A-B) or x in (C-B)
=> (x in A and x not in B) or (x in C and not in B)
=> x not in B and (x in A or x in B)
=> x in (AUC)-B.

 

[TangerineCloud]

Your equation is f(x)=x^(1/4).

We take the derivative by sticking the exponent up in front as a coefficient and subtracting 1 from the exponent. So

f'(1)=(1/4). So the slope of the tangent line is (1/4).

Our point is (1,1) and our slope is (1/4). Plug numbers into point-slope form, (y-y1)=m(x-x1).
(y-1)=(1/4)(x-1)
y = (1/4)x + (3/4).

Hm, I understand everything except the bolded. Why does plugging 1 into f'(x)=(1/4)x^(-3/4) give us the slope?

 

[Orbis Tabula]

Hm, I understand everything except the bolded. Why does plugging 1 into f'(x)=(1/4)x^(-3/4) give us the slope?

It's from the definition of the derivative, f'(x). The derivative of a function f will tell you the slope of f at any point you choose to plug in. That's the reason we study it. So if you want to find the slope of an unusual curve at a point, you can take the derivative of the curve, plug in the x value into this derivative, f'(x), and you will get the slope at x in the original function as an output.

 

[TangerineCloud]

It's from the definition of the derivative, f'(x). The derivative of a function f will tell you the slope of f at any point you choose to plug in. That's the reason we study it. So if you want to find the slope of an unusual curve at a point, you can take the derivative of the curve, plug in the x value into this derivative, f'(x), and you will get the slope at x in the original function as an output.

Gah, I can't believe I forgot this. I ended up completely over-thinking this problem, but I understand it now. Thank you both so much.

The derivative is the rate of change at any given point in the function. If you put a number in the derivative's formula, it will give you the slope of the tangent line at that point in the function. That's because tangent lines show the rate of change at a given point in the original function.

I hope that's clear.

edit: nevermind, they got to it first

Ah, that's actually a helpful explanation. I've been focusing so much on getting the right answer at the expense of actually understanding why I'm doing the various operations that lead up to it. I'll try keep this in mind. Thanks!

 

[ItWasMeantToBe19]

Hm, I understand everything except the bolded. Why does plugging 1 into f'(x)=(1/4)x^(-3/4) give us the slope?

The derivative is the rate of change at any given point in the function. If you put a number in the derivative's formula, it will give you the slope of the tangent line at that point in the function. That's because tangent lines show the rate of change at a given point in the original function.

I hope that's clear.

edit: nevermind, they got to it first

 

[MisterLuffy]

Here's one direction. If you get this you should be able to do the other way yourself:

x in (A-B)U(C-B) => x in (A-B) or x in (C-B)
=> (x in A and x not in B) or (x in C and not in B)
=> x not in B and (x in A or x in B)
=> x in (AUC)-B.

Thanks for your help.

 

[Anustart]

I have no clue how to divide polynomials.

How the fuck is (x^2-x-12)/(x+1) = x-2 for the oblique asymptote? How is this done? Sure, x+1 goes in x times to get x^2 + x, which would leave a -12, how the FUCK does x+1 go into that 2 times?

 

[alatif113]

I have no clue how to divide polynomials.

How the fuck is (x^2-x-12)/(x+1) = x-2 for the oblique asymptote? How is this done? Sure, x+1 goes in x times to get x^2 + x, which would leave a -12, how the FUCK does x+1 go into that 2 times?

x+1 | x^2-x-12 : x goes into x^2 x times. Multiply x+1 by x
x+1 | x^2+x : subtract
---------------
x+1 | -2x-12 : x goes into -2x, -2 times. Multiply x+1 by -2
x+1 | -2x-2 : subtract
---------------
x+1 | -10 : x goes into -10, 0 times. Remainder is -10

Solution: x-2 remainder -10 or x-2-10/(x+1)

 

[Anustart]

x+1 | x^2-x-12 : x goes into x^2 x times. Multiply x+1 by x
x+1 | x^2+x : subtract
---------------
x+1 | -2x-12 : x goes into -2x, -2 times. Multiply x+1 by -2
x+1 | -2x-2 : subtract
---------------
x+1 | -10 : x goes into -10, 0 times. Remainder is -10

Solution: x-2 remainder -10 or x-2-10/(x+1)

Yeah my problem is the +x and -x that combine to 2x instead of what I assumed would be 0.

 

[MisterLuffy]

I need help with this problem: prove algebraically that ( A ∪ C ) - ( B - A ) = A ∪ ( C - B )

Here's what I have so far: ( A ∪ C ) - ( B - A ) = ( A ∪ C ) - ( B n A^c) = ( A ∪ C ) n [B n A^c]^c = ?
I don't know if I did it right or if I'm in the right direction.

 

[Therion]

I need help with this problem: prove algebraically that ( A ∪ C ) - ( B - A ) = A ∪ ( C - B )

Here's what I have so far: ( A ∪ C ) - ( B - A ) = ( A ∪ C ) - ( B n A^c) = ( A ∪ C ) n [B n A^c]^c = ?
I don't know if I did it right or if I'm in the right direction.

You're headed in the right direction. For the next step you probably want De Morgan.

 

[MisterLuffy]

You're headed in the right direction. For the next step you probably want De Morgan.

( A ∪ C ) - ( B - A ) = ( A ∪ C ) - ( B n A^c) = ( A ∪ C ) n [B n A^c]^c = (AUC) n (B^cUA) = ?

It would make sense if if was (AUC) n (AUB^c) = AU(CnB^c) = AU(C-B)

What can I do about (AUC) n (B^cUA)?

 

[upandaway]

Complex numbers question: I just wanna know if I'm anywhere near right with this
http://i.imgur.com/U2zAHau.jpg
(It says "find all solutions of ___")

I'm sorta doing it blind without learning from anyone so I have no clue.

I need help with this problem: prove algebraically that ( A ∪ C ) - ( B - A ) = A ∪ ( C - B )

I did it for practice so here's my result (dunno if it helps)
http://i.imgur.com/Kxn6SlS.jpg

 

[boviscopophobic]

Complex numbers question: I just wanna know if I'm anywhere near right with this
http://i.imgur.com/U2zAHau.jpg
(It says "find all solutions of ___")

I'm sorta doing it blind without learning from anyone so I have no clue.

The reasoning looks pretty good overall, but you have a sign error in Im(z) which propagates through the entire calculation. Also, the treatment of 2 pi k at the end isn't quite correct. gamma (the argument of z^4) is what should be 2 pi periodic.

 

[Therion]

( A ∪ C ) - ( B - A ) = ( A ∪ C ) - ( B n A^c) = ( A ∪ C ) n [B n A^c]^c = (AUC) n (B^cUA) = ?

It would make sense if if was (AUC) n (AUB^c) = AU(CnB^c) = AU(C-B)

What can I do about (AUC) n (B^cUA)?

It looks to me like you have the complete answer already. I'm not sure exactly what you're asking about unless it's B^cUA = AUB^c. Unions and intersections both have a commutative property that allow you to switch the order whenever it is convenient.

 

[MisterLuffy]

It looks to me like you have the complete answer already. I'm not sure exactly what you're asking about unless it's B^cUA = AUB^c. Unions and intersections both have a commutative property that allow you to switch the order whenever it is convenient.

Oh, I forgot you can do that , then cool I finished a problem. Thanks for answering my questions.

I did it for practice so here's my result (dunno if it helps)
http://i.imgur.com/Kxn6SlS.jpg

I've been wondering what the front slash means. Is it (B/A) = (B-A)? I'm not sure if it's right since I have to analyze the notation with my answers to see if its correct. Thanks for trying, I'm saving that image just in case.

 

[Casanova]

Nevermind, Fuck it.

 

[MisterLuffy]

I need help on the other problem: proof algebraically (A-B)U(B-A) = (AUB)-(AnB)

I know the first step is doing the differences laws.

(A-B)U(B-A) = (AnB^c)U(BnA^c) = ?

I really don't know what to do right after that. I was wondering if someone can give me hints or help me on knowing the next step.

 

[fannybandit]

Thanks never learned that trick but it seems obvious now.


I missed a lecture last week and I'm trying to do my calculus homework but I'm not sure of the method. It's about finding the sum of infinite series.

The first one is
[series from n=1 to infinity] of n/(n+12)

I just evaluated the limit of the sequence like you would normally do, multiply the fraction by 1/n.

If you do that you get 1/(1+12/n) and 12/n goes to zero as n approaches infinity.

So then for the sum of the series I put in my answer as 1 but it's wrong.

What am I doing wrong?

what you did was not finding the sum, you were finding the limit of the funcgtion n/(n+12) as n approaches infinity which in general is not equal to the sum. to find the sum, you have to find the limit of the partial sums. so you have to find the limit (if it exists) as n goes to infinity of the function: sum of k/(k+12) where 1 <= k <= n

 

[upandaway]

The reasoning looks pretty good overall, but you have a sign error in Im(z) which propagates through the entire calculation. Also, the treatment of 2 pi k at the end isn't quite correct. gamma (the argument of z^4) is what should be 2 pi periodic.

Gotcha, thanks. I think I got it now.

I've been wondering what the front slash means. Is it (B/A) = (B-A)? I'm not sure if it's right since I have to analyze the notation with my answers to see if its correct. Thanks for trying, I'm saving that image just in case.

Yeah everything in B that isn't in A.

I need help on the other problem: proof algebraically (A-B)U(B-A) = (AUB)-(AnB)

I know the first step is doing the differences laws.

(A-B)U(B-A) = (AnB^c)U(BnA^c) = ?

I really don't know what to do right after that. I was wondering if someone can give me hints or help me on knowing the next step.

I solved it too (got a test Thursday so these are nice practice) but I only know how to do it using a variable, I don't know what your 3rd sentence means, so I can't really help besides uploading my answer again

 

[Therion]

I need help on the other problem: proof algebraically (A-B)U(B-A) = (AUB)-(AnB)

I know the first step is doing the differences laws.

(A-B)U(B-A) = (AnB^c)U(BnA^c) = ?

I really don't know what to do right after that. I was wondering if someone can give me hints or help me on knowing the next step.

I didn't work it out entirely, but I think that if you completely distribute what you have now, you should be able to reassemble the pieces to get your answer (after taking care of terms like AUA^c, etc).

Edit: Sorry, I was on my way out when I posted before and didn't notice I was working from right to left whereas you were were working left to right. If you're going left to right the post below mine looks good.

 

[concordia]

(A-B)U(B-A) = (AnB^c)U(BnA^c) =(AnB^c)U(BnA^c) U (AnA^c)U(BnB^c)=An(A^cUB^c)UBn(A^cUB^c)=AUB n (A^c U B^c )=AUB n (A n B)^c