The Math Help Thread

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Oct 30, 2007
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how can {2x} represent the distance from x to the nearest integer?
It doesn't. It represents the distance from 2x to the nearest integer. It would be the same zigzaggy function, with a higher amplitude; i.e. each of the peaks wouldn't be at .5 but at 1.
 
Oct 30, 2007
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-COOLIO- said:
ooooOOOOOOOOOOOOOOOOOOOOOHHHH
Yeah, what you are studying has to do with step functions, which can get pretty neat. You're problem set probably ends up describing several types but maybe without the names (the greatest integer function, the least integer function, the fraction-part function, and most importantly the unit step function or Heaviside function are common ones).
 

-COOLIO-

The Everyman
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The Crimson Blur said:
Yeah, what you are studying has to do with step functions, which can get pretty neat. You're problem set probably ends up describing several types but maybe without the names (the greatest integer function, the least integer function, the fraction-part function, and most importantly the unit step function or Heaviside function are common ones).
yeah but now that i get what the question is saying ill bitch slap them.
 

Earthstrike

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Umm, I think the comments about g(x) may be wrong.

If {x} is the distance to the nearest integer then {2x} would be the distance to the nearest integer after x is multiplied by 2.

e.g. x = .25 | 2x = .5 | {2x} = .5
e.g. x = .10 | 2x = .20| {2x} = .20
e.g. x = .70 | 2x = 1.4| {2x} = .40

In other words the amplitude wouldn't increase. Instead what you'd get would be a compression of f (x) along the x axis. (each "mountain" is 0.5 units wide.)

Again this is just the impression I get. I don't think {2x} is well defined in your problem. I'm assuming we're supposed to infer that the {} brackets imply the operation of taking distance from the nearest integer.
 
Oct 30, 2007
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Earthstrike said:
Umm, I think the comments about g(x) may be wrong.

If {x} is the distance to the nearest integer then {2x} would be the distance to the nearest integer after x is multiplied by 2.

e.g. x = .25 | 2x = .5 | {2x} = .5
e.g. x = .10 | 2x = .20| {2x} = .20
e.g. x = .70 | 2x = 1.4| {2x} = .40

In other words the amplitude wouldn't increase. Instead what you'd get would be a compression of f (x) along the x axis. (each "mountain" is 0.5 units wide.)

Again this is just the impression I get. I don't think {2x} is well defined in your problem. I'm assuming we're supposed to infer that the {} brackets imply the operation of taking distance from the nearest integer.
Yea, I am way wrong lol. All it does is make the period = .5. Every max happens at x = (+/-).25*n and every min happens at (+/-).5*n where n is any integer.

Should have looked at it more closely!
 

-COOLIO-

The Everyman
Jun 9, 2007
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Earthstrike said:
Umm, I think the comments about g(x) may be wrong.

If {x} is the distance to the nearest integer then {2x} would be the distance to the nearest integer after x is multiplied by 2.

e.g. x = .25 | 2x = .5 | {2x} = .5
e.g. x = .10 | 2x = .20| {2x} = .20
e.g. x = .70 | 2x = 1.4| {2x} = .40

In other words the amplitude wouldn't increase. Instead what you'd get would be a compression of f (x) along the x axis. (each "mountain" is 0.5 units wide.)

Again this is just the impression I get. I don't think {2x} is well defined in your problem. I'm assuming we're supposed to infer that the {} brackets imply the operation of taking distance from the nearest integer.
i actually took his meaning, to mean your meaning.

so ya im not touching the aplitude, just compressing it horizontally.

thanks all.
 
Aug 27, 2006
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-COOLIO- said:
if im making a math thread i should definitely put this link somewhere:

http://www.khanacademy.org/
Oh shit these videos are awesome. I already passed all of my math courses for my program, but on some things I scraped by and I would like to have a better understanding and these videos are perfect. Great stuff.
 

roosters93

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Can someone help me with this question? I've done something like it before, but for some reason just can't figure out what way to go about it.

Question:

The line 6x-ky-4=0 passes through the point (3,2). Find the value of k.
 

Roude Leiw

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roosters93 said:
Can someone help me with this question? I've done something like it before, but for some reason just can't figure out what way to go about it.

Question:

The line 6x-ky-4=0 passes through the point (3,2). Find the value of k.
put in x=3 and y=2, and solve for k?
 

Ydahs

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Any here ever understood vectors? I have a Maths exam in a month and a bit (doing IB) and it's the only topic which I'm really struggling with.

I'm not really asking for help, just wondering if anyone else struggled with it.

The khanacademy seems awesome, though it doesn't cover vectors!
 
Apr 13, 2006
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Ydahs said:
Any here ever understood vectors? I have a Maths exam in a month and a bit (doing IB) and it's the only topic which I'm really struggling with.

I'm not really asking for help, just wondering if anyone else struggled with it.

The khanacademy seems awesome, though it doesn't cover vectors!
yes
 

giga

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Oct 28, 2006
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-COOLIO- said:
if im making a math thread i should definitely put this link somewhere:

http://www.khanacademy.org/
Why didn't I have this when I was taking underclassmen courses. Fuck.

Sal received his MBA from Harvard Business School where he was president of the student body. He also holds a Masters in electrical engineering and computer science, a BS in electrical engineering and computer science, and a BS in mathematics from MIT where he was president of the the Class of 1998. While at MIT, Sal was the recipient of the Eloranta Fellowship which he used to develop web-based math software for children with ADHD. He was also an MCAT instructor for the Princeton Review and volunteered teaching gifted 4th and 7th graders at the Devotion School in Brookline, MA.
 

Evolved1

make sure the pudding isn't too soggy but that just ruins everything
Sep 24, 2006
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far left of center
revolverjgw said:
I really needed something like Kahnacademy. I'm back to school after not being in a math course for 7 years, not concerning myself with it since, and not remembering ANYTHING. I needed a refresher, and more.
Download his free adaptive math program... it is fucking amazing. I cannot recommend it enough. The way it charts your progress and constantly gives you new stuff... and it's all tied into his video demos in case you get hung up.

It has helped a lot. First for review... but I've been going through some of the higher lvl stuff just for my own curiosity and now I'm pretty confident to tackle some of that at university, whereas before I was a little freaked out.
 

-COOLIO-

The Everyman
Jun 9, 2007
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im suppose to solve this inequality for x:

(x+2)(2x-3) < 1

so far ive turned it into

2x^2 + x - 7 < 0

but im not suppose to use a calculator to solve this quetion so i figure id just write

neg infinity < x < 7/2x+1

but when you solve an inequality you're only suppose to have x in one spot right?

but 2x^2 + x - 7 < 0 is unholy. there's no way x is an easy to spot number without a calculator right?
 

Fjolle

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Sep 13, 2006
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-COOLIO- said:
im suppose to solve this inequality for x:

(x+2)(2x-3) < 1

so far ive turned it into

2x^2 + x - 7 < 0

but im not suppose to use a calculator to solve this quetion so i figure id just write

neg infinity < x < 7/2x+1

but when you solve an inequality you're only suppose to have x in one spot right?

but 2x^2 + x - 7 < 0 is unholy. there's no way x is an easy to spot number without a calculator right?
Nah, i'm tired but you cant solve it like that without calculator. It will be 1/4(-1+/- sqrt(57)).
 

-COOLIO-

The Everyman
Jun 9, 2007
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I FINALLY ACED A 137 QUIZ

ahahahahahahahaaaa


AHAHAHAHAHAHHAHAHAHAAAAAAAAAAAAHAHAHAHAHAH


AHHHHHHHHHHHHHAHAHAAHAHAHHAAHHAHAHAHAHAHAAHA
AHAHHAHAHAHAHAHAHAHAHAAHHAHAHAHAHAHAHAHAHAHA
 

TimesEunuch

Member
Aug 13, 2007
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-COOLIO- said:
im suppose to solve this inequality for x:

(x+2)(2x-3) < 1

so far ive turned it into

2x^2 + x - 7 < 0

but im not suppose to use a calculator to solve this quetion so i figure id just write

neg infinity < x < 7/2x+1

but when you solve an inequality you're only suppose to have x in one spot right?

but 2x^2 + x - 7 < 0 is unholy. there's no way x is an easy to spot number without a calculator right?
In general, you can find the roots of the quadratic equation ax^2+bx+c (i.e. the values where it equals zero) from x=( -b +/- sqrt(b^2-4ac) ) / ( 2a ). No calculator needed. Once you have the roots, and factorise the quadratic, it should be easy to see what values of x make the inequality hold (is this clear, or could you use more explanation?)
 

Guled

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Nov 2, 2007
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I really really hate calculus, now I got that out of the way can anyone here help me derive this?
f(t) = 4 sqrt(t) - (4/sqrt(t))
 

Dipper145

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Dec 17, 2008
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Guled said:
I really really hate calculus, now I got that out of the way can anyone here help me derive this?
f(t) = 4 sqrt(t) - (4/sqrt(t))
should be
f'(t)=2(t)^(-1/2) + 2t^(-3/2)
if I did the math right and understood the question.

with square roots turn the square roots into exponents, and then go from there.
sqrt(t) = t^1/2 /sqrt(t) = (t)^-1/2
 

RSTEIN

Comics, serious business!
Mar 22, 2007
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giga said:
Sal received his MBA from Harvard Business School where he was president of the student body. He also holds a Masters in electrical engineering and computer science, a BS in electrical engineering and computer science, and a BS in mathematics from MIT where he was president of the the Class of 1998. While at MIT, Sal was the recipient of the Eloranta Fellowship which he used to develop web-based math software for children with ADHD. He was also an MCAT instructor for the Princeton Review and volunteered teaching gifted 4th and 7th graders at the Devotion School in Brookline, MA.
pfft. that's nothing. This morning I took a piss and brushed my teeth at the same time.
 

GaimeGuy

Volunteer Deputy Campaign Director, Obama for America '16
Jun 18, 2004
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*goes back to trying to figure out how to decrypt an RSA ciphertext given only the following information:

#1: My encryption key
#2 My decryption key
#3: Someone else's encryption key
#4: an encrypted message which was encrypted using #3
#5: The modulus, n, which is common to my encryption scheme and the other person's scheme as well.

Also, I'm not allowed to factor the modulus, and I don't know the decryption keyfor this messge :(

*
 

Guled

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Nov 2, 2007
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3 questions:

- find y' and y'' for y = ln(sec(9x) + tan(9x))

- find y' and y'' for y = e^(4ex)

- Find the equations of the tangent line and normal line to the given curve at (9, 3 / 13)
y = sqrt(x) / (x + 4)

a thousand blessings to those who can help me out
 

hemtae

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Nov 4, 2007
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Guled said:
3 questions:

- find y' and y'' for y = ln(sec(9x) + tan(9x))

- find y' and y'' for y = e4ex

- Find the equations of the tangent line and normal line to the given curve at (9, 3 / 13)
y = sqrt(x) / (x + 4)

a thousand blessings to those who can help me out
well for the second one, the way you wrote it, y-prime would just be e4e and y-double prime would be 0.
 

Guled

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Nov 2, 2007
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hemtae said:
well for the second one, the way you wrote it, y-prime would just be e4e and y-double prime would be 0.
my bad, i fixed it. its meant to be y = e^(4e^(x))
i figured out y' which is (4e^x) (e^(4e^(x)) ), so all i need is y''
 

Ri'Orius

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Aug 2, 2008
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Guled said:
3 questions:

- find y' and y'' for y = ln(sec(9x) + tan(9x))

- find y' and y'' for y = e^(4ex)

- Find the equations of the tangent line and normal line to the given curve at (9, 3 / 13)
y = sqrt(x) / (x + 4)

a thousand blessings to those who can help me out


No, seriously. It's the chain rule, and it's what you need.

Presumably you know (or can look up) the derivatives of the component functions, right? e^x -> e^x, ln(x) -> 1/x, etc.?
 

hemtae

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Nov 4, 2007
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Ri'Orius said:
http://www.hipsterwave.com/wp-content/uploads/2009/03/yo-dawg-derive-while-you-derive-700.jpg

No, seriously. It's the chain rule, and it's what you need.

Presumably you know (or can look up) the derivatives of the component functions, right? e^x -> e^x, ln(x) -> 1/x, etc.?
He'll also need the Quotient rule for the third one
 
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