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The Math Help Thread

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J

Justin Timberlake

Member
how can {2x} represent the distance from x to the nearest integer?
Click to expand...

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.
 
J

Justin Timberlake

Member
-COOLIO- said:
ooooOOOOOOOOOOOOOOOOOOOOOHHHH
Click to expand...

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-

-COOLIO-

The Everyman
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).
Click to expand...
yeah but now that i get what the question is saying ill bitch slap them.
 
E

Earthstrike

Member
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.
 
J

Justin Timberlake

Member
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.
Click to expand...

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-

-COOLIO-

The Everyman
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.
Click to expand...

i actually took his meaning, to mean your meaning.

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

thanks all.
 
R

roosters93

Member
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.
 
R

Roude Leiw

Member
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.
Click to expand...

put in x=3 and y=2, and solve for k?
 
Y

Ydahs

Member
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!
 
Tornado Condor

Tornado Condor

Member
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!
Click to expand...

yes
 
G

giga

Member
-COOLIO- said:
if im making a math thread i should definitely put this link somewhere:

http://www.khanacademy.org/
Click to expand...
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.
 
R

revolverjgw

Member
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.
 
Evolved1

Evolved1

make sure the pudding isn't too soggy but that just ruins everything
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.
Click to expand...

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-

-COOLIO-

The Everyman
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

Fjolle

Member
-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?
Click to expand...
Nah, i'm tired but you cant solve it like that without calculator. It will be 1/4(-1+/- sqrt(57)).
 
-COOLIO-

-COOLIO-

The Everyman
I FINALLY ACED A 137 QUIZ

ahahahahahahahaaaa


AHAHAHAHAHAHHAHAHAHAAAAAAAAAAAAHAHAHAHAHAH


AHHHHHHHHHHHHHAHAHAAHAHAHHAAHHAHAHAHAHAHAAHA
AHAHHAHAHAHAHAHAHAHAHAAHHAHAHAHAHAHAHAHAHAHA
 
T

TimesEunuch

Member
-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?
Click to expand...
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?)
 
G

Guled

Member
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))
 
D

Dipper145

Member
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))
Click to expand...
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
 
R

RSTEIN

Comics, serious business!
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.
Click to expand...

pfft. that's nothing. This morning I took a piss and brushed my teeth at the same time.
 
G

GaimeGuy

Volunteer Deputy Campaign Director, Obama for America '16
*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 :(

*
 
G

Guled

Member
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

hemtae

Member
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
Click to expand...

well for the second one, the way you wrote it, y-prime would just be e4e and y-double prime would be 0.
 
G

Guled

Member
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.
Click to expand...
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''
 
R

Ri'Orius

Member
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
Click to expand...

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.?
 
hemtae

hemtae

Member
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.?
Click to expand...

He'll also need the Quotient rule for the third one
 
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