Support NeoGAF

[The One Who Knocks]

Didn't know we had a math thread! I need serious help with this,




How does e^x=e^-x e^2x? (I know x=x)

The rule they use is highlighted in the picture but it doesn't make sense to apply it here. I guess my question is how did they know to apply that rule here

First, your question e^{x} = e^{-x} * e^{2x} is by the laws of exponents highlighted there.

e{-x} * e^{2x} = e^{-x + 2x} (by law of exponents) = e^{x} (by -x+2x=x)


So one way to do that question (in quite a long manner, if you don't factor out immediately):

Start: You have (e^{x} - e^{-x}) / (e^x) (*)

First recall that for exponents: 1/(a^b) = a^(-b)
-> (*) can be written as (e^{-x})(e^{x} - e^{-x}) (**)

Second recall that multiplication is distributive
->(**) can be written as (e^{-x})(e^x) - (e^{-x})(e^{-x}) (***)

Third recall that for exponents (x^a)(x^b) = x^(a+b)
->(***) can be written as e^{-x+x} - e^{-x-x} (****)

Fourth by addition
->(****) can be written as e^{0} - e^{-2x} (5*)

Fifth by recalling that we define x^0 = 1 for all x in R
->(5*) can be written as 1 - e^{-2x} (6*)

Sixth by recalling that x^{-y} = 1/(x^{y})
->(6*) can be written as 1 - (1/(e^{2x}) (7*)

Seventh by taking a common denominator, and recalling that (x^a / x^a) = 1 (with x=/=0 if a=/=0)
->(7*) can be written as (e^{2x} / e^{2x}) - (1 / e^{2x}) (8*)

Eighth, by recalling (a/b)+(c/b) = (a+c)/b
->(8*) can be written as (e^{2x} -1) / (e^(2x))

Now putting it all together:

(e^{x} - e^{-x}) / (e^{x}) = (e^{2x} -1) / (e^(2x)) and done.

Alternatively, you can do it how it's done there.
It's quicker there since you're immediately factoring the e^{-x}, but you need to be able to notice that we are subtracting by an e^{-x}, and thus want to break the e^{x} term apart. As we want a -x here (to factor it out), we want to figure out what 'a' for, a real number, satisfies x= -x + ax (which is 2) and thus by using x=-x+2x is the desired way to 'split' e^x is e^x = e^{-x} * e^{2x}

Alternatively again, what you could also try and do to simplify it is you can factor out the e^{x} term to get it to cancel with the denominator.
Then, as e^{-x} = e^{x-2x} = (e^{x})(e^{-2x})
If you do this, then we get (1 - e^{-2x}) = (e^{2x} - 1)/(e^{2x}) as we wanted.

EDIT: And way too late here.

EDIT 2:

I have a question pertaining to math courses in general as a computer science major. My school requires a weirdly low amount of math courses to graduate in computer science. All that is required is elementary statistics, calc 1 and discrete mathematics. Would it be beneficial for me to take higher level calc classes, perhaps even proofs or linear algebra?

I spoke to my advisor and he just gave me some wishy washy "well it's not required but if you'd like to that's fine" type answer. I also really enjoy math and have loved my calc class and discrete math class and did really well in them so it's not something I dread doing or anything like that.

Any insight would be greatly appreciated!

It seems crazy to me that linear algebra, at least of some form, would not be required for a Computer Science degree. Even if it's not in an extremely rigorous manner, it's kind of odd (very much so, actually) that you'd do none. I guess if its not required it's perfectly fine not to do it, but I'd imagine it would be beneficial at least on some level (matrix manipulation and understanding how to find inverses at the very least). If you enjoy it, linear algebra is a very pervasive and wide-reaching domain. My recommendation (as I don't know how your university would teach it) is if that if you've not enrolled in a 'proofs' class, that you definitely become familiar with them on your own or take whatever module serves as the introduction to proofs in your university before you enrolled in any maths modules where they play a big role (depending on how your university tackles linear algebra, this could be that module).

 

[Allonym]

I have a question pertaining to math courses in general as a computer science major. My school requires a weirdly low amount of math courses to graduate in computer science. All that is required is elementary statistics, calc 1 and discrete mathematics. Would it be beneficial for me to take higher level calc classes, perhaps even proofs or linear algebra?

I spoke to my advisor and he just gave me some wishy washy "well it's not required but if you'd like to that's fine" type answer. I also really enjoy math and have loved my calc class and discrete math class and did really well in them so it's not something I dread doing or anything like that.

Any insight would be greatly appreciated!

Man that's some bullshit, Rutgers requires you to take up to and including Linear Algebra to graduate with a degree in Computer Science. The shit is crazy, I'm happy for you though, at least not everyone who wants to a degree in CS has to suffer.

 

[Bacon]

Man that's some bullshit, Rutgers requires you to take up to and including Linear Algebra to graduate with a degree in Computer Science. The shit is crazy, I'm happy for you though, at least not everyone who wants to a degree in CS has to suffer.

I'm just worried that I'm going to be at a disadvantage compared to people who graduate from a more rigorous program such as yours.

 

[Himynameischris]

First, your question e^{x} = e^{-x} * e^{2x} is by the laws of exponents highlighted there.

e{-x} * e^{2x} = e^{-x + 2x} (by law of exponents) = e^{x} (by -x+2x=x)


So one way to do that question (in quite a long manner, if you don't factor out immediately):

Start: You have (e^{x} - e^{-x}) / (e^x) (*)

First recall that for exponents: 1/(a^b) = a^(-b)
-> (*) can be written as (e^{-x})(e^{x} - e^{-x}) (**)

Second recall that multiplication is distributive
->(**) can be written as (e^{-x})(e^x) - (e^{-x})(e^{-x}) (***)

Third recall that for exponents (x^a)(x^b) = x^(a+b)
->(***) can be written as e^{-x+x} - e^{-x-x} (****)

Fourth by addition
->(****) can be written as e^{0} - e^{-2x} (5*)

Fifth by recalling that we define x^0 = 1 for all x in R
->(5*) can be written as 1 - e^{-2x} (6*)

Sixth by recalling that x^{-y} = 1/(x^{y})
->(6*) can be written as 1 - (1/(e^{2x}) (7*)

Seventh by taking a common denominator, and recalling that (x^a / x^a) = 1 (with x=/=0 if a=/=0)
->(7*) can be written as (e^{2x} / e^{2x}) - (1 / e^{2x}) (8*)

Eighth, by recalling (a/b)+(c/b) = (a+c)/b
->(8*) can be written as (e^{2x} -1) / (e^(2x))

Now putting it all together:

(e^{x} - e^{-x}) / (e^{x}) = (e^{2x} -1) / (e^(2x)) and done.

Alternatively, you can do it how it's done there.
It's quicker there since you're immediately factoring the e^{-x}, but you need to be able to notice that we are subtracting by an e^{-x}, and thus want to break the e^{x} term apart. As we want a -x here (to factor it out), we want to figure out what 'a' for, a real number, satisfies x= -x + ax (which is 2) and thus by using x=-x+2x is the desired way to 'split' e^x is e^x = e^{-x} * e^{2x}

Alternatively again, what you could also try and do to simplify it is you can factor out the e^{x} term to get it to cancel with the denominator.
Then, as e^{-x} = e^{x-2x} = (e^{x})(e^{-2x})
If you do this, then we get (1 - e^{-2x}) = (e^{2x} - 1)/(e^{2x}) as we wanted.

EDIT: And way too late here.

Hey not too late!, thanks that's awesome, yeah they noticed ahead of time so that's why they were able to use that trick I guess, i just didn't see that and it was driving me crazy trying to figure out how they knew to do that but I got it now.

Thanks again!

 

[Allonym]

I'm just worried that I'm going to be at a disadvantage compared to people who graduate from a more rigorous program such as yours.

I think most of the people taking these courses aren't really retaining things. I know the class averages are supposed to be very low, so they grade on a crazy curve. So, I think you'll be fine compared to most.

 

[myco666]

Man that's some bullshit, Rutgers requires you to take up to and including Linear Algebra to graduate with a degree in Computer Science. The shit is crazy, I'm happy for you though, at least not everyone who wants to a degree in CS has to suffer.

Linear algebra is beneficial for a lot of fields in computer science. Obviously not every CS student needs to learn it but you are basically closing doors for yourself if you leave it out.

 

[M.D]

General question about functions and graphs

If I'm looking at y=f(x) on a graph, why does y=f(x+1) means the function is moving left instead of right on the X axis? and for y=f(x-1) the function moves right?

 

[RainbowIslands]

General question about functions and graphs

If I'm looking at y=f(x) on a graph, why does y=f(x+1) means the function is moving left instead of right on the X axis? and for y=f(x-1) the function moves right?

Think of x as your input and f(x) your output. If you put in 1, then the output is f(x+1) = f(1+1)= f(2), the f(x) value of one number to the right. Try f(2+1), f(3+1), etc. Effectively, you shifted the whole graph to the left by 1 unit because each input is spitting out the f(x) value one spot to its right.

Similar argument for f(x-1)...

 

[eldudebro]

Hello everyone.

Hope I can get help with the following problem on entropy.

I am literally staring at this with no idea what to do and the notes I have aren't helping!

Any hints/solutions ?

 

[Koren]

Any hints/solutions ?

I'll try...

I'll assume that, by "color entropy of the ball", they mean the color entropy of the balls in the bag... I find the wording a bit akward, but it may just be a translation problem. I can't see what it could be.


So if you draw a ball from the bag, you have a probability 1/2 of drawing a red, 1/4 of drawing a blue, and 1/4 of drawing a green.

You just have the formula to compute the entropy S: it's the sum of -p log(p) over all cases, where p is the probability of each case.

So S = - (1/2) * log(1/2) - (1/4) * log(1/4) - (1/4) * log(1/4)

Thus S = (log(2) + log(4)) / 2

"log" can technically be any log: ln, log2, log10... Usually it's log2 in information theory. It's more complex in physics, but I'll avoid that. You'll have to check which one you're suppose to use.

Now, if you remove a green ball, probabilities change. Just compute the probability for each color, and use the formula again...

The change in probability is explained by the fact you know more (decrease) or less (increase) about the content of the box.

 

[ZBR]

So i finished the semester with a 92% (same grade I got on the cumulative final also)! I was thinking of taking Calculus 1 over the summer but I did some research on the instructor that runs the course and apparently said person isn't that great. Overall, I'm waiting until the fall semester to begin. I was wondering if there are any tips or things I can do to help me get ahead over the summer?

Thank You,
ZBR

 

[stanley1993]

According to Chegg, the span of this matrix {{0,0,0,0},{0,0,0,0},{1,1,2,0},{-2,1,2,-2}} is span({-2,0,1,3} & {-2,2,0,3}). How did they arrive to this conclusion? For example:
I am stuck at: w+x+2y=0 & -2w+x+2y-2z=0.
How do I go from that to knowing:
w=-2
x=0,2
y=1,0
z=3

I will provide more context if needed.

 

[Koren]

According to Chegg, the span of this matrix {{0,0,0,0},{0,0,0,0},{1,1,2,0},{-2,1,2,-2}} is span({-2,0,1,3} & {-2,2,0,3}). How did they arrive to this conclusion? For example:
I am stuck at: w+x+2y=0 & -2w+x+2y-2z=0.
How do I go from that to knowing:
w=-2
x=0,2
y=1,0
z=3

I will provide more context if needed.

Well, span is unique, but the two 4D vectors that create it are not, so there isn't a single solution.

There also isn't a single way to reach the result... It depends on what tools you know.

A way to solve it is to use Gauss-Jordan. Switching rows 1 and 3, rows 2 and 4, and removing twice row 1 to row 2 gives

| 1 1 2 0 |
| 0 3 6 -2 |
| 0 0 0 0 |
| 0 0 0 0 |

Thus
w + x + 2y = 0
3x + 6y -2z = 0

You can go farther (divide row 2 by three, then remove row 2 to row 1) and get

| 1 0 0 2/3 |
| 0 1 2 -2/3 |
| 0 0 0 0 |
| 0 0 0 0 |

Thus
w + 2/3 z = 0
x + 2y - 2/3 z = 0

Given the shape, you can freely choose any y and z you want, and compute w and x by
w = -2/3 z
x = -2y + 2/3 z


Since the matrix is rank 2, you need two independant vectors to describe the span. You can freely choose two independant pairs for y and z then compute the remaining w and x using the previous equations.

If you take (y=0 and z=3), you get (-2, 2, 0, 3)
If you take (y=1 and z=3), you get (-2, 0, 1, 3)

It's interesting to take z=3 because of the -2/3, to avoid non-integers, but there's no need.

with (y=0, z=3) and (y=1, z=0) you get span ( (-2, 2, 0, 3), (0, -2, 1, 0) )

with (y=0, z=1) and (y=1, z=0) you get span ( (-2/3, 2/3, 0, 1), (0, -2, 1, 0) )

All those answers are equivalent.


It may seems complex, and if so, I apologize, but it heavily depends on what you know about linear algebra and tools :/

 

[stanley1993]

Well, span is unique, but the two 4D vectors that create it are not, so there isn't a single solution.

There also isn't a single way to reach the result... It depends on what tools you know.

A way to solve it is to use Gauss-Jordan. Switching rows 1 and 3, rows 2 and 4, and removing twice row 1 to row 2 gives

| 1 1 2 0 |
| 0 3 6 -2 |
| 0 0 0 0 |
| 0 0 0 0 |

Thus
w + x + 2y = 0
3x + 6y -2z = 0

You can go farther (divide row 2 by three, then remove row 2 to row 1) and get

| 1 0 0 2/3 |
| 0 1 2 -2/3 |
| 0 0 0 0 |
| 0 0 0 0 |

Thus
w + 2/3 z = 0
x + 2y - 2/3 z = 0

Given the shape, you can freely choose any y and z you want, and compute w and x by
w = -2/3 z
x = -2y + 2/3 z


Since the matrix is rank 2, you need two independant vectors to describe the span. You can freely choose two independant pairs for y and z then compute the remaining w and x using the previous equations.

If you take (y=0 and z=3), you get (-2, 2, 0, 3)
If you take (y=1 and z=3), you get (-2, 0, 1, 3)

It's interesting to take z=3 because of the -2/3, to avoid non-integers, but there's no need.

with (y=0, z=3) and (y=1, z=0) you get span ( (-2, 2, 0, 3), (0, -2, 1, 0) )

with (y=0, z=1) and (y=1, z=0) you get span ( (-2/3, 2/3, 0, 1), (0, -2, 1, 0) )

All those answers are equivalent.


It may seems complex, and if so, I apologize, but it heavily depends on what you know about linear algebra and tools :/

So P and P^-1 are not unique? D and A in the equations A=PDP^-1 and D=(P^-1)AP are unique right?
What about this question...
What are the eigen vectors for this matrix?
|2 0 0 4|
|0 2 0 0|
|0 0 -2 0|
|0 0 0 -2|

EDIT: Answered my own question.
also

I have {1,0,0,-1} & {0,0,1,0} for eigen of -2. RREF of the matrix given above:
{{4,0,0,4},{0,4,0,0},{0,0,0,0},{0,0,0,0}}
However, Chegg has these eigen vectors:
{0,0,1,0} & {-1,0,0,1}
I think those are the same vectors, right? Just with one of them multiplied by -1. So any matrix multiplied by a constant, be it neg or pos, is the same matrix?

I would post pictures, but I've made so many photobucket accounts. I don't want to make another one and forget my password.


Thanks for all the help already. I think I just need to watch a good video on how to do all this, but I can't find any that goes over everything that I'm having trouble with.

 

[Koren]

So P and P^-1 are not unique? D and A in the equations A=PDP^-1 and D=(P^-1)AP are unique right?

Not sure I understand...

If A is your matrix and D is a diagonal matrix, there's no D an P that gives A=PDP^-1.

Here, det(A) = 0 (and that's useful here, or the span would be the singleton {(0,0,0,0)}, not a 2D-plane)

And even for matrices A whose det(A)=0 (not the case here), you would need additional constraints to have unique D and P (just multiply P by -1 and you get another solution), so I really don't understand the question.

What about this question...
What are the eigen vectors for this matrix?
|2 0 0 4|
|0 2 0 0|
|0 0 -2 0|
|0 0 0 -2|

EDIT: Answered my own question.
also

I have {1,0,0,-1} & {0,0,1,0} for eigen of -2. RREF of the matrix given above:
{{4,0,0,4},{0,4,0,0},{0,0,0,0},{0,0,0,0}}
However, Chegg has these eigen vectors:
{0,0,1,0} & {-1,0,0,1}
I think those are the same vectors, right?

They are not the same vectors, but the eigenspace for eigenvalue -2 is a 2D plane which is generated by both eigenbases
{0,0,1,0} & {-1,0,0,1}
and
{0,0,1,0} & {1,0,0,-1}

There's an infinity of eigenbases...
{1,0,1,-1} & {4,0,2,-4}
is correct, too. Any two independant linear combinations of the two vectors of a eigenbasis is still an eigenbasis.

So any matrix multiplied by a constant, be it neg or pos, is the same matrix?

Err, definively not?

What "level" of linear algebra are you currently studying? Is it a new field for you?

I would post pictures, but I've made so many photobucket accounts. I don't want to make another one and forget my password.

Maybe you can use imgur without an account?

Thanks for all the help already.

You're welcome... I hope I'll be able to actually help a bit.

I think I just need to watch a good video on how to do all this, but I can't find any that goes over everything that I'm having trouble with.

I'm not even sure I could suggest good books in my own languages (learned all this 20 years ago now :/) so finding a good video is hard...

Still, it would be interresting to know which kind of stuff you're working one. At first, I thought it was about solving linear systems for pratical purposes, but seeing eigenvectors and basis changes now makes me wonder.

Edit: in the first example, you look for an eigenbasis for the eigenspace of eigenvalue 0, so they are actually related, but my first post was using a totally different, though related, way to get it.

 

[M.D]

[-(m+3)]^2 equals (m+3)^2

that is to say -(m+3)^2 = (-m-3)*(-m-3)=(m+3)^2? also equals (-m-3)^2? which equals to [(-m)^2-2*(-m)*3+3^2] ?

Thanks

 

[kgtrep]

[-(m+3)]^2 equals (m+3)^2

that is to say -(m+3)^2 = (-m-3)*(-m-3)=(m+3)^2? also equals (-m-3)^2? which equals to [(-m)^2-2*(-m)*3+3^2] ?

Thanks

Yes, you're correct. I'd use parentheses more carefully to illustrate your ideas.

For example, you can write,

[-(m + 3)]^2 = [-(m + 3)] * [-(m + 3)] = (-1)^2 * (m + 3)^2 = (m + 3)^2,

and

[-(m + 3)]^2 = (-m - 3) * (-m - 3) = (-m)^2 + 2(-m)(-3) + (-3)^2 = m^2 + 6m + 9.

 

[M.D]

Yes, you're correct. I'd use parentheses more carefully to illustrate your ideas.

For example, you can write,

[-(m + 3)]^2 = [-(m + 3)] * [-(m + 3)] = (-1)^2 * (m + 3)^2 = (m + 3)^2,

and

[-(m + 3)]^2 = (-m - 3) * (-m - 3) = (-m)^2 + 2(-m)(-3) + (-3)^2 = m^2 + 6m + 9.

Thanks for the tips!

Shouldn't 2(-m)(-3) have the 2 as minus and 3 as plus tho? I thought the whole point is (a-b)^2 takes the minus into account a^2-2ab+b^2

 

[kgtrep]

Thanks for the tips!

Shouldn't 2(-m)(-3) have the 2 as minus and 3 as plus tho? I thought the whole point is (a-b)^2 takes the minus into account a^2-2ab+b^2

You're very welcome.

Both approaches are correct. I was thinking of the formula (a + b)^2 = a^2 + 2ab + b^2.

What you want to substitute for a and b is up to you!

 

[Himynameischris]

So I want to make sure I did this right because it feels wrong

Express W=4i-3j in terms of Magnitude and Direction.

I got 5 for magnitude and -253.74 (rounded up) for direction. Shouldn't the direction be positive?

For 0tan^-1(-.075) I got (-36.869897645844021296855612559093)

I think that's where I'm messing up.

 

[ibyea]

So I want to make sure I did this right because it feels wrong

Express W=4i-3j in terms of Magnitude and Direction.

I got 5 for magnitude and -253.74 (rounded up) for direction. Shouldn't the direction be positive?

For 0tan^-1(-.075) I got (-36.869897645844021296855612559093)

I think that's where I'm messing up.

Well, think about what quadrant your direction is in. Your arctan calculation is certainly correct.

 

[Himynameischris]

Well, think about what quadrant your direction is in. Your arctan calculation is certainly correct.

Right, it's the direction from the origin not the distance, ok I gotta remember that.

Thanks!

 

[Himynameischris]

Ok I need help again...

After I multiply the ^5 I get

Or should I convert the 5/8pi to degrees first and then multiply ^5?

In the books example it's given as degrees already and then they multiply the exponent.

I'm so lost on these...the teacher didn't have time to go over them in class and he just gave us a work sheet with some questions and bid us good luck :/

 

[All Possible Ways]

Ok I need help again...

After I multiply the ^5 I get

Or should I convert the 5/8pi to degrees first and then multiply ^5?

In the books example it's given as degrees already and then they multiply the exponent.

I'm so lost on these...the teacher didn't have time to go over them in class and he just gave us a work sheet with some questions and bid us good luck :/

It makes no difference. Just notice however that 25 pi/8 actually winds around the circle more than once (1.5 times plus pi/8), so it is actually a bit more than 180 degrees, in the third quadrant.

 

[M.D]

Would like some help with this

ABC (trinagle), AB=AC

AB > y= -8x-12
AC > y= -4/7x+20/7

A (-2,4)
C (5,0)

Distance of AC = square root 65 (should equal the distance of AB)

I need to find both options for point B. I tried to express x or y using the distance formula taking into account the fact that the distance between AB and AC is the same, but it's not working or I'm doing something wrong

Thanks!

Edit:

Answer should be

B (-1,-4) or ( -3,12)

 

[kgtrep]

Would like some help with this

ABC (trinagle), AB=AC

AB > y= -8x-12
AC > y= -4/7x+20/7

A (-2,4)
C (5,0)

Distance of AC = square root 65 (should equal the distance of AB)

I need to find both options for point B. I tried to express x or y using the distance formula taking into account the fact that the distance between AB and AC is the same, but it's not working or I'm doing something wrong

Thanks!

Edit:

Answer should be

B (-1,-4) or ( -3,12)

Let (x, y) denote the coordinates of the point B. Since AB = AC implies AB = sqrt(65), we know from the distance formula that x and y must satisfy the following equation:

(1) (x + 2)^2 + (y - 4)^2 = 65.

We also know that the point B lies on the line y = -8x - 12. Hence, x and y satisfies a second equation:

(2) y = -8x - 12.

We can substitute (2) into (1) to get a quadratic equation in terms of x. Solve for x, then for y.

After you have solved for the two possible solutions, I'd graph the problem to visualize what's going on. You would draw 1 circle and 2 lines, and locate points A, B, and C.

 

[M.D]

Let (x, y) denote the coordinates of the point B. Since AB = AC implies AB = sqrt(65), we know from the distance formula that x and y must satisfy the following equation:

(1) (x + 2)^2 + (y - 4)^2 = 65.

We also know that the point B lies on the line y = -8x - 12. Hence, x and y satisfies a second equation:

(2) y = -8x - 12.

We can substitute (2) into (1) to get a quadratic equation in terms of x. Solve for x, then for y.

After you have solved for the two possible solutions, I'd graph the problem to visualize what's going on. You would draw 1 circle and 2 lines, and locate points A, B, and C.

Thanks! worked out nice and easy ;p

How would you go about drawing that? I used desmos to visualize it by finding the 2 possible equations for BC which turned out to be

y=2/3x-10/3 and y=-3/2x+15/2

 

[Koren]

You can try Geogebra...

Looking for a link, I found an online version, but I'm not sure it's working.
https://www.geogebra.org/geometry

 

[kgtrep]

Thanks! worked out nice and easy ;p

How would you go about drawing that? I used desmos to visualize it by finding the 2 possible equations for BC which turned out to be

y=2/3x-10/3 and y=-3/2x+15/2

First, draw points A and C.

We know that AB = AC, so point B must lie on the circle whose center is A and radius is sqrt(65). In other words, without additional information, there can be infinitely many solutions for B.

However, we do know something more about B: It lies on the line y = -8x - 12. Hence, find where this line intersects the circle. From the drawing, we see that we will get 2 solutions.

 

[M.D]

You can try Geogebra...

Looking for a link, I found an online version, but I'm not sure it's working.
https://www.geogebra.org/geometry

First, draw points A and C.

We know that AB = AC, so point B must lie on the circle whose center is A and radius is sqrt(65). In other words, without additional information, there can be infinitely many solutions for B.

However, we do know something more about B: It lies on the line y = -8x - 12. Hence, find where this line intersects the circle. From the drawing, we see that we will get 2 solutions.

Thanks for this! Seeing this makes quite the difference than just dealing with algebra

 

[Himynameischris]

Can someone help me explain the books process with this?

I get how they got -2B-3C but shouldn't they have subtracted -3 from 5? So it should be -2B-3C=2 and then why X EQ 1? EQ 1 is just 1+1+1=0,

Then Eq 3 is the same issue, I get how they got -4B-3C but shouldn't it be 5 not 7? And again they multiply by EQ1...

And then the last part EQ 3 + (-2) X Eq 2...I don't get how they end up with 3C=3....


Any help would be greatly appreciated as my final is on Monday :/

 

[cpp_is_king]

Can someone help me explain the books process with this?

I get how they got -2B-3C but shouldn't they have subtracted -3 from 5? So it should be -2B-3C=2 and then why X EQ 1? EQ 1 is just 1+1+1=0,

Then Eq 3 is the same issue, I get how they got -4B-3C but shouldn't it be 5 not 7? And again they multiply by EQ1...

And then the last part EQ 3 + (-2) X Eq 2...I don't get how they end up with 3C=3....


Any help would be greatly appreciated as my final is on Monday :/

Did you forget to post the original system of equations?

 

[Himynameischris]

Ok I get it now :/ I'm so dumb, we are multiply by 0 so when we add the 4th column...we add 0.

I think I've been at this too long

 

[M.D]

Is this the correct way to prove f(x) is an odd function?

 

[FortuneFaded]

Is this the correct way to prove f(x) is an odd function?

.

 

[BlueLiquid]

Is this the correct way to prove f(x) is an odd function?

You're correct in showing that f(-x) = -f(x) but you forgot that (-x)^3 = -x^3. Thus you should have f(-x) = 4x^3 = -f(x).

 

[kgtrep]

Is this the correct way to prove f(x) is an odd function?

Yep, you're doing great!

In addition to what BlueLiquid said, I recommend writing that your proof is true "for all (real numbers) x." If it were true for only one number x, your result may not be as interesting.

 

[M.D]

.

You're correct in showing that f(-x) = -f(x) but you forgot that (-x)^3 = -x^3. Thus you should have f(-x) = 4x^3 = -f(x).

Yep, you're doing great!

In addition to what BlueLiquid said, I recommend writing that your proof is true "for all (real numbers) x." If it were true for only one number x, your result may not be as interesting.

Thanks first of all

So the full solution should be

My answer is correct for -f(x) because -(-4x^3) results in 4x^3? Is that correct?

I recommend writing that your proof is true "for all (real numbers) x." If it were true for only one number x, your result may not be as interesting.

Good point!

 

[Daria]

Not an equation question but still related to math. I attempted and failed Business Calc last semester after coming from college algebra but I'm tempting to change majors and Calc I is required.

I'm trying to figure out how I can prepare for the Fall so will any of what I cover in BC be relevant in Pre-Calc or are these two different areas? The types of things I'm thinking of are proofs because we never had to do those in BC. Prof said BC was a mix of Calc I, II, III. For context, I went from intermediate algebra > college algebra > BC. Never have I taken a trig class or anything of that sort.

 

[Bacon]

Not an equation question but still related to math. I attempted and failed Business Calc last semester after coming from college algebra but I'm tempting to change majors and Calc I is required.

I'm trying to figure out how I can prepare for the Fall so will any of what I cover in BC be relevant in Pre-Calc or are these two different areas? The types of things I'm thinking of are proofs because we never had to do those in BC. Prof said BC was a mix of Calc I, II, III. For context, I went from intermediate algebra > college algebra > BC. Never have I taken a trig class or anything of that sort.

You probably won't have to do proofs in calc 1. But there's lots of good khan academy videos to get you up to speed on functions and things of that nature that will be helpful in calc 1.

 

[BlueLiquid]

Thanks first of all

So the full solution should be

My answer is correct for -f(x) because -(-4x^3) results in 4x^3? Is that correct?

That's right!

 

[Risible]

I thought we had an ongoing Math Help thread but I couldn't find it, anyone have any tips on what it might be named?

I'm trying to solve "One root of 6x^2-13x+c is -(1/3). What is the other root?" and I'm drawing a blank. Could someone help me out here?

 

[Goldfishking]

I thought we had an ongoing Math Help thread but I couldn't find it, anyone have any tips on what it might be named?

I'm trying to solve "One root of 6x^2-13x+c is -(1/3). What is the other root?" and I'm drawing a blank. Could someone help me out here?

If one answer is -1/3 you can factorised the equation to get (3x+1)(2x-5). So the other solution is...

 

[Cosmo Clock 21]

This is actually a systems of equations question disguised as a quadratics question. Tell us what you've tried to do so far.

 

[CrudeDiatribe]

Math help thread.

Where are you having trouble?

Spoiler

Solve for C.

Spoiler

Find the roots.

 

[danthefan]

Plug the value you have back in and solve for C.

Then just solve the quadratic like any other.

 

[Risible]

Thanks all. I was on the cusp and kept making a stupid error, you guys cleared it up. Thanks for your help!

 

[Tater]

I thought we had an ongoing Math Help thread but I couldn't find it, anyone have any tips on what it might be named?

I'm trying to solve "One root of 6x^2-13x+c is -(1/3). What is the other root?" and I'm drawing a blank. Could someone help me out here?

If you look at the quadratic formula, you've got values for a, b, and x. From there, you can rewrite the quadratic formula to solve for c.

Once you have c, you can go back to the standard quadratic formula and plug that in to get the other root. Make sure that one of the roots is still 1/3 to sanity check your answer.

If you're having trouble rewriting the quadratic formula, post back here with what you have. When I did it, I got a "reasonable" fraction under 10.

 

[Koren]

I thought we had an ongoing Math Help thread but I couldn't find it, anyone have any tips on what it might be named?

I'm trying to solve "One root of 6x^2-13x+c is -(1/3). What is the other root?" and I'm drawing a blank. Could someone help me out here?

If a second degree equation ax²+bx+c=0 has two roots x1 and x2, you can write

ax² + bx + c = a(x-x1)(x-x2)

Developping the second member gives ax² + (-x1-x2)*a*x + a*x1*x2

thus the sum of the roots is -b/a and the product of the roots is c/a

It's a REALLY important result that makes a LOT of problems easier. The result is well known, so you don't have to prove it, except in lower classes.

here, a=6 and b=-13, thus -b/a = 13/6

So the second root is x2 = -b/a - x1 = 13/6 - (-1/3) = 15/6

You actually don't have to substitute x with the first root, solve for c, and recompute the roots, it's correct, but a waste of time. In fact, even proving the result x1+x2 = -b/a is shorter than the subsitute-solve-find roots solution.

One you have a root, getting the other is trivial :
x2 = - (b / a) - x1
or
x2 = c / (a * x1)

Edit: if you need c, you know that x1*x2 = c/a so c = a*x1*x2 = 6 * (-1/3) * (15/6) = -15/3.

Again, no need for solving.

 

[M.D]

Can anyone explain this please? This is a partial solution for a question from an exam

The last part which leads to the solutions is solving this equation once with plus and once with minus before the quadratic equation on the left, which leads to 2 quadratic equations with the solutions 4,3 and 4,1

The other way to solve this is through a polynomial equation but this is obviously a lot of easier

The only thing this reminds me of is equations with 2 absolute values where you would check for 4 possible scenarios of ++, +-, -+ and --

Could anyone explain why is it possible to solve it this way instead of a polynomial equation? Thanks!