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The Math Help Thread

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D

Deleted member 607798

Unconfirmed Member
M.D said:
GgzfmYz.jpg


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!
Click to expand...

The term inside the parenthesis could be a negative term, which is why we must check for both cases. For example, the solution for x^2 = 36 could be 6 or -6.
 
S

sackninja

Member
Okay, so I have a question, wondering if i did this right. We have to find the area between two curve e^x and e^-x and the line x =0.75. So I integrated e^x - e^-x, with the limits 0 and 0.75. I got an equation that looked like this.

(e^0.75 -e^-0.75/-1) -(e^0-e^0/-1)

When I put this into the calculator, i got 0.589366... We were asked for 4 decomal places so I said 0.5894units^2.

Is there any mistakes in that, because it doesn't seem right to me.
 
M

M.D

Member
Win said:
The term inside the parenthesis could be a negative term, which is why we must check for both cases. For example, the solution for x^2 = 36 could be 6 or -6.
Click to expand...

So if you were to let the exponent and square root cancel out each other instead of squaring both sides at the beginning, you would lose a possible solution? because then you would only check for when the equation on the left is a plus?
 
N

Number_6

Member
M.D said:
So if you were to let the exponent and square root cancel out each other instead of squaring both sides at the beginning, you would lose a possible solution? because then you would only check for when the equation on the left is a plus?
Click to expand...

To me it looks like they didn't actually have to square both sides. They did more work than necessary.

But yes, you need to consider both + and - on the left expression after you let the radical cancel the square. Square root of a square behaves exactly like absolute value. Depending on the text, it is how absolute value is defined. So you must consider two cases when solving.
 
B

Bear

Member
sackninja said:
Okay, so I have a question, wondering if i did this right. We have to find the area between two curve e^x and e^-x and the line x =0.75. So I integrated e^x - e^-x, with the limits 0 and 0.75. I got an equation that looked like this.

(e^0.75 -e^-0.75/-1) -(e^0-e^0/-1)

When I put this into the calculator, i got 0.589366... We were asked for 4 decomal places so I said 0.5894units^2.

Is there any mistakes in that, because it doesn't seem right to me.
Click to expand...

Your answer is right. All you need to do here is Integrate the two terms separately and taking the sum to get the function's integral so your approach is completely right.

Just as a side note, this is a great example of how numerical methods can be used to to check your answers. You don't need to know this but it's a nice tool if you are dealing with well behaved expressions like this (i.e. ones that don't change dramatically or unpredictably over the values of x being integrated). You can easily approximate the value should be using something basic like the rectangle method, in this case you just need to evaluate a couple of points to get a pretty good approximation.

e.g. Approximating integral from x=0 to x=0.75 using 3 rectangles of width 0.25:
f(x)dx = 0.25*[ f(0.125) + f(0.375) + f(0.625) ]
f(x)dx = 0.25*[ 0.25056 + 0.76770 + 1.3329 ]
f(x)dx = 0.06266 + 0.1919 + 0.5878
f(x)dx = 0.5878

This is an extremely coarse approximation but in this case, even just 3 points give something pretty close to the answer you have. As you add more terms, you get a better approximation so can serve a nice sanity check (e.g. 10 terms gives 0.58922 and 50 gives 0.58936). It's great if you know how to program since evaluating more terms becomes really easy, but for simple functions a few terms are often enough to get something within a significant digit or two.

This is by no means a replacement for learning how to integrate but it's nice to be able to check answers by evaluating a few terms and seeing if your answers are at least in the same ballpark. This isn't always a viable approach but it's great for functions that are as well behaved as this one.
 
S

sackninja

Member
Bear said:
Your answer is right. All you need to do here is Integrate the two terms separately and taking the sum to get the function's integral so your approach is completely right.

Just as a side note, this is a great example of how numerical methods can be used to to check your answers. You don't need to know this but it's a nice tool if you are dealing with well behaved expressions like this (i.e. ones that don't change dramatically or unpredictably over the values of x being integrated). You can easily approximate the value should be using something basic like the rectangle method, in this case you just need to evaluate a couple of points to get a pretty good approximation.

e.g. Approximating integral from x=0 to x=0.75 using 3 rectangles of width 0.25:
f(x)dx = 0.25*[ f(0.125) + f(0.375) + f(0.625) ]
f(x)dx = 0.25*[ 0.25056 + 0.76770 + 1.3329 ]
f(x)dx = 0.06266 + 0.1919 + 0.5878
f(x)dx = 0.5878

This is an extremely coarse approximation but in this case, even just 3 points give something pretty close to the answer you have. As you add more terms, you get a better approximation so can serve a nice sanity check (e.g. 10 terms gives 0.58922 and 50 gives 0.58936). It's great if you know how to program since evaluating more terms becomes really easy, but for simple functions a few terms are often enough to get something within a significant digit or two.

This is by no means a replacement for learning how to integrate but it's nice to be able to check answers by evaluating a few terms and seeing if your answers are at least in the same ballpark. This isn't always a viable approach but it's great for functions that are as well behaved as this one.
Click to expand...

Thank you so much, the question was in an exam and was sure I got it wrong, I actually did it a second time, but left the first answer down so I should get full marks. Really put my mind at ease.
 
K

Koren

Member
M.D said:
GgzfmYz.jpg


Can anyone explain this please? This is a partial solution for a question from an exam
Click to expand...
I'm curious to know where it comes from exactly, because it's somehow wrong...

The first line implies the second one, but the second one doesn't imply the first, so any solution you find this way can be wrong.

Since the left term is obviously positive, the right one should be too, so each line, starting with the second, should have "and x < 4".

If you find solutions that are greater that 4 at the end, they're actually not solutions.

For example, if the left part was simply 2, x=6 is not solution of 2 = 4-x, because 2 <> 4 - 6

But x=6 is solution of 2² = (4-x)² because (2)² = (4 - 6)²

Writing things like this can be error-prone, be careful.
 
M

M.D

Member
Number_6 said:
To me it looks like they didn't actually have to square both sides. They did more work than necessary.

But yes, you need to consider both + and - on the left expression after you let the radical cancel the square. Square root of a square behaves exactly like absolute value. Depending on the text, it is how absolute value is defined. So you must consider two cases when solving.
Click to expand...

Koren said:
I'm curious to know where it comes from exactly, because it's somehow wrong...

The first line implies the second one, but the second one doesn't imply the first, so any solution you find this way can be wrong.

Since the left term is obviously positive, the right one should be too, so each line, starting with the second, should have "and x < 4".

If you find solutions that are greater that 4 at the end, they're actually not solutions.

For example, if the left part was simply 2, x=6 is not solution of 2 = 4-x, because 2 <> 4 - 6

But x=6 is solution of 2² = (4-x)² because (2)² = (4 - 6)²

Writing things like this can be error-prone, be careful.
Click to expand...


Rq0zh6B.jpg


This is the solution as it appears in the exam
I'm studying at a college but in a pre-academy environment so for a lot of the stuff we're taught we end up using techniques to simplify things for the time being

If I understood you correctly, we're essentially doing what you described by taking the solutions to the quadratic equations and testing if they are true or false for the original equation

If for example one of the solutions I got was 5, I'd end up writing it off after it would test false for the original equation
 
K

Koren

Member
M.D said:
If for example one of the solutions I got was 5, I'd end up writing it off after it would test false for the original equation
Click to expand...
Indeed...

Each time a line implies the next one but the opposite is not true, you have a larger set of solutions.

Some of the solutions you get at the end may not be actual solutions of the original equation. So you need to test them.

In this case, all solutions are <4, so all of them are correct.



But consider the equation

sqrt( (x²-6x+8)² ) = x-4

If you put each side at the power of two, you get

(x²-6x+8)² = (x-4)²

So the exact equation as before... You thus get the solutions 1, 3 and 4

But only 4 is actually correct. 3 and 1 aren't solutions:

sqrt( (1²-6+8)² ) = 3 when 1-4 = -3

sqrt( (3²-6*3+8)² ) = 1 when 3-4 = -1



It's not an issue if you notice that you've used "this equation implies that equation" (thus increasing the set of solutions) instead of "this is an equivalent of that" (the set of solutions remains the same). You just have to test each solution at the end.

Well, it can be an issue when it increase a lot the number of solutions... you may not be able to test all of them.

Dumb case :

x-3 = pi

you use sine on each side:

sin(x-3) = 0

All x = 3+k*pi (for all integers k) are solutions. Now you just have to test an infinity of solutions ;) Only k=1 is correct.

At least, using x -> x² here only multiplies at most by two the number of solutions...



Some examples... If you transform

Code: 
this:              into this:

x = y        <=>   ax+b = ay+b               (assuming a<>0), the solutions are the same


x = y         =>   x² = y²                   (you increase the number of solutions)


x² = y²      <=    x = y                     (avoid this, you lose solutions)

x² = y²      <=>   x = y  or  x = -y         (correct way to do it, the solutions remains the same)


x = y        <=    x / y = 1                 (avoid this, you lose solutions)

x = y        <=>   x / y = 1  or  x = y = 0  (correct way to do it)


sqrt(x) = y   =>   x = y²                    (you increase the number of solutions)

sqrt(x) = y  <=>  x = y²  and  y > 0         (the solutions are the same)


You can always increase the number of solutions if you don't forget to check them at the end, but if you can avoid it, it's always better.

But never decrease the number of solutions. That's the reason why removing the (...)² on each side produced TWO equations in your example. If you only consider the left side, you forget that (-2)² = (+2)²
 
N

Noaloha

Member
Apologies if the following question is unclear! I'm limited to basic approaches to math and whatever I recall from school twenty years ago. With that said,

I have a problem which I suspect miiiiight require mathematical muscle to get past. It feels a little convoluted to explain, so bear with me. I'll rely mainly on a couple of diagrams and hope that they're sufficient to demonstrate the value (or formula) which I'm looking for:


General: I want to use a game engine to create the illusion of a tower wall scrolling by. Step 1 of my approach is to create a wall tile on the right-most edge of the tower, with its image's x-scale at zero, and gradually return that scale to '1' as it approaches the centre of the screen, then reduce it back to zero as it carries on scrolling to the left. I hope this diagram (with arbitrary values punched in) helps illustrate that:

circle2m6pcd.png
Click to expand...

(Once I've figured this part out, the rest of the wall tiles can be inserted and - I think - they'll all behave nicely and the total 'diameter' of the full tower from create point (right wall) to destroy point (left wall) will stay constant.)

I need to put something in the code 'image_xscale = ???' where ??? starts at 0, tends toward 1, then back to 0. It can't do this with linear scaling obvs as that wouldn't give the illusion of a tower wall rotating around. The scaling needs to be fast at first and slowing as it approaches 1.

My gut feeling is that the value of the scaling is related to the way these horizontal lines scale according to a fixed increment on the y-axis:

circle166qlc.png


However, another part of me feels like I should just be dumping some variation of a sin(x) or something in there and it'll work perfectly.

To further break down what I assume are the important variables: an image with width 30 will be created at x=520 (the room's x-value). It will have image_scale=0 (ie. an actual rendered width of zero, rather than its non-scaled width of 30). The image's x-value will reduce at a constant rate (moving the image left in the room). At x=300 (the centre of the room) it will have image_scale=1 (full size, no scaling). It continues left and now its image scaling is returning toward zero at x=80.

Question then: can anyone tell me the necessary formula for how to scale a value as above as, at a linear speed, it approaches a fixed value and then moves away from that fixed value ?

(I hope this makes sense!)
 
K

Koren

Member
The solution is indeed related to trigonometric functions.

Not sure what you want exactly, though... You have a picture that goes around the cylinder and you want a (x,y) -> (x',y) mapping?

Edit: assuming you have a "texture" image (on the left) and you want to map it on a cylinder, using orthogonal projection afterwards, so the result is on the right...
9P7q6K9.png

(forgive the basic design and pixellated result, I used a basic generator and nearest neighbor interpolation to do this faster)

The formula would be something like:

x_texture = x0_texture + R * arcsin( (x_image - xc) / R )

where R is the radius of the cylinder (half-width), xc the column for the center of the cylinder on screen, x0 the coordinates in your texture of the column that will be on the center of the cylinder.

You'll need a texture image that goes from x0_texture - R*pi/2 to x0_texture + R*pi/2

The local x-scaling factor would be (1-(x-xc)²/R²)^(-0.5)
 
N

Noaloha

Member
Koren said:
The solution is indeed related to trigonometric functions.

Not sure what you want exactly, though... You have a picture that goes around the cylinder and you want a (x,y) -> (x',y) mapping?
Click to expand...

I think? Ha, I can't comment much on proper notation sorry. But yes, if that means what it looks like it means (an image with an x,y determining its 'size' having its x value modified) then yah.

I totally get it if my description is lacking or difficult to follow. I'll try again.

I need some function to apply to a width_scaling number. In that diagram there's an image that is 30 pixels wide. The image will be rendered on screen by multiplying its fixed width (30) with a 'width_scaling' value which will range from 0 to 1 (so, 0 pixels wide to 30 pixels wide).

The space which the image occupies will be the 'room'. The room is 600 pixels wide. Any value 'x' for the purposes of this description only corresponds to the horizontal position in that room. The leftmost edge of the room is x=0, the rightmost x=600. If our image has x=300, we're saying that is in the centre of the room, horizontally (not that its image is now 300 pixels wide).

Our image is created at x=520, with image_scaling=0. It will linearly reduce its own x value at a fixed rate, such that the image travels left on the screen. At creation, the image has image_scaling=0. This image_scaling value will tend, um... inverse exponentially(?) toward 1 as the image moves toward the centre of the room. At the centre, x=300, image_scaling=1 (no squishing of the image's width). The image will carry on traveling left as its x decreases, this time the value image_scaling will decrease exponentially back toward 0 as the image's position appraoches x=80. Image_scaling=0 at this point.

What then is the function for image_scaling which matches the visual illusion of a rotating cylinder (a poster stuck on the wall of rotating drum, for example, when viewed directly from the side).

(I'm worried I made the question worse. :p )


EDIT: Thanks for that reply. I see now that I missed an important detail, sorry.

It isn't one single texture. Let's say the visible space is 600 wide. And there is a cylinder with width 440 in the centre (giving 'empty space' of 80 to the left and right). The image I'm manipulating is only, say, 40 in width. Basically a tile. I should have just said tile initially, ha. I will create the illusion of the entire cylinder by repeating this tile such that, together, they cover the whole 440 diameter of the cylinder.

EDIT2: Threw down a quick gif of the single tile being created at a rightmost wall, traveling toward the left and being destroyed. Obviously the width_scaling is very wrong, hence the request for guidance. Just a single oscillation is what I'm looking for, wrapped around the centre point. Link to .gif

(My intuition tells me that, as long as I can solve this single tile, I should be able to fill the 'wall of the cylinder' with an appropriate number of evenly-spaced tiles and the cylinder's total diameter will stay constant. Does that seem right?)
 
K

Koren

Member
Noaloha said:
It isn't one single texture. Let's say the visible space is 600 wide. And there is a cylinder with width 440 in the centre (giving 'empty space' of 80 to the left and right). The image I'm manipulating is only, say, 40 in width. Basically a tile. I should have just said tile initially, ha. I will create the illusion of the entire cylinder by repeating this tile such that, together, they cover the whole 440 diameter of the cylinder.
Click to expand...
It's not really an issue... You just have to use a modulo in the previous formula.

Assuming a 40x40 tile, a 600x400 result, and a 440 (R=220) cylinder centered on column 300. And let's assume you want the center of a tile on column 300.

For all (x, y) (x being the column, beware... it depends on what kind of stuff you're playing with) :

if x<80 of x>520, nothing

if 80<x<520, you use the coordinates
( (R*arcsin((x-xc)/R + Twidth/2) %Twidth + Twidth) %Twidth, y%Theight)
in the tile where
R = 220
xc = 300
Twidth = 40
Theight = 40

The double-modulo is just a trick so that you always get positive numbers... If you tell me which language you want to use, it'll be easier...
 
K

Koren

Member
A sidenote: the scaling factor you're after is (1-(x-xc)²/R²)^(-0.5)

(it's 1 when x=xc, 0 when x=xc+R or xc-R) but if you scale linearly each tile, you'll get issues... like tiles not connecting, since you'll basically get a *flat* tile on the cylinder.

Though you can correct it by a small factor that I can compute if you wish that will correct the issue (it'll work with sufficiently small tiles, or you'll get artifacts, though, because what you'll get is a polygonal cylinder)
 
N

Noaloha

Member
Excellent! Thanks for the assistance. I'll have a toy around with it later today hopefully and punch in the appropriate values. Fingers crossed it all just slots neatly into place.
 
K

Koren

Member
Noaloha said:
Excellent! Thanks for the assistance. I'll have a toy around with it later today hopefully and punch in the appropriate values. Fingers crossed it all just slots neatly into place.
Click to expand...
You're welcome... I hope it'll help.

Throw me an MP is you have another question or if something is not working in case I miss the thread update (or if you want a practical example)

Edit: if you want to do something realtime with this, a common tric is to precompute the scaling factor for all x in a table or a texture, and use a look-up table, it's faster (by far).
 
N

Noaloha

Member
Koren said:
if you want to do something realtime with this, a common tric is to precompute the scaling factor for all x in a table or a texture, and use a look-up table, it's faster (by far).
Click to expand...

I think, overall, the question falls simply under the umbrella of 'I-woke-up-with-a-math-problem-which-is-an-itch-at-the-back-of-my-mind'. I'm no mathematician nor am I a game dev, hobby or otherwise. But I do occasionally find myself with little game ideas and I do have access to a copy of GameMaker 1.4, so every so often I find myself opening it up and just trying to figure a thing out, with no grander purpose.

Yesterday I woke up wondering about whether the illusion of a side-scrolling tower could be done in a 2d game with flat sprites using something like I described above. Today, to rid myself of the distraction, I decided to try and find out. And that's kinda the be-all and end-all, haha.

Your comment is interesting to me though, as I was also wondering about how expensive computationally this would be; would it be at all practical even, assuming it works. The alternative approach I'd thought of was to essentially animate the tiles as pre-stretching and shrinking and then move the sprites linearly. Which'd be a hideous amount of work on the sprite-making side. Having those image_scaling values available as simple look-ups in a table (since they're entirely fixed) sounds like a much more sensible approach!
 
K

Koren

Member
Noaloha said:
I think, overall, the question falls simply under the umbrella of 'I-woke-up-with-a-math-problem-which-is-an-itch-at-the-back-of-my-mind'. I'm no mathematician nor am I a game dev, hobby or otherwise. But I do occasionally find myself with little game ideas and I do have access to a copy of GameMaker 1.4, so every so often I find myself opening it up and just trying to figure a thing out, with no grander purpose.
Click to expand...
Well, if you're interested into the details, I can explain... I'll just need to upload a figure.

The basic result would be this:
https://media.giphy.com/media/3o7bubSxbeWIiP1Li0/giphy.gif
(just an animated gif built in Python with a sprite and the above formula)

No experience on GameMaker, though, most of my hobby game programming is currently done in C++/SQL2...

Noaloha said:
Your comment is interesting to me though, as I was also wondering about how expensive computationally this would be; would it be at all practical even, assuming it works. The alternative approach I'd thought of was to essentially animate the tiles as pre-stretching and shrinking and then move the sprites linearly. Which'd be a hideous amount of work on the sprite-making side. Having those image_scaling values available as simple look-ups in a table (since they're entirely fixed) sounds like a much more sensible approach!
Click to expand...
The look-up table is indeed a common approach to speed-up the issue.

But at the same time, today, most 2D renderers are actually 3D renderers, so you could just use actual 3D coordinates of a cylinder and nullify the Z-one.

Just using a x-based scaling factor may be useful if you wanted to design a mode-7 SNES game, it's taylored for this kind of stuff (and indeed, there's many games that used this kind of rolling cylinders) even if the lack of anisotropic filtering of the sprite doesn't help...
 
N

Noaloha

Member
Koren said:
The basic result would be this:
https://media.giphy.com/media/3o7bubSxbeWIiP1Li0/giphy.gif
(just an animated gif built in Python with a sprite and the above formula)
Click to expand...

Ooh, that gif you posted is exactly what I was seeing in my head and what I was trying to create. And my itching question of "does it create an effective replication of a rotating cylinder?" has a satisfying affirmative answer. Also neat that it has the optical illusion I expected, wherein you can force it to be the internal or external side of a cylinder. (I assume with some more deft math, the tile's image could be skewed in such a way to taper vertically toward the outer edge, increase or decrease, to force a perspective.)

Just seeing it has actually scratched the itch; I no longer need to replicate it myself. Thanks!

Which is good, as I didn't manage to insert the formula into the code effectively. I'm loathe to query too much further though as, at that point, I'd simply be straying into how-do-I-code territory rather than this thread's focus on the straight math. That said, if you could explain the '%' symbol in your notation, that'd be great. What does that represent as a math function? (GameMaker didn't recognise it as a symbol for a formula.)
 
K

Koren

Member
Noaloha said:
I assume with some more deft math, the tile's image could be skewed in such a way to taper vertically toward the outer edge, increase or decrease, to force a perspective.
Click to expand...
Yes... It's a planar projection in this case, but you can do for example a pinhole one to get the effect you want.

In fact, parallax scrollings in old 2D games are using exactly this trick.


About your idea, I think the oldest game that used that idea is Nebulus
https://www.youtube.com/watch?v=faPS312Tthk

It's not uncommon, it was actually a semi-common trick on 2D 16-bits games.


With mode7, it's easier to get both of your effects in the opposite direction. Castlevania IV (see 3:10) :
https://www.youtube.com/watch?v=8e3Lhy2IrmI

Axelay, too, used it to give the feeling of travelling over a round world:
https://www.youtube.com/watch?v=0jFsGTuXdFs&t=825s

Though when it's flat, it's easier to get the effect in the vertical direction. Skyblazer:
https://www.youtube.com/watch?v=RCKoO0UEIFU


Noaloha said:
That said, if you could explain the '%' symbol in your notation, that'd be great. What does that represent as a math function? (GameMaker didn't recognise it as a symbol for a formula.)
Click to expand...
Maybe it's "mod" in GameMaker.

Basically, it's the remainder of the (integer) division.

17 mod 3 is 2 because 17 is 5*3+2
34 mod 6 is 4 because 21 is 6*5+4

When you write a % b (or a mod b) It's as if you remove b from a till it's lower than b
17 -> 14 -> 11 -> 8 -> 5 -> 2
34 -> 28 -> 22 -> 16 -> 10 -> 4

It's handy to tile things (anything that repeats, basically)
 
S

sam777

Member
I am actually good at maths but a friend is stuck on this question and I can't fgure it out, any help would be appreciated.

A new-born baby spends approximately 15 hours asleep each day for the first month and then 16 hours each day for the next two months. As a fraction, how much of a 3 month year old's life has been spent asleep (where 1 month is equalivalent to 4 weeks)?
 
C

cpp_is_king

Member
sam777 said:
I am actually good at maths but a friend is stuck on this question and I can't fgure it out, any help would be appreciated.

A new-born baby spends approximately 15 hours asleep each day for the first month and then 16 hours each day for the next two months. As a fraction, how much of a 3 month year old's life has been spent asleep (where 1 month is equalivalent to 4 weeks)?
Click to expand...

15*28 + 16*28*2 = 28*(15+32) = 28*47 hours of his life asleep.

In 3 months there are 28*3*24 hours total

28*47/28*3*24 = 47/72
 
F

FortuneFaded

Member
sam777 said:
I am actually good at maths but a friend is stuck on this question and I can't fgure it out, any help would be appreciated.

A new-born baby spends approximately 15 hours asleep each day for the first month and then 16 hours each day for the next two months. As a fraction, how much of a 3 month year old's life has been spent asleep (where 1 month is equalivalent to 4 weeks)?
Click to expand...

To calculate the fraction, you only need to look at the average day. Since it was 15 hours for 1 month, then 16 for 2, then average is 15.6666 or 47/3 (should be easy to see why). Then to calculate the fraction, we have average hours of sleep as the numerator and the average hours in a day (easy since all days have 24 hours) as the denominator. So then we have


47/3 / 24 = 47/72
 
S

sam777

Member
cpp_is_king said:
15*28 + 16*28*2 = 28*(15+32) = 28*47 hours of his life asleep.

In 3 months there are 28*3*24 hours total

28*47/28*3*24 = 47/72
Click to expand...

FortuneFaded said:
To calculate the fraction, you only need to look at the average day. Since it was 15 hours for 1 month, then 16 for 2, then average is 15.6666 or 47/3 (should be easy to see why). Then to calculate the fraction, we have average hours of sleep as the numerator and the average hours in a day (easy since all days have 24 hours) as the denominator. So then we have


47/3 / 24 = 47/72
Click to expand...

Thanks a lot.
 
S

shockthrill

Banned
I know this is a Math Help thread, but i didn't see anywhere to post general mathematics question.

One question is who would you consider the greatest mathematician of all time and also the 20th century

all time the candidates tend to be: Archimedes, Newton, Gauss, Euler

20th century it's Hilbert and Grothendieck

Von Neumann was probably the greatest pure genius to ever live, seriously he was not of this planet..but in terms of achievements, I don't know if he matches up to the greats

Eugene Wigner's quote about Von Neumann is important

"I have known a great many intelligent people in my life. I knew Planck, von Laue and Heisenberg. Paul Dirac was my brother in law; Leo Szilard and Edward Teller have been among my closest friends; and Albert Einstein was a good friend, too. But none of them had a mind as quick and acute as Jansci John von Neumann. I have often remarked this in the presence of those men and no one ever disputed me.

... But Einstein's understanding was deeper even than von Neumann's. His mind was both more penetrating and more original than von Neumann's. And that is a very remarkable statement. Einstein took an extraordinary pleasure in invention. Two of his greatest inventions are the Special and General Theories of Relativity; and for all of Jansci's brilliance, he never produced anything as original.
 
H

Hypertrooper

Member
Lebesgue created the measure theory on his own which is fundamental for stochastic and some statistic stuff. So I would be adding him to the list of 20th century people.

Like Lebesgue, Shannon delivered the foundation for information theory and the motivation for coding theory. So add him to the list.

We could add Turing, too, if you count theoretical computer science as a

Well. If you are a topologist, you want to mention Hausdorff and Borel?

There had been tons of great mathematician in the 20th century.
 
S

shockthrill

Banned
Hypertrooper said:
Lebesgue created the measure theory on his own which is fundamental for stochastic and some statistic stuff. So I would be adding him to the list of 20th century people.

Like Lebesgue, Shannon delivered the foundation for information theory and the motivation for coding theory. So add him to the list.

We could add Turing, too, if you count theoretical computer science as a

Well. If you are a topologist, you want to mention Hausdorff and Borel?

There had been tons of great mathematician in the 20th century.
Click to expand...

That is true, there are so many to name

I know its often dubious to try and "rank" mathematicians and scientists in some way or another. but in terms of the 20th century, I honestly believe Grothendieck was above the rest. His power of abstraction and the ability to look at math in a general way really lead to modern algebraic geometry which he essentially build single handily

Grothendieckportra_3107171c.jpg
 
M

M.D

Member
Can anyone help me with this? I'm wondering if there's any other of way of solving this

I put this in some online calculator and the way it solved is by defining alpha/2 as t (or u, or whatever you're used to) and turned sin(alpha) and cos(alpha) into sin(2t) and cos(2t) since 2*alpha/2 would get you back to alpha and using the identities for sin(2alpha) and cos(2alpha) and so on


a1PReEk.png



This is the original one

6j0YzhC.png
 
K

kgtrep

Member
M.D said:
Can anyone help me with this? I'm wondering if there's any other of way of solving this

I put this in some online calculator and the way it solved is by defining alpha/2 as t (or u, or whatever you're used to) and turned sin(alpha) and cos(alpha) into sin(2t) and cos(2t) since 2*alpha/2 would get you back to alpha and using the identities for sin(2alpha) and cos(2alpha) and so on


a1PReEk.png



This is the original one

6j0YzhC.png
Click to expand...


Because I saw alpha, 2*alpha, and 3*alpha (I will just write "a" for alpha), I thought maybe I should go for symmetry about 2a.


We can simplify the LHS as follows:

( cos(a) - cos(3a) ) / ( sin(a) + 2sin(2a) + sin(3a) )

= ( cos(2a - a) - cos(2a + a) ) / ( sin(2a - a) + 2sin(2a) + sin(2a + a) )

= 2sin(2a)sin(a) / ( 2sin(2a) * ( cos(a) + 1 ) )

= sin(a) / ( cos(a) + 1 )


You get the 2nd equality by using sum formulas for sine and cosine, then canceling common terms.
 
R

RazorbackDB

Member
Hey GAF, I'm starting second semester next week and is a bit math heavy, plus the CS courses I'll be taking seem project heavy so I just thought I should prepare a bit beforehand this semester. Assuming I don't have any schedule conflict I'll be taking Multivariable Calculus, Numerical Methods/Analysis and Statistics and Probability.

I already took Multivariable Calculus last semester but after the first couple of bad grades I just dropped it and focused on other stuff, and it ended up being the only course I didn't pass so I guess it was a good decision(?). The other ones is all new to me.

Anyway I'm just looking stock up on some resources so if anyone has some good material, links, books, youtube videos, etc. to share besides the usual stuff (Khan, PatrickJMT, Paul's Notes), I'll appreciate it a lot.
 
_Aaron_

_Aaron_

Member
RazorbackDB said:
Hey GAF, I'm starting second semester next week and is a bit math heavy, plus the CS courses I'll be taking seem project heavy so I just thought I should prepare a bit beforehand this semester. Assuming I don't have any schedule conflict I'll be taking Multivariable Calculus, Numerical Methods/Analysis and Statistics and Probability.

I already took Multivariable Calculus last semester but after the first couple of bad grades I just dropped it and focused on other stuff, and it ended up being the only course I didn't pass so I guess it was a good decision(?). The other ones is all new to me.

Anyway I'm just looking stock up on some resources so if anyone has some good material, links, books, youtube videos, etc. to share besides the usual stuff (Khan, PatrickJMT, Paul's Notes), I'll appreciate it a lot.
Click to expand...

MIT video lectures: https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/

Does your course recommend any particular text books?
 
M

M.D

Member
Simple question I'm getting the wrong answer to

How many times does the number 1 appear a single time from 100 to 999?

My thinking is I have 10 numbers (0-9) and 3 cases

1. 991
2. 199
3. 919

1 has to be used in each case and it leaves me with 9 other options for the two remaining digits, so (9*9*1)*3 = 243, but apparently the answer should be 225

What am I getting wrong?
 
T

the-pi-guy

Member
M.D said:
Simple question I'm getting the wrong answer to

How many times does the number 1 appear a single time from 100 to 999?

My thinking is I have 10 numbers (0-9) and 3 cases

1. 991
2. 199
3. 919

1 has to be used in each case and it leaves me with 9 other options for the two remaining digits, so (9*9*1)*3 = 243, but apparently the answer should be 225

What am I getting wrong?
Click to expand...

Your first digit can't be 0. You're including numbers like 091.

This'll give you 225.
 
C

cpp_is_king

Member
Does anyone know of an online tool where I can enter a list of numbers followed by a "sum", and I want to find out if any subset of the list adds up to the sum? I can write my own program to do it, but I'd rather use an existing website for convenience and the ability to do it from my phone.
 
T

Two Words

Member
I'm about to start a grad-level computer graphics course. How much of a refresher should I do on Linear Algebra. I've basically forgotten everything but the basics.
 
M

M.D

Member
Can someone please explain something in regard to logs - when is it not allowed to solve a problem (just as an example) by equating log10 (x^2) to 2log10(x) as part of your solution since you could lose possible solutions by doing so? X>0 in the first case but X>0 or X <0 for the second
 
K

Koren

Member
M.D said:
Can someone please explain something in regard to logs - when is it not allowed to solve a problem (just as an example) by equating log10 (x^2) to 2log10(x) as part of your solution since you could lose possible solutions by doing so? X>0 in the first case but X>0 or X <0 for the second
Click to expand...
Well,

log10(x^2) = 2 has 2 solutions, x = 10 or -10
2 log10(x) = 2 has only a single solution, x = 10

Thus they can't be the same.

You can write log10(x^2) = 2 log10( sqrt( x^2 ) )

The issue is that sqrt(x^2) is |x|, not x.

So log10(x^2) = 2 log10( |x| )
 
T

theinternetman

Member
Dreadnought said:
I need some help CalculusGaf. How am I supposed to do the g(x) problems a, b, c if theres no equation? Where do I get it from the graph? Also did I do d - g correctly?
Click to expand...

It looks like d through g are correct to me.

And you're right, to do a,b,and c you should use the graph. These questions ask "what is the value of the function g(x) when x is a particular value" for a few different values. You can use the graph to find the y coordinate corresponding to the different values of x for a,b, and c.
 
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