Himynameischris
Member
Ohhhhh yeah :/
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The Math Help Thread
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Ohhhhh yeah :/
I have this equation
(x+yi)(x-yi)+(x+yi)+1=i
The results are
If x is an imaginary number, doesn't z = 'imaginary number' + 'real number' * i thus meaning z is not a complex number anymore? I thought the end result is z equaling a real number + imaginary number (by multiplying with i) so maybe this is where I got it wrong?
Markitron
Is currently staging a hunger strike outside Gearbox HQ while trying to hate them to death
Probably been asked to death already so apologies, can anyone recommend an online resource/YT channel for calculus?
I haven't done it in well over a decade and realised recently that I have almost completely forgotten it.
I haven't done it in well over a decade and realised recently that I have almost completely forgotten it.
I have this equation
(x+yi)(x-yi)+(x+yi)+1=i
The results are
If x is an imaginary number, doesn't z = 'imaginary number' + 'real number' * i thus meaning z is not a complex number anymore? I thought the end result is z equaling a real number + imaginary number (by multiplying with i) so maybe this is where I got it wrong?
It seems like x is a complex number here, not imaginary. Did you not get information on whether x and y are real or complex as given in the equation? This equation does not use the conventional use for x/y in the context of complex numbers so you have to solve it in terms of assuming both x and y are complex. A tip for these sorts of equations is that the real part and the imaginary part of both sides of the equation has to be equal, so you can separate it into a system of equations after some simplification. It might be easier to solve if you substitute x = a + bi and y = c + di, so you can work with real variables, or it might not be easier. It depends.
eot
Banned
I have this equation
(x+yi)(x-yi)+(x+yi)+1=i
The results are
If x is an imaginary number, doesn't z = 'imaginary number' + 'real number' * i thus meaning z is not a complex number anymore? I thought the end result is z equaling a real number + imaginary number (by multiplying with i) so maybe this is where I got it wrong?
Like Leezard said, you can't assume 'x' and 'y' are real, they could be anything unless otherwise stated. For this equation I would expand the expression:
(x+yi)(x-yi)+(x+yi)+1=i
x^2 + y^2 + x + iy + 1 = i
From this point you can guess y=1 and see what happens, because this removes the explicitly imaginary terms. Then you're left with:
x^2 + x + 2 = 0
which is just a normal second degree equation, with a complex solution
Okay, someone explain this to me:
https://youtu.be/ZffZvSH285c?list=PLD6DA74C1DBF770E7&t=489
I get everything this dude is saying up until this point. When theta turns into a 90 degree angle, I'm not understanding how you're supposed to figure out what either it's sin or cos are. Sin is the opposite side/hypotenuse, but...at a 90 degree angle, the only opposite side is the hypotenuse! Same with cos.
What's the deal?
https://youtu.be/ZffZvSH285c?list=PLD6DA74C1DBF770E7&t=489
I get everything this dude is saying up until this point. When theta turns into a 90 degree angle, I'm not understanding how you're supposed to figure out what either it's sin or cos are. Sin is the opposite side/hypotenuse, but...at a 90 degree angle, the only opposite side is the hypotenuse! Same with cos.
What's the deal?
John Caboose
Member
In the same video twenty seconds earlier:
https://www.youtube.com/watch?v=ZffZvSH285c&feature=youtu.be&list=PLD6DA74C1DBF770E7&t=429
The cos and sin values are just the x and y coordinates of where the line intersects the circle.
Khan Academy has been a great resource, it probably still is.
Yes, I saw the whole video and I understand that, but I can only visualize it with acute angles. I mean, I guess I can just accept that it's easy to just take the x and y coordinates, but I want to understand why.
John Caboose
Member
I understand, and you have exactly the right attitude. I'm not the right person to explain it, a math professor at uni probably could.
What follows is instead more of a general answer on the "why" question. Some things in math (and science) are "by definition" which means the answer is ultimately "it's like that because we say it's like that". They can also be "axiomatic" in that some truths are so basic that we can not explain them in the form of anything else.
Watch this excellent interview with Richard Feynman where he says this in words better than mine:
https://www.youtube.com/watch?v=36GT2zI8lVA
EDIT: Actually I want to take a crack at it. Look at this picture.
When you make theta larger and larger, making the intersection come ever closer to the top there, you can see how the sin(theta) line becomes closer and closer to overlapping with the hypothenuse (that is always of length one because that is the radius of the unit circle). Eventually they overlap completely, at which point it becomes one.
Apply this visualization to the cos(theta) part as well, as it gets closer and closer you can see that it eventually becomes zero.
Hopefully that helps a little bit for you to visualize it. The triangle sort of folds in over itself once one of the lines that makes up the triangle becomes length zero.
Look at the definitions again: sin(theta) = opposing side / hypothenuse. In the unit circle the hypothenuse is of length 1 and as such sin(theta) = opposing side / 1 = opposing side.
cos(theta) = adjacent side / hypothenuse = adjacent side / 1 = adjacent side.
When adjacent side becomes length zero, cos is also zero. When opposing side becomes length one then sine becomes one!
Markitron
Is currently staging a hunger strike outside Gearbox HQ while trying to hate them to death
I'll check it out thanks.
I understand, and you have exactly the right attitude. I'm not the right person to explain it, a math professor at uni probably could.
What follows is instead more of a general answer on the "why" question. Some things in math (and science) are "by definition" which means the answer is ultimately "it's like that because we say it's like that". They can also be "axiomatic" in that some truths are so basic that we can not explain them in the form of anything else.
Watch this excellent interview with Richard Feynman where he says this in words better than mine:
https://www.youtube.com/watch?v=36GT2zI8lVA
EDIT: Actually I want to take a crack at it. Look at this picture.
When you make theta larger and larger, making the intersection come ever closer to the top there, you can see how the sin(theta) line becomes closer and closer to overlapping with the hypothenuse (that is always of length one because that is the radius of the unit circle). Eventually they overlap completely, at which point it becomes one.
Apply this visualization to the cos(theta) part as well, as it gets closer and closer you can see that it eventually becomes zero.
Hopefully that helps a little bit for you to visualize it. The triangle sort of folds in over itself once one of the lines that makes up the triangle becomes length zero.
Look at the definitions again: sin(theta) = opposing side / hypothenuse. In the unit circle the hypothenuse is of length 1 and as such sin(theta) = opposing side / 1 = opposing side.
cos(theta) = adjacent side / hypothenuse = adjacent side / 1 = adjacent side.
When adjacent side becomes length zero, cos is also zero. When opposing side becomes length one then sine becomes one!
Okay, this helps a lot more. Thanks!
Well, I'd argue that in this case, cos and sin first and foremost linked to complex exponential or series (so very much related to your graph and really nice explanation)...
It's a very natural concept linked to waves, as you can see when you look at Fresnel diagrams. The definition isn't as random as it may seem.
The thing is, you first encounter cosine and sine use in triangles, but that's mostly a byproduct.
Those are two solution among an infinite numer of solutions, unless I'm mistaken... Unless there's something else in your problem...I have this equation
(x+yi)(x-yi)+(x+yi)+1=i
The results are
Problem is...
you can choose y=0, y=2, y=i or y=42+sqrt(3)i
All those choices will give you two solutions for x... There's no real reason to want the explicit imaginary parts to disappear.
There's an infinite number of (x,y) that solve the equation, since you basically try to solve 2 equations
Re(x^2 + y^2 + x + iy + 1 - i) = 0
Im(x^2 + y^2 + x + iy + 1 - i) = 0
in a 4D space ( C² is R^4 )
So yes, by choosing y=1 (which means adding 2 equations, Re
= 1 and Im = 0, thus a system of 4 equations for 4 unknowns), you get the two results above, but the two results above don't make sense if you don't have something that says y=1 first...Either there's something missing from the problem, or the given solution are just wrong (well, incomplete).
Edit:
A bit more details:
The root of (x+yi)(x-yi)+(x+yi)+1=i are
(y = anything, x = -0.5+z where z²= -y²-3/4 + i (y-1) )
So
(x = -0.5 + i sqrt(7)/2, y = 1) and (x = -0.5 - i sqrt(7)/2, y = 1) are solutions
But
(x = -i, y = 0)
(x = -1-i, y = 0)
(x = i, y = -i)
(x = -1-i, y = -i)
are solutions, too (I could write an infinite list of solutions)
The easiest to check is the first...
(x+yi)(x-yi)+(x+yi)+1=i becomes i*i+i+1 = i, which is obviously true.
So while
are indeed solutions,
it's NOT the result of the equation, just a tiny, tiny part.
Himynameischris
Member
Is there another way to figure this out other than looking at the graph?
Our teacher only told us that it's when v and a have the same signs....but other than (9/4, ∞ I can't figure out the other part, the graph is too small.
It's suppose to be (0,?)U(9/4, ∞
Our teacher only told us that it's when v and a have the same signs....but other than (9/4, ∞
I can't figure out the other part, the graph is too small. It's suppose to be (0,?)U(9/4, ∞
Stumpokapow
listen to the mad man
Something is speeding up when it has positive acceleration, velocity with positive slope, or concave up distance -- or negative acceleration, velocity with negative slope, and concave down distance. You can think of these two cases as speeding up forwards and speeding up backwards relative to the origin point. All of these can be seen by visual inspection.
The exact points are unclear because of the resolution of the graph; but presumably to select that graph you had to do something with an equation before question g?
Have you learned calculus? The acceleration function is the second derivative of the distance function (or the first derivative of the velocity function. When the distance function has a positive second derivative (positive acceleration), then the particle is speeding up. You need only one of the distance, velocity, and acceleration functions to derive the rest (subject to a constant initial value for each).
The exact points are unclear because of the resolution of the graph; but presumably to select that graph you had to do something with an equation before question g?
Have you learned calculus? The acceleration function is the second derivative of the distance function (or the first derivative of the velocity function. When the distance function has a positive second derivative (positive acceleration), then the particle is speeding up. You need only one of the distance, velocity, and acceleration functions to derive the rest (subject to a constant initial value for each).
Himynameischris
Member
Edit: I figured it out, I just did my own graph, :/
It's 3. Well that was easier than I thought.
Thanks for the help, when you mentioned the graph I just graphed v(t) and a(t) on my own.
It's 3. Well that was easier than I thought.
Thanks for the help, when you mentioned the graph I just graphed v(t) and a(t) on my own.
Stumpokapow
listen to the mad man
I'm not near either paper or a computer where I can solve this, but think about this as a puzzle.
You know from high school how to use algebra to find when a function is exactly zero. You have a function for velocity and a function for acceleration.
First, consider the velocity function. Solve the zeroes (take the FOC of the distance function). Create a series of intervals from -infinity to your first zero, from your nth zero to your n+1th zero, and from your final zero to +infinity. Test one point in each interval to see if the velocity is positive or negative. Now do the same for acceleration (take the SOC of the distance function). Which intervals have both positive velocity and positive acceleration (speeding up forwards), or negative velocity and negative acceleration (speeding up backwards)?
You know from high school how to use algebra to find when a function is exactly zero. You have a function for velocity and a function for acceleration.
First, consider the velocity function. Solve the zeroes (take the FOC of the distance function). Create a series of intervals from -infinity to your first zero, from your nth zero to your n+1th zero, and from your final zero to +infinity. Test one point in each interval to see if the velocity is positive or negative. Now do the same for acceleration (take the SOC of the distance function). Which intervals have both positive velocity and positive acceleration (speeding up forwards), or negative velocity and negative acceleration (speeding up backwards)?
Himynameischris
Member
Oh that really helps, that way I don't have to rely on a graph, thanks!
eot
Banned
Yes, you're right. You need at least one constraint for every unknown. I figured that was a bit beyond what he was asking though.
Is there another way to figure this out other than looking at the graph?
Our teacher only told us that it's when v and a have the same signs....but other than (9/4, ∞
It's suppose to be (0,?)U(9/4, ∞
The graph that's marked as correct doesn't even seem right to me. The velocity function is negative for a bit, which means that the position function, s, should be decreasing on that same interval. It never appears to decrease, however.
Linkstrikesback
Member
Thanks!
Any idea on how to solve this? Infinite Geometric Series
Answer is x = -3
1) 2/5 + 2/5^3 + 2/5^5 + 2/5^7... = 2/5 + (2/5)*(1/5^2) + (2/5)*(1/5^2)^2 + (2/5)*(1/5^2)^3....
Geometric sum = 5/12
2) x/5^2 + x/5^4 + x/5^6... = x/5^2 + (x/5^2)*(1/5^2) + (x/5^2)*(1/5^2)^2...
Geometric sum = x/24
3) x/24 + 5/12 = 7/24 = x/24 + 10/24
x = -3
Yup, none of the figures looks right. I blame shitty plotting.
Very possibly.
That's the reason I would like to see the original question. Because It can't be just what we see, and the two answers he has can't be the correct answer.
I developped the thing a bit, because I felt choosing a y and stopping with the results you get is somehow wrong, even if the results match what he had. Especially if someone that read it think you should find the y that make the quadratic equation in x without imaginary elements.
Is there a math way of working this out?
I got the answer by narrowing down the few possibilities and then the dad likes this number clue took me in the direction of beer / lager, ended up being Heineken 1873, which also happened to be the year the RMS Atlantic sank killing over 500 people.
I find out today my deduction was right, but without knowing the age of beer or a 140yr old disaster i fail to see how it equates to a math problem, something tells me there must be a mathematical solution in there somewhere for it to be math homework but i just cannot see it.
So have at it.
Use the clues to find the missing number!
This is a number with 4 digits.
The thousand and hundred digits added together make 9.
It is not a good number to remember.
The tens and units digits added together make 10.
There are an odd number of tens.
My dad likes this number.
The number is more than 1000 but less than 2000
I got the answer by narrowing down the few possibilities and then the dad likes this number clue took me in the direction of beer / lager, ended up being Heineken 1873, which also happened to be the year the RMS Atlantic sank killing over 500 people.
I find out today my deduction was right, but without knowing the age of beer or a 140yr old disaster i fail to see how it equates to a math problem, something tells me there must be a mathematical solution in there somewhere for it to be math homework but i just cannot see it.
So have at it.
Use the clues to find the missing number!
This is a number with 4 digits.
The thousand and hundred digits added together make 9.
It is not a good number to remember.
The tens and units digits added together make 10.
There are an odd number of tens.
My dad likes this number.
The number is more than 1000 but less than 2000
cpp_is_king
Member
Not familiar with this problem, so I'll just type my thoughts as I read through each sentence, and see where it leads.
C = 2k + 1, where k = 0, 1, 2, 3, or 4
Not sure what the last two mean, but let's start with what we have.
A + B = 9
-> 1000A + 100B = 900A + 100(A + B) = 900A + 900
C + D = 10
-> 10C + D = 9C + (C + D) = 9C + 10
Substituting back in, we now have:
N = 900A + 900 + 9C + 10 = 900A + 9C + 910
Back to the clues:
A = 1
-> N = 900 + 910 + 9C = 1810 + 9C
C = 2k + 1
-> N = 1810 + 9(2k + 1) = 1810 + 18k + 9 = 1819 + 18k
Now, we know k = 0, 1, 2, 3, or 4.
k = 0 -> N = 1819
k = 1 -> N = 1837
k = 2 -> N = 1855
k = 3 -> N = 1873
k = 4 -> N = 1891
It says it's not easy to remember, so this probably rules out 1819.
So it's either 1837, 1855, 1873, or 1891. I'm guessing that "It's not easy to remember" also means that it can't have any repeated digits. That rules out 1855 and 1891. So either 1837 or 1873. Not sure how to choose between these two.
Maybe it's 1873 because the kid is 18 and the dad is 73 years old?
Edit: Just saw your comment about Heineken, completely missed that part the first time I read your post lol
N = 1000A + 100B + 10C + D
A + B = 9
C + D = 10
(Does this mean the 10s digit is odd?)
C = 2k + 1, where k = 0, 1, 2, 3, or 4
A = 1
?
Not sure what the last two mean, but let's start with what we have.
A + B = 9
-> 1000A + 100B = 900A + 100(A + B) = 900A + 900
C + D = 10
-> 10C + D = 9C + (C + D) = 9C + 10
Substituting back in, we now have:
N = 900A + 900 + 9C + 10 = 900A + 9C + 910
Back to the clues:
A = 1
-> N = 900 + 910 + 9C = 1810 + 9C
C = 2k + 1
-> N = 1810 + 9(2k + 1) = 1810 + 18k + 9 = 1819 + 18k
Now, we know k = 0, 1, 2, 3, or 4.
k = 0 -> N = 1819
k = 1 -> N = 1837
k = 2 -> N = 1855
k = 3 -> N = 1873
k = 4 -> N = 1891
It says it's not easy to remember, so this probably rules out 1819.
So it's either 1837, 1855, 1873, or 1891. I'm guessing that "It's not easy to remember" also means that it can't have any repeated digits. That rules out 1855 and 1891. So either 1837 or 1873. Not sure how to choose between these two.
Maybe it's 1873 because the kid is 18 and the dad is 73 years old?
Edit: Just saw your comment about Heineken, completely missed that part the first time I read your post lol
Thanks for the reply, well that shows me that there is a mathematical way around it.
With the two left it may be that a number comprised of two pairs of small / large digits like 1837 would be easier to remember than a mirrored number of small / large, large / small like 1874
cpp_is_king
Member
Could be like an accelerated class. Coincidentally, I'm actually teaching my own 8 year old about equations, variables, and stuff like this right now.
The hardest part is getting him to pay attention, but if he does he understands everything. He wouldn't be able to do this specific question (yet), but I could get him there in a couple more months.
The "maths" are easy enough, but the fact that you need to know about beer/ship...
But take the clues out of order:
It's between 1000 and 2000. So it's 1xxx or 2000.
The thousands and hundreds proves it's 18xx
The odd tens says it's 181x 183x 185x 187x or 189x
The tens and units gives 1819 1837 1855 1873 or 1891.
In the right order, it's quite trivial. The last part, though... Google the numbers?
Hello. Can someone please help me with a stats question?
so i'm working on a small research project where i'm trying to determine whether or not abnormal values collected in the daytime also correlate to a higher likelihood of abnormal values collected at nighttime. i was told that i would need to use a chi square test to make this determination?
so far i have my table in excel set up like:
person............daytime value.....nighttime value
1....................normal...............abnormal
2....................normal...............normal
...300
help is much appreciated. haven't taken stats in about 6 years unfortunately.
so i'm working on a small research project where i'm trying to determine whether or not abnormal values collected in the daytime also correlate to a higher likelihood of abnormal values collected at nighttime. i was told that i would need to use a chi square test to make this determination?
so far i have my table in excel set up like:
person............daytime value.....nighttime value
1....................normal...............abnormal
2....................normal...............normal
...300
help is much appreciated. haven't taken stats in about 6 years unfortunately.
cpp_is_king
Member
Im not formally trained in stats, so I cant comment on how or even whether to apply a chi square distribution, but to me this sounds like aa textbook case of conditional probability. Let P(A) be the probability of abnormality in the day and P(B) be the probability of abnormality at night. Then you would be looking for P(B|A) and comparing it to P(B). Normally this means finding out how many standard deviations P(B|A) is away from P(B), Maybe the chi squared cones into play when doing this step?
Hm I see and is there a way to test this using excel? That's currently where all my data is right now
cpp_is_king
Member
Code:
A B C
------------------------------------------------
1 | P(B)
2 | P(C)
3 | P(B ∩ C)
4 | P(C | B)
5 | person daytime value nighttime value
6 | 1 normal abnormal
7 | 2 normal normal
8 | 3 abnormal normalB1: =COUNTIF(B6:B8, "abnormal") / ROWS(B6:B8)
B2: =COUNTIF(C6:C8, "abnormal") / ROWS(C6:C8)
B3: =COUNTIFS(B6:B8, "abnormal", C6:C8, "abnormal") / ROWS(B6:B8)
B4: =B3 / B1
I haven't tested this out, but I think it should work. The idea is to compare the values in B4 and B2. If there's a correlation, B4 should be higher significantly (in the statistical sense) higher than B2
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