solving equations but they get increasingly more impossible?
ฝัง
- เผยแพร่เมื่อ 20 พ.ค. 2024
- We will solve 4 impossible-looking equations, sqrt(x)+sqrt(-x)=2, ln(x)+ln(-x)=0, e^x+e^(-x)=0, and sin(x)+sin(-x)=2. From verifying the domains of the functions to finding real solutions, we go step-by-step through each equation. And just when you think you've got it figured out, we hit you with one more equation that's sure to blow your mind! Get ready to flex your mental muscles and solve some of the trickiest equations out there. #math #equation #blackpenredpen
0:00 sqrt(x)+sqrt(-x)=2
3:27 ln(x)+ln(-x)=0
6:10 e^x+e^(-x)=0
10:29 sin(x)+sin(-x)=2
Here's the video for Q4 • sin(x)+sin(-x)=2
🛍 Shop math t-shirts & hoodies: blackpenredpen.creator-spring...
10% off with the code "WELCOME10"
----------------------------------------
💪 Support the channel and get featured in the video description by becoming a patron: / blackpenredpen
AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls Jon Daivd R meh Tom Noa Overloop Jude Khine R3factor. Jasmine Twinkletoes
----------------------------------------
#maths #equations #blackpenredpen
Putting these no solution questions as a pre-cal bonus question saying "graph the function" would be humorous when the people who don't know what to do leave it blank and then you mark it correct.
😆
I would still manage to fail by drawing some bullshit curve 🤣
@@janfilby7086 😂😂
@@blackpenredpen I got x equals 2 divided by i when I squared it..you didn't mention that? Why not?
@@leif1075 If you multiply 2/i by i/i, you get -2i. They are equivalent.
The last one has no solution at all because sin(-x) is equal to -sin x so the equation sin x+sin(-x)=2 is the same as saying sin (x)-sin(x)=2 now we can cancle the two sines and we get the equation 0=2 and it has no solution.
thanks, i forgot that sin(-x)=-sin(x)
so it seems that sin(x)+sin(-x)=/=2
Exactly what I got! But, in a slightly different way.
So, we have sin(x) + sin(-x) = 2
We know that sin(a) + sin(b) = 2sin(a+b/2)cos(a-b/2)
So, applying this, we get,
2sin(x-x/2)cos(x+x/2) = 2
2sin(0)cos(x) = 2
0=2
The fact that sin(x) is an odd function struck me much later, and now I feel I wasted time doing all this😂
@@notmuchgd9842 Don't think that's what is intended. Instead, it would still be sin(x) + sin(-x) = 2 (this comes from how we defined the separate functions) but f(x) doesn't have a solution.
What you're saying is like saying x^2 =/= -1 just because you can't find a solution in the Real realm. So, maybe the other function has a solution in a set of numbers we haven't discovered yet👀
Is sin(x) odd also in complex field? I always forget XD
@@alexsoft55 Pretty sure yes. Sin and tan are odd and cos is even, even in the complex world
Here's an alternate take for the exponential equation.
That parabola-like curve is a catenary; and eˣ + e⁻ˣ = 2cosh(x); twice the hyperbolic cosine. But from Euler's formula, (circular) cosine can be written
cos(x) = ½(e ͥˣ + e⁻ ͥˣ) = cosh(ix); likewise, because cos and cosh are even functions,
cosh(x) = cos(-ix) = cos(ix)
So eˣ + e⁻ˣ = 2cosh(x) = 0, means that
2cos(ix) = 0 = 2cos(-ix)
But we know where cosine is 0:
-ix = (n+½)π ; x = (n+½)iπ
And that's another way to solve this one.
Fred
PS. Great idea, this set of problems!
Thank you!!
@@blackpenredpen 😇
Hey Fred,
How did you get from squareroot(x) + squareroot(-x) = 2 to e^x + e^-x = 2 ? Thanks!
Fred one more question:
What math topic should i study to be able to make these connections you made between e and cosine and that other one coshine? I want to be able to learn that topic and answer as you did!
@@MathCuriousity Thanks for the questions. I think I can speak for bprp as well as myself, that it's encouraging to hear from those who are genuinely interested in learning. We must all strive for that!
"How did you get from √x + √-x = 2 to eˣ + e⁻ˣ = 2 ?"
I didn't. The former was Q#1 in the video; the latter is Q#3, which is why I referred to it as "the exponential equation."
"What math topic should I study ... coshine?"
-- BTW, it isn't "coshine;" it's cosh, which is short for "hyperbolic cosine."
I'm not sure what name that topic might go by today, but in my school days, it would be either Algebra 2, Advanced Algebra, Complex Algebra, or Analysis, the last of which nowadays goes by the name, "pre-calculus."
This is unreal 😂
Pretty complex too in parts!
Good observation Dr Pun-yam
Imagine that.
Should this video be called unreal tournament?
For #3, it’s interesting that if you substitute e^i(pi) = -1 too soon or too late, you get stuck with a tautology where x = anything.
For #4, you can analyze the series expansions and see that no integer values of n produce any totals that coincide between the two series. The first series is pi times … -1.5, 0.5, 2.5… and the second series is pi times … -0.5, 1.5, 3.5….
@@Paul-222 So I solved that using sin(x) = ((e^ix)-(e^-ix))/2i and got 0=4i, then I saw in the other comments that the oddness of sin gives you 0=2 immediately and I felt dumb for overcomplicating so much, but I'm glad to see someone else made it even more complex.
It looks like sin(x)+sin(-x) = 2 has no solutions since sin(x) is an odd function, even with using Euler's formula it leads to a contradiction of 0=2.
From what I can tell, Euler's formula is just how to separate the even and odd parts of the exponential as two coordinates, just specialized for the imaginary case, even if there are two other cases which are just as interesting.
j² = 1: e^jφ = coshφ + jsinhφ (hyperbola)
ε² = 0: e^εφ = 1 + εφ (flat line)
i² = -1: e^iφ = cosφ + isinφ (circle)
you just apply sin(-x) = -sin(x) and that's it
@@angeldude101 why j as imaginary unit ):
@@tobyayres5901 You'll need to more precisely define "imaginary" in that question. I said "j² = +1", which is very much not the imaginary unit you're familiar with. This is a _hyperbolic_ number, not a _complex_ number.
@@angeldude101 is that hard and when u have to study it
-a highschool student
If you, like me, are bothered by the fact that the process in part 1 doesn't yield both solutions, then read on. The reason is because a number has two square roots, and complex numbers don't have a preferred one of the two like positive real numbers do - you can't just take the "positive" square root because most complex numbers don't have a purely real, positive square root. More specifically, the problem is in the step √[-x] = i√x. Check this out:
√[-(-x)] = i√[-x] = i*i*√x = -√x.
So √x = -√x, which is absurd. The way to resolve this is that √[-x] may be either i√x or -i√x, depending on which square root of a number is being chosen.
Please see my solution to the first equation in the comments for clarification on this point.
Mathematician: "Infinity is not a number, therefore this equation has no solution"
Me, a physicist: "I have no such weakness"
Mathematicians sometimes use infinity as a number. But they're careful to specify WHICH infinity they're using and how it plays with the finite numbers.
For the e based expression, I used cosine def in terms of e so we get (e^x + e^-x)/2 = 0 = cos (-ix)
Take the inv cos from both sides to get pi/2 = -ix and solve for x to get -i*pi/2
This is honestly one of my favorite videos of yours - it’s very clear and concise but still enough content to fill a 10-minute video! All of these sections are different but have the same theme so it still feels like one video - my rating is (pi^2 + 1)/10
On a scale of 1-10 that's a LOW score :(
If you divide by 10, then yes
@@ajl4878 Ah, I see it now, it's _out of_ 10 lol
@@jumpman8282 yea lol
That's 10.86 out of 10 lol
İ can do that! İn the last question, it have no solution. Because if we try to do that, we get 0=2 and that was an issue.
2:45 the second solution can also be found by observing that taking out a factor of i and taking out a factor of -i are both equally valid starting points ((-i)^2 = i^2 = -1). So in reality, you needed to take out a factor of +-i rather than just +i
Very interesting video, I wish I had a calc 2 professor like you. I had a hard time passing that course. But I still love math, and I enjoy watching your videos!
**gets paper ready for last question**
**realises sine is a strictly odd function**
nvm
sin(x)+sin(-x)=2 sin(x)-sin(x)=2 0=2. There are no solutions.
yop
the last one has no real nor complex solution, because if we use the complex definition of sinx, we have (e^(ix)-e^(-ix))/2i+(e^(-ix)-e^(ix))/2i = 2, then e^(ix) - e^(-ix) + e^(-ix) - e^(ix) = 4i, which LHS cancels out to 0, we have 0 = 4, no solution
Right, only you forgot an i in the definition of sin(x) = (e^ix - e^-ix) /2i
@@meurdesoifphilippe5405 *cough cough* no one saw that
Or you could use the fact that sin(x) is odd, meaning sin(x) + sin(-x) = 0 for all complex x. No calculation required.
@@logicxd1836 See what? ;p
The domain remain same as it is an example of the 2 graphs same ranges too.
Common domain is: all real numbers.
Common range is:[-1,1]
But they are just image of one another if rotated across y axis by 180°.
This also proves that sin(x) is a odd function.
The solution of e^x + e(-x) = 0 is straightforward if thinking the two terms as two vectors in the complex plane, having the same module. For simmetry, their phases should be pi/2 and -pi/2
.. and so:
1) Re(x)=Re(-x)
2) Im(x)=i*pi*(2n+1/2)
3) Im(-x)=-i*pi*(2m+1/2)
Where n and m can be any integer.
Since Re(z)=-Re(z) for any complex z, 1) implies Re(x)=0.
Since the same goes for the imaginary part of the complex number, i.e Im(-z)=-Im(z), 2) and 3) can be combined to:
4) Im(x)=i*pi*(2n+1/2)=i*pi*(2m+1/2)
So n and m needs to be equal, and the final solution is (still): x=0+i(1+4n)pi/2 with n being any integer.
Answers (in Cartesian form):
1. x = -2i, x = 2i
2. x = -i, x = i
3. x = 1/2 i (2 π n + π), n element Z
4. False
Love looking all these. I'm looking these years later but they are so great videos
e^x + e^-x = 0
"It's almost like cosh" It is exactly 2cosh(x). cos(x) = cosh(ix), and we know plenty of places where cos(x) = 0.
sin(x) + sin(-x) = 2
Where cosine is the even part of the exponential, sine is the odd part, so sin(-x) = -sin(x). So sin(x) + sin(-x) = sin(x) - sin(x) = (1-1)sin(x) = 0 ≠ 2 for all x. It isn't just never 2, it's never _not 0._
I have no idea where to start with the last problem, it's so odd tbh
Pun intended?
Yup
8:45 can you use the Lambert W function if you multiply both sides by x? That yields an extraneous solution x=0 for one of the branches but I was wondering if this method is viable here if you know what you’re doing
I was curious about this as well
ok so multiply both sides by x≠0 :
xeˣ = -xe⁻ˣ
now use the lambert W function :
x = -x ⇒ x = 0
but we assumed that x cannot be 0.
Hence it’s not a good thing to use here
@@rshawty it's fine to use it, just get rid of the x=0 solution because we already clarified that x≠0 for the equation we're trying to solve
@@farklegriffen2624 ok but you can clearly see that’s useless
@@rshawty It isn't useless. The Lambert W function does not allow you to conclude that xe^x = -xe^(-x) implies x = -x. It only allows you to conclude that x = W(-xe^(-x)), and you must remember this is multivalued.
Holding the poceball gives him the ultimate mathematical power
The answer you gave ,I understood but x=-2i ,so there must be a conjugate of the complex solution ,
So it's equtions are:
x=2i,x=-2i.
8:40
e^x + e^-x = 0
e^x = -e^-x
x = ln(-e^-x)
x = ln(-1) + ln(e^-x)
x = pi*i+2n*pi + -x
2x = pi*i + 2n*pi
x = n*pi + pi/2 * i
4:39 How to draw a Cartesian plane properly. Also, how to describe the inside of a black hole.
f(x) = ln(x) + ln(-x)
Petition to make I dont like to be on the bottom, I like to be on the top merch.
You make me smile with each video.
Thank you
Glad to hear. Thank you.
9:40 I love that so much
For the first equation we can square both sides twice to get a quadratic equation when we solve it we'll get ±2i directly
If you replace the x by z and assume complex solutions, would z = +/-2i be considered valid? I squared both sides, isolated SQRT(-z²) and squared again to find the two candidates
Junior Math: 1+1
Highschool Math: ax^3+bx^2+xc+d
University Math: I don't know anymore. - Ho Lee Fuk
6:21 6:37 But ♾and -♾ are both numbers! They are also solutions
7:36 that's a hyperbolic cosine function: 2*cosh(x)
He said “it’s almost like a cosh function”, but yeah, it actually IS a cosh function
Great examples!!!
Hey, i’m in algebra1 right now and i have been thinking about testing out of some math classes in the future and siblings/friends have told me that pre calc is just algebra2 a bit more advanced trigonometry and a few more concepts? If i end up doing well in algebra 2 and i feel confident with my algebra skills would it be a good idea to test out of pre calc?
I have also been wondering if it is a good idea to test out of geometry. I have been studying a lot and I so far have learned about some trig (just basics sin, cos, tan, unit circle etc). I have been averaging roughly 75-90% on some online practice tests.
I also plan on going for some sort of stem major at a top school (like Caltech). The careers I have thought about so far are astrophysics, or quantum mechanics. I know that is a long ways a way but I think its kinda fun to think about and I hear about people not knowing what to do when they hit 12th grade so it cant hurt.
I might be wrong, but I've always thought it wasn't correct to use SQRT and LN with negative numbers, thought there are complet solutions. So is it "correct" to write SQRT(-1)=i and ln(-e)=1+iPi for example? IDK if it's only to avoid confusion, but all math teachers keep using the definition of i as i²=-1, but never SQRT(-1)=i
7:47 you said it yourself it's literally a cosh function, exactly 2cosh(x) 😆
No entiendo nada lo que dice pero sí entiendo todo lo escribe , excelente canal 👌👌 , saludos desde Perú.
Last one is always 0 regardless of x. Sin is an odd function so you can move the - to the outside. The sin(x) terms then cancel, leaving the equation 0=2. Therefore, sin(x)+sin(-x)=2 has no solutions at all.
Can you write in text what you are saying when you say "if you ever feel the need to graph the __________" (axis meeting point, red dot) around 1:19? I can't make out what you are saying. Thank you so much for great videos!
Wait for that e^x + e^-x question at the end can't you just leave it as e^iπ and take the ln to get 2x = iπ so x can be iπ/2 cuz i am not sure why you added 2πn, i understand basic calculus so i understand how Euler's number works but that 2πn is it really necessary to add ?
for the third one, that is e^x + e*(-x) = 0, when we get to the step: 2x = ln(-1), why cant we write it as: 2x = ln(I^2) = 2ln(i), therefore x = ln(i)?
(idk that much about logs or calculus or anything of that sort, so if I make a mistake please excuse me)
Hey blackpenredpen! I am a huge fan, will do explain this problem for me please?
When is the product of X1 and X2 maximum given that the function is f(x)=(X-X1)(X-X2) (derivative is not allowed but I'd like you to do it as if it is at first). I have a divine understanding of parabolas and I only got closer to the answer but never actually got it. I'd appreciate it you would explain that for me, thank you.
im in absolutely no mood for writing down the solution after banging my head on the last problem for an hour, but yeah, so i first did Euler's form, then i didn't simplify the iota terms i.e.- i didn't write e^i³x as e^-ix and proceeded then i wrote the inverses as fractions and took LCM, i substituted e^ix as a, then proceeded, and with some trivial calculations i got my answer as -ln(0)/i i know it was a pretty simple method but I JUST BROKE MATHS !
sin(x)+sin(-x)=sin(x)-sin(x)=0 so 0=2
simple cheat:
e^x+e^-x=0 divide both sides by 2 and know that cos(x)=(e^xi+e^-xi)/2
so cos(x)i=0
i have also made math problem witch i think is hard:
proof that:
2ln((2cos(ln(i))+sqrt(2cos(ln(-1))-2))/2)=pi
after getting e^2x = -1
I just set x = i(theta) /2
then original function can change to e^i(theta) = -1, theta = pi + 2npi
then x = i(pi + 2npi) /2
I love how he was able to cheer me up enough to make me smile in the first 5 minutes
Love your videos 🦉
On the surface,
sin(-x) = -sinx, therefore
sinx + sin(-x) = sinx - sinx = 0
therefore sinx + sin(-x) != 2 for all x in R
QED
But I know nothing about the complex sine function, so I'll leave to the smartheads to figure it out 🤓
If sine retains it's oddness as a complex function, I guess there's no complex solutions either.
Edit: Actually, I just checked and the Taylor expansion of the sine function has only odd degree terms, so yeah, it retains it's oddness. So no solutions at all! Postscripty QED.
Here's another way for number 2, when we get to ln(-x^2) = 0, can't we do ln(-1) + ln(x^2) = 0, and ln(-1) is i*pi because of e^(i*pi) so in other words i*pi + ln(x^2) = 0; ln(x^2) = -i*pi; x^2 = e^(-i*pi) = (e^(i*pi))^(-1) = (-1)^(-1) = 1/(-1) = -1; so therefore x = +-sqrt(-1), aka +-i
Isnt last one sin(π/2+2πk) which always will give one and sin(-3π/2+2πk) which will also end on the positive side of the circumference as its going to the negatives and then the postives? Reasoning like that it would be 1+1 =2
I thought I was crazy with the last question! thanks everyone for confirming my suspicions.
For those who wonder why 𝑒^𝑥 = 0 doesn't have any complex solutions:
If it did have a complex solution then there would exist two real numbers 𝑎 and 𝑏 such that 𝑒^(𝑎 + 𝑏𝑖) = 0
But 𝑒^(𝑎 + 𝑏𝑖) = 𝑒^𝑎⋅𝑒^(𝑏𝑖) = 𝑒^𝑎(cos(𝑏) + 𝑖 sin(𝑏)), and there is no real value 𝑎 such that 𝑒^𝑎 = 0
and also no real value 𝑏 such that cos(𝑏) = sin(𝑏) = 0
If it can be written as an infinite polynomial, wouldn’t we get infinitly many sol. I think i see the falicy in my logic but…
@@schizoframia4874 that's an interesting question. Like, what happens to the roots of the taylor polynomial as the number of terms approaches infinity
@@alejrandom6592 thanks
That eˣ=0 has no solution is equivalent to the well known fact that ln 0 is not defined.
@@schizoframia4874 Calling it an infinite polynomial is inaccurate. TH-camrs often say it is an infinite polynomial for the sake of analogy, but the problem with analogies is that they are imperfect and flawed and not an accurate description of what is happening. They are there to aid your intuition, not to give you an accurate understanding.
For the third one we should go for the definition e^ix
It's still no sol I think. Correct me if I'm wrong
@@HershO. It's the 4th that has no solution, because (hyperbolic) sine is odd no matter what you through at it. You're basically solving sin(x) - sin(x) = 2. The third one is perfectly fine and it's just saying 2cosh(x) = 2cos(ix) = 0.
@@angeldude101 oh sorry I thought they were talking about the 4th one
Can you do a series about problem solving involving exponential growth/decay? Thank you!!!
Problem:
Rhyz and Zhayn lives in an island-town with population of 2000 people. They came back from vacation to the island but they catch the highly-contagious COVID-19. A week after their return to their town, they infected 6 more people.
a. How many will be affected after another week (assuming no health protocols have been practiced.)? Their public health center decides to isolate their town once 30% or more of their people are infected.
b. After how many weeks will the public health center isolate the town?
On the first question how do you know which square root of the complex numbers to take, (which is the principal solution)?
Please see my solution to the first equation in the comments for clarification on this point.
Hey btw a good idea for a video is the integral from - to + infinity of sechx is π
For the equation e^x = -e^-x could you not just multiply both sides by x giving you xe^x = -xe^-x then take the Lambert W function: W(xe^x) = W(-xe^-x) giving you x = -x therefore x = 0 ?
A neat trick with the last one, if we can argue there is no real component to x, it follows x = i*|x| so this just becomes 2cos(|x|) = 0 ie. |x| = (n+1)*pi/2
There neither real nor imaginary solution because for any x, sinx+sin(-x) =sinx-sinx=0
Would you please talk about this topic?
As we all know for x approches infinity:
(1+1/x)^x=e and
(1-1/x)^x=1/e
Multiply above equation both side will give us:
(1-(1/x)^2)^x=1
... as if the value of (1/x)^2=0
Can we then define that when x approches infinity (1/x)^2=0?
To your last question - yes
me watching this as if i understand whats going on😂
When I taught algebra in college, I always checked my solutions to make sure they weren't extraneous. I would like to have seen that here, especially on the second equation.
f(i) = ln(i) + ln(-i)
f(i) = iπ/2 - iπ/2
f(i) = 0, therefore i is a solution of f(x) = 0
Since it's symmetrical, -i is a solution too.
sir what is the integration of 1/(sinx+cosx)
@7:46 - considering cosh(x) = (e^x + e^(-x))/2, it's exactly like a cosh function.
I suppose he wasn't completely wrong. It's not _exactly_ cosh(x), but rather 2cosh(x).
For the first one, sqrt(x) + sqrt(-x) = 2, how can you find 2 with the solutions?
Like, if you take 2i, you get:
sqrt(2i) + sqrt(-2i)
sqrt(2i) + i×sqrt(2i)
(1+i) × sqrt(2i)
And then I'm stuck, and the same with -2i. How do I get 2 from that?
sqrt(2i) is either 1+i or -1-i since (1+i)^2 =1^2+2i+i^2 =2i
while sqrt(-2i) is either 1-i or -1+i
If you define sqrt(2i)=1+i and sqrt(-2i)=1-i the equation works but I think those different square roots can cause problems in other equations.
1:49, that's when you lose the other solution, sqrt(-1) is i OR -i, so you should get two equations
Exactly. Without that assumption the solution is not complete.
For the first equation, noticing the symmetry is the more elegant way to find the second solution, but you could also just pull a -i out of sqrt(-x), since (-i)^2=-1 as well, and then solve sqrt(x)-i*sqrt(x)=2.
Or, if a complex number is a root of a polynomial, then its conjugate is a root too
Clever but sort of misdirection on ex.1 converted two separate equations into one simultaneous equation. Different animals
😊
Using the evenness of the sine function, sin(-x) is the same as -sin(x).
The equation sin(x) + sin(-x) = 2 becomes: sin(x) - sin(x) = 2 => 0 = 2 [->
When you consider the complex definition for sine, sin(x)=[e^(ix)-e^(-ix)]/2i, then it’s clear when you plug in sin(-x) that sin(-x)=-sin(x) ∀x∈ℂ, therefor sin(x)+sin(-x)=0≠2 ∀x∈ℂ
Fun fact: There's nothing more impossible than impossible because it's already impossible, and if something was posed as impossible and it was done means it's actually possible
The last example has actually +- sign because of the symmetry mentioned in the first example
Somebody knows where you can buy the framed e from the background?
A doubt. Why e^ipi+2npi? When without the 2npi the answer would be ipi/2?
maybe you guys think i am crazy but please correct me. for me my first intuition direktly was that you have to turn the whole plane by 180deg so both of the graphs would interfere to zero. but to turn the whole coordinate by 180deg you got to do that in an eytra dimension eg the imaginary plane.
In the first equation, dont you have to introduce + and - i, when pulling the i^2 from under the root?
No you do not, for example if you simplify sqrt(8)= sqrt(4)*sqrt(2)= 2sqrt(2) not -2sqrt(2). You consider the + or - signs when your solving for the value of x when taking even numbered roots. Keep in mind the + or - signs come from the absolute value of x. |x|=sqrt(x^2)
Yes, you do. That is why blackpenredpen did not get the solution 2i originally--he left out the + or -.
@@moeberry8226 That does not work for complex numbers. There is no obvious preferred principal square root, since there is no positive or negative among complex numbers. The standard for "principal square root" is to take the root with positive real part, and if both roots have zero real part then take the one with positive imaginary part.
@@rorydaulton6858 I understand that I was giving ano a very easy example with respect to the reals. So it can be shown more clearly. But 100 percent your right there is no principle square root that’s preferred when in the complex world. But in this case we are not solving or finding a principle root at the part Ano is talking about. We take sqrt(-1) to be just i and at the end as blackpenredpen showed we have symmetry along with the fact of the conjugate root theorem which states if a+bi is a zero then a-bi is also.
@@moeberry8226 couldn't i do the following:
sqrt(-1) = sqrt((-1)(-1)(-1)) = sqrt(i*i*(-1)(-1)) = sqrt((-i)(-i)) = sqrt((-i)^2) = -i
but since: sqrt(-1) = i
I have to consider both +/-?
For the last one after writing e^x = -e^-x , multiple by x on both sides and take lambert w on both sides you get x=-x => x = 0 . But 0 isnt the answer so where does this go wrong cuz I feel like the steps seem legit
Omg I was so laughing after the end. Dude, you are awesome in humour the same in math xd
I have the best solution for the third one:
e^x + e^-x = e^-i i x + e^i i x = 0
this is already one of eulers identities but ill write it out anyway:
cos(-ix) + i sin (-ix) + cos(ix) + i sin(ix) = 2 cos(ix)=0
cos(ix)=0
ix = pi/2 + n pi
x=-i (pi/2+n pi)
Alternatively, just use cosh identities: e^(x) + e^(-x) = 2cosh(x) = 2cos(ix). But this is a nice way to derive the same thing with more fundamental identities, without needing cosh.
@@nG27227 I didnt know that identity but pretty interesting :)
7:46 In fact, double the cosh function!
Without watching, I see that it is c(1+i)=2, where c=√2. So the left equation is a line through 1+i from the origin. A 45 degree line in the 1st and 3rd quadrant of a graph. The line obviously goes through y=2 at x=2. So the point at which the equation equals two will occur at √2+√2i.
I was so wrong. I was taking the absolute value of the left side. Ignore my ideas, the video is good.
lets consider the essence of sin(x)=-sin(-x), expand sin x as its taylor series, thus we can see sin(x)=-sin(-x) if (-x)^k=-(x)^k and k is odd. but i think quaterions dont obey rule since they break distributive law. so the answer might be a quaterion
you and rednilebluenile are best two people. both make videos of my fav subject
sin(x)=1 x=π/2+2nπ n∈ℤ
sin(-x)=1 x=3π/2+2mπ m∈ℤ
π/2+2nπ=3π/2+2mπ
1/2+2n=3/2+2m
1/4+n=3/4+m
n=m+1/2
Considering n,m∈ℤ, we can conclude there are no real solutions.
But what about sin of complex numbers?
solution for last one:
sinx + sin(-x) = 2
the angle -x lies in the 4th quadrant, and here sin is negative, hence the equation can be reformed as
sinx + (-sin(x)) = 2, giving
sinx-sinx=2, i.e 0=2
Hence the solution set is {ϕ}
I was feeling really smug about solving these mentally, but had to break out the whiteboard for the last one. I was getting quite frustrated until I decided to accept what I thought were failures as proof that it's impossible.
i immediately noticed e^x + e^(-x) is just 2cosh(x)
What were your subject combination in bachelor
I encourage you to continue, and I hope you solve it this limite: lim_(x->0) ((x^π - π^x)/(x^e - e^x))^(1/sin(x))
for exp(x)+exp(-x), I would have substituted x = iy and then you have a 2cos(y) which obviously has many zeros.
Can anyone explain why in the first equation (√x + √-x = 2) the process didn't yield also the 2i result? It's fine saying the function is symmetrical, so we need to add it, but shouldn't the methodical way produce both results?
it's because you can't turn √-x into i√x if x is negative, (2i is negative, -2i is not) therefore the only answer that you can get is -2i
Good bro👍🤘
Square root of X raised to the power of 2 is module of X so that's why you get ±2i.
I know how to find the complex solutions. But what confuses/fascinates me is, what's the point of those solutions?
La primera ecuación no me da igual a 2 con las soluciones x=+-2i... solo funciona con -2i
sin(-x) = - sin(x)
sin(x) + sin(-x) = 2
sin(x) - sin(x) = 2
0 = 2
False
We can use the fact that sine is odd and we have 0=2 so we will get contradiction - no solution in both real and complex