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.
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!
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."
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.
@Venkata Bhamidipati well not in a matmatical set as is prove in a very old book that the set of the intersection of 2 diffrent numbers {a}, {b} is the empty set as the set solution to the question 0 = 2 is the intersection of {0}, {2} which is the empty set which you can say is a number but then you wouldnt be able to do simple arithmatic
I did it on complex numbers, ended up with 0=4i then I realized sin(x) is an odd function and that makes the initial equation just 0=2. I got pretty mad lol
@@alexsoft55 Yes, sin is an odd function for complex numbers as well. You can see this by looking at the Taylor expansion of sin: sin(x) = x - x³/3! + x⁵/5! … Notice that if you use that for sin(-x) that since all the powers of x in the expansion are odd all the signs flip and the formula is otherwise the same, meaning -sin(x) = sin(-x).
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
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.
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
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)
@@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.
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.
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
@@samuelgomes4288 yes, taking the square root of a complex number is ambiguous, or in other words taking the square root on complex numbers is not a function. Every complex number has two square roots (which might be the same in case of 0). It is a common misconception to say that i is the square root of -1, i is ONE of the square roots of -1, the other one is obviously -i.
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.
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.
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
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!
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
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 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.
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._
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.
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.
@@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.
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.
At 1:17, what are we going to get? It sounds like "the woodshed" or "the wart's head", I can't tell what word it is. I love that he concludes the entire function is just a point.
Я знаю причину по которой у вас не было второго ответа (2i) в первой решении, вы посчитали, что √-1 = i, но ведь на самом деле √-1 = ±i, если бы вы это учли, то как раз бы получили второй ответ
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
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.
For the last one, the equation will always gives 0 = 2, because the sine function is even. I also tried using complex exponential form to confirm this result and It should be correct
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 !
for all inputs, real and complex, sin(z) = -sin(-z) sin(z) + sin(-z) = 0 identically for all complex inputs. Hence the last equation has no solution at all over the reals or the complex numbers
@@MathCuriousity the reason is, sin(z) is entirely analytic over the complex plane, hence we don’t need to deal with any problems such as branches - and hence still maintains the odd property (u can even see this using a Taylor Expansion in z, which has infinite radius of convergence) for all z in the complex plane. Hence for the exactly the same reason as no real solutions x, there are no complex solutions z either
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
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
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?
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
@@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.
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?
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 ?
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.
For the first one, I think you forgot to put the absolut value of x, after doing (√x)². From that point, you'd get both 2i and -2i values of x, instead of having to analyze it from its beggining to find the other one.
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
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.
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.
I mean the sine one question as the sine waves on +&-y axes nullify each other to give: f(x)=sin(x)+sin(-x) = f(x)=0 Anyways that was obvious the other way as well cuz sin(-x) = -sin(x)
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)
@@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 +/-?
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
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?
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.
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
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!
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
The first one has another take (Simplified) :- Let the equation √x+√-x=2 be true Therefore, √x+√x*i=2 As the LHS is a complex no. While the RHS is a real no. Therefore the equation is false (Proved)
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.
Not any complex solutions, since spherical magnitude is defined as r² = x² + y². Hyperbolic magnitude however is r² = x² - y², meaning we need to use the split-complex numbers where j²=1, j≠±1, |±j|² = -1. Edit: it occurs to me that that's only magnitude squared, so the actual magnitude would be imaginary rather than negative... Sorry about that.
my teachers say it’s really really bad when you write square root of a negative number or of a complex number, because in this case it’s not a function, its a multivaluated function and √x+√(-x) is a set. So for the first one wouldn’t it be better to write the question as “find x so that 2∈{√x+√(-x)} “ or something like that ?
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)
Sin(-x)=-sin(x) Basically Sin(x)+Sin(-x)=Sin(x)-Sin(x) So it's equal to zero. That means we're stating that: 0=2 This is Contradiction, as such, this equation has no solutions.
Not exactly this, but it's simmilar, the step near end sqrt(x)²=x is correct, because x is complex and you don't have to take absolute value. Now, exactly this but in reverse happened here sqrt(-x)=isqrt(x), so it's saying √-1=i... But actually it can be -i as well, because (-i)²=-1 too, so it should be sqrt(-x)=±isqrt(x)
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.
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."
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
Is sin(x) odd also in complex field? I always forget XD
@Venkata Bhamidipati well not in a matmatical set as is prove in a very old book that the set of the intersection of 2 diffrent numbers {a}, {b} is the empty set as the set solution to the question 0 = 2 is the intersection of {0}, {2} which is the empty set which you can say is a number but then you wouldnt be able to do simple arithmatic
I did it on complex numbers, ended up with 0=4i
then I realized sin(x) is an odd function and that makes the initial equation just 0=2. I got pretty mad lol
@@alexsoft55 Yes, sin is an odd function for complex numbers as well. You can see this by looking at the Taylor expansion of sin:
sin(x) = x - x³/3! + x⁵/5! …
Notice that if you use that for sin(-x) that since all the powers of x in the expansion are odd all the signs flip and the formula is otherwise the same, meaning -sin(x) = sin(-x).
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
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.
This is unreal 😂
Pretty complex too in parts!
Good observation Dr Pun-yam
Imagine that.
Should this video be called unreal tournament?
Wow
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
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
**gets paper ready for last question**
**realises sine is a strictly odd function**
nvm
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.
Thanks, that was bothering me.
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
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.
Thanks for the comment. I was wondering why algebra doesn't return the two solutions at first
@@samuelgomes4288 yes, taking the square root of a complex number is ambiguous, or in other words taking the square root on complex numbers is not a function. Every complex number has two square roots (which might be the same in case of 0). It is a common misconception to say that i is the square root of -1, i is ONE of the square roots of -1, the other one is obviously -i.
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.
Me, sadly putting the Riemann sphere back into my backpack
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.
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
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!
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.
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
I learned this almost 15 years ago and I am satisfied with my answer being 2i. It was a joy to solve this!
Holding the poceball gives him the ultimate mathematical power
Petition to make I dont like to be on the bottom, I like to be on the top merch.
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.
I have no idea where to start with the last problem, it's so odd tbh
Pun intended?
Yup
Junior Math: 1+1
Highschool Math: ax^3+bx^2+xc+d
University Math: I don't know anymore. - Ho Lee Fuk
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._
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.
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
sin(x)+sin(-x)=2 sin(x)-sin(x)=2 0=2. There are no solutions.
yop
For the first equation we can square both sides twice to get a quadratic equation when we solve it we'll get ±2i directly
You make me smile with each video.
Thank you
Glad to hear. Thank you.
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.
I love how he was able to cheer me up enough to make me smile in the first 5 minutes
@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).
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.
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
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)
04:21 a number bigger than zero and smaller than zero: impossible!
4,294,967,295: hold my beer
9:40 I love that so much
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
At 1:17, what are we going to get?
It sounds like "the woodshed" or "the wart's head",
I can't tell what word it is.
I love that he concludes the entire function is just a point.
origin
i immediately noticed e^x + e^(-x) is just 2cosh(x)
Я знаю причину по которой у вас не было второго ответа (2i) в первой решении, вы посчитали, что √-1 = i, но ведь на самом деле √-1 = ±i, если бы вы это учли, то как раз бы получили второй ответ
Love looking all these. I'm looking these years later but they are so great videos
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
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.
No entiendo nada lo que dice pero sí entiendo todo lo escribe , excelente canal 👌👌 , saludos desde Perú.
For the last one, the equation will always gives 0 = 2, because the sine function is even. I also tried using complex exponential form to confirm this result and It should be correct
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 !
for all inputs, real and complex, sin(z) = -sin(-z) sin(z) + sin(-z) = 0 identically for all complex inputs. Hence the last equation has no solution at all over the reals or the complex numbers
I understand why there are no reals, as x can never equal -x logic. But why is there no complex?
@@MathCuriousity the reason is, sin(z) is entirely analytic over the complex plane, hence we don’t need to deal with any problems such as branches - and hence still maintains the odd property (u can even see this using a Taylor Expansion in z, which has infinite radius of convergence) for all z in the complex plane. Hence for the exactly the same reason as no real solutions x, there are no complex solutions z either
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
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.
Be careful, we can only say √(-x) = i√x when x≥0 or Im(x)
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
I thought I was crazy with the last question! thanks everyone for confirming my suspicions.
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 [->
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?
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
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.
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?
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 ?
7:47 you said it yourself it's literally a cosh function, exactly 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.
The last example has actually +- sign because of the symmetry mentioned in the first example
For the first one, I think you forgot to put the absolut value of x, after doing (√x)². From that point, you'd get both 2i and -2i values of x, instead of having to analyze it from its beggining to find the other one.
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
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.
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.
"if you ever feel like the need to graph the origin, go ahead and just use this function"
Last question literally demonstrates *destructive interference* 💀
I mean the sine one question as the sine waves on +&-y axes nullify each other to give:
f(x)=sin(x)+sin(-x) =
f(x)=0
Anyways that was obvious the other way as well cuz
sin(-x) = -sin(x)
In the very end he proves that he is a true math teacher!
3:23 sqrt(x)^2 = +- x
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 +/-?
Hey btw a good idea for a video is the integral from - to + infinity of sechx is π
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
e^x and e^-x is also a symmetric function, why is the negative of that not also a solution?
It is, it's implied by the n which can also be negative
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
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 :)
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
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!
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
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
if x is naiural,this function will always be equal to y+yi because y=√x and yi=√-x
Omg I was so laughing after the end. Dude, you are awesome in humour the same in math xd
A doubt. Why e^ipi+2npi? When without the 2npi the answer would be ipi/2?
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))
Love your videos 🦉
The first one has another take
(Simplified) :-
Let the equation
√x+√-x=2 be true
Therefore,
√x+√x*i=2
As the LHS is a complex no.
While the RHS is a real no.
Therefore the equation is false
(Proved)
7:46 In fact, double the cosh function!
you and rednilebluenile are best two people. both make videos of my fav subject
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.
Square root of X raised to the power of 2 is module of X so that's why you get ±2i.
Can you make a video using the quartic formula ?? 😬😬😬
Does |x|=-9 have any complex solutions ? If it dosent can’t we just invent something like the complex numbers we did for radicals ??
It doesn't, since absolute value is defined as a distance from the origin. Negative distance doesn't make any sense.
Not any complex solutions, since spherical magnitude is defined as r² = x² + y².
Hyperbolic magnitude however is r² = x² - y², meaning we need to use the split-complex numbers where j²=1, j≠±1, |±j|² = -1.
Edit: it occurs to me that that's only magnitude squared, so the actual magnitude would be imaginary rather than negative... Sorry about that.
my teachers say it’s really really bad when you write square root of a negative number or of a complex number, because in this case it’s not a function, its a multivaluated function and √x+√(-x) is a set. So for the first one wouldn’t it be better to write the question as “find x so that 2∈{√x+√(-x)} “ or something like that ?
Your teacher is spot on!
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)
for exp(x)+exp(-x), I would have substituted x = iy and then you have a 2cos(y) which obviously has many zeros.
Great examples!!!
Sin(-x)=-sin(x)
Basically
Sin(x)+Sin(-x)=Sin(x)-Sin(x)
So it's equal to zero.
That means we're stating that:
0=2
This is Contradiction, as such, this equation has no solutions.
For the first one, couldnt you just say there are 2 solutions because sqrt(x²)=|x|? Just curious
Not exactly this, but it's simmilar, the step near end sqrt(x)²=x is correct, because x is complex and you don't have to take absolute value. Now, exactly this but in reverse happened here sqrt(-x)=isqrt(x), so it's saying √-1=i... But actually it can be -i as well, because (-i)²=-1 too, so it should be sqrt(-x)=±isqrt(x)
@@madghostek3026 Ooh, that makes sense! Thanks for the explaination!
doesn’t the last equation give 0 = 2 ??