which equation has no solutions: sin(z)=2, sin(z)=i, tan(z)=2, tan(z)=i?
Вставка
- Опубліковано 27 вер 2024
- which equation has no solutions: sin(z)=2, sin(z)=i, tan(z)=2, tan(z)=i? Sign up for a free account at brilliant.org/... and try their daily challenges now. You can also get a 20% off discount for their annual premium subscription so you can get access to ALL of their awesome designed courses!
thanks for Calvin for the clever way: Another way to see that tan z = \pm i has no solution, is to recall that sin^2 z + cos^2 z = 1. So, if tan z = \pm i, then sin z = \pm i cos z, and sin^2 z = - cos^2 z, so sin^2 + cos^2 z = 0, which is a contradiction
sin(z)=2, • Math for fun, sin(z)=2
sin(z)=i , • how to solve sin(x)=i?
Does tan(z) = i have a solution?
If you enjoy my videos, then you can click here to subscribe www.youtube.co...
T-shirts: teespring.com/...
my site: blackpenredpen...
twitter: / blackpenredpen
Patreon: / blackpenredpen
Blue Microphone: amzn.to/2DAbHYL
Another way to see that tan z = \pm i has no solution, is to recall that sin^2 z + cos^2 z = 1. So, if tan z = \pm i, then sin z = \pm i cos z, and sin^2 z = - cos^2 z, so sin^2 + cos^2 z = 0, which is a contradiction.
Thanks to Calvin for the clever way!
Wanna win a εδ-hoodie?
See the post
ua-cam.com/channels/_SvYP0k05UKiJ_2ndB02IA.htmlcommunity?lb=UgwO8tnhF7RQXDDM30N4AaABCQ
what is "\pm" ?
NoName plus/minus
you can also derive this by looking at the integral expression for arctan(z)=integral (1/(1+t^2)), but this mean that 1+t^2=/0, so t^2 cannot equal -1 (which it is i and -i). You can say that arctan(i) diverges.
--> look at the graph of arctanh(x)
tan(x)=i
sin(x)/cos(x)=i
sin(x)=i*cos(x)
i*sin(x)=-cos(x)
cos(x) + i*sin(x) = 0
Euler formula
e^(i*x)=0
But there is no complex solution to this equation
Wow much simpler
Wow, very nice!!
x--->∞*i would be the solution, but infinity is not a number
@@lilyyy411 So what you're saying is there's no solution lol
Simple way. Why can't I think about that lol
mom: what do you wanna be when you grow up
Me: I want an imaginary friend
blackpenredpen:
I want an imaginary friend, too!
@@blackpenredpen You sir are a real legend.
@@blackpenredpen lol what should we name him?
@@JaydentheMathGuy
I will name him "arctani"
You sure you're not from the shining.....?
*tan(x)=2 or any real number always has a real solution. because Range of tan(x)=(-infinity, +infinity)*
Yup. Tan(x) also represents slope, which should always have a corresponding angle
i sucked, i was looking at this and thinking of plucking complex formula of tangent. lol
Something I found by accident without a background in complex trig:
The integral of 1/((x^2)-1).
Solve by partial fraction decomposition.
Solve by u substitution for u=ix.
Set the solutions equal to each other.
Make sure +C isn't a problem.
Divide both sides by i.
Substitute each x for -ix.
Complex Inverse Tangent.
Just square both sides and add 1
You'll get that sec z=0, which has no solutions
Oh yes yes
nice
You need to prove that this inequality still holds in the complex plane...
How do you know sec z=0 has no solution?
@@gigachad6844 sec(z) = 1/cos(z), so if sec z = 0 then 1/cos(x) = 0, which is not possible
my guess before watching or working it out on paper is D. reason is that i know A and C are possible, and B seems possible because sin(ix) = i*sinh(x) so sinh(x) just has to be 1.
i was right tho i was expecting there to be some deeper reason why its not worth trying to define arctan(i). anyway, nice video
@@nathanisbored arctan(z) has a pole on i and -i, thats why its not worth it, check out complex graphs on wolfram
Another simple solution:
Tanh(iz)=iTan(z)
If Tan(z) = i or -1
Tanh(iz) = -1 or 1
And Tanh can't reach 1 or -1
6:00 super effect!
Is there any way to assign a meaningful value to arctan(i), maybe by using something other than complex numbers?
Ryan Bellafiore No, and that is for the same reasons there is no meaningful value to log(0) by extending the number system. These expressions are not merely undefined, but undefinable.
Angel is right. Some undefined values, like the square roots of negative numbers, can be given definitions that play well with the rest of math. Other undefined values, like division by zero, cannot. Arctan(i) falls in the latter category. You can try to create a new number that's defined in this way, but you'll run into contradictions you just can't iron out.
@@General12th division by 0 can be given a value in different systems though. the extended complex numbers allow for division by 0.
@@angelmendez-rivera351 actually, there IS a set which arctan(i) is defined, it's the adherence of C
@@leofisher1280 can you explain further? Have any links I can look into? Are you talking abt hyperreals/infinitesimals? Bc those are NOT 0
Random question, does it mean that tan-1(z) has singularities at z = +-i? If so, what type?
yes it does, if arctan(z) tends to i it goes to positive complex infinity but the real part does not diverge and as it tends to -i it goes to negative complex infinity but the real part does not diverge.
you can look at the graph of i*arctanh(x) to see this for your self.
@@jeremy.N So then what is the real part if it doesn't diverge?
@@ZipplyZane the real part has an absolute value of pi/4, the sign depends on from where you approach it.
@@ZipplyZane also damn, this is 2 years old, you think ill remember? xD
you could just integrate dt/(1+t²)
1/(1+t²)=1/(t+i) - 1/(t-i) =int= ln((x+i)/(x-i))
Hey bro I love your maths trick.Can you provide quick solution to sin100.sin120.sin140.sin180
Tan(z) = i => sin(z)/cos(z) = i => sin(z) = icos(z) => cos(z) = -isin(z)
Now consider e^(iz) = cos(z) + isin(z)
We know cos(z) = -isin(z) so we substitute that in and get that e^(iz) = -isin(z) + isin(z) = 0, meaning z = -iln(0), but ln(0) is undefined, therefore z has no solutions
Couldn’t you also use the fact that the derivative of arctangent is 1/(1+x^2) and inputting in i would give you an undefined answer?
y=|x| has a value at 0 but no defined derivative at 0
@@BigDBrian yea, but if u plug 0 into derivative of a |x| you get 0/0, in arctan situation you get 1/0 form
Well, sqrt(x) is an example. It has a value but not derivative at 0
Just because a function has an undefined derivative at some point doesn't mean the function itself is undefined. The Weierstrass function, for example, is defined at every point (in fact it's continuous) but has no derivative anywhere.
Good point.
@BPRP
It could have solution if you consider different type of complex number z which are much different than *i* like *j* or *k*
You think there is a solution in quaternions?
You have several videos of calculus,do some of linear transformations please.,could the next video be these problem please?.be T:R^3 a R^3 the linear transformation give by T (x,y,z)=(3x+z,x+y,-x).define if it s possible a linear transformation T1:R^3 a R^3 such as T (T1 (0,2,1))=(-1,1,3);T1 (0,1,0)=(1,0,1) y Nu (T1) ≠{0}.
Do you know my friend Dr. Peyam? He is the go to for linear algebra. Go check out his channel! : )
What is iverse tangent aproaching when z aproaches i
sometimes my professor says this equation(some equation) has no solution without trying to calculate it, and I have no idea why...
Sin/cos = i
Sin²+cos²=0
But, for any x
sin²+cos²=1
Hence no x is possible for Sin²+cos²=0
Hence tanZ=i have no solution
here because i had no idea how to solve this in your poll :D
Finding this hard to get. Intuitively, if sin(z) = i has a solution, then why shouldn't cos(z) = i have a solution? And since tan = sin/cos, then tan(z) = i should have a solution too. So I worked out an expression for cos (z) using sin squared + cos squared = 1. Then it was simple to compute tan (z) and the answer I got was i/(sqrt 2). Of course, I had to divide by cos(z) which obviously is 0 when Z = 90 degrees, but in the real world, nobody would say that tan (theta) has no solution because when theta is 90 degrees the answer is infinity. Hope this makes sense. Apologies if there is a mistake somewhere - I got my BSc (admittedly in Physics) way back in 1971 ! Really like the channel btw!
I'm pretty sure your calculation is wrong because tan^1(i/sqrt(2)) = 0.8813..., though I can't really tell where because I find it difficult to understand. Either way, try to calculate tanh^1(±1).
tan(z)=+/-i
tan^2(z)=-1
1+tan^2(z)=0
sec^2(z)=0
sec(z)=0
1/sin(z)=0
Which is a contradiction
Why are you standing with a PokeBall
that’s the microphone
Dude this video straight up crashed my computer when you started pulling out the magic!
Here's a list of equations I know has no solution, even over the complexes:
1. tan(z) = +-i (mentioned in the video)
2. cot(z) = +-i (same thing as tangent)
3. sec(z) = 0 and csc(z) = 0
4. sech(z) = 0 and csch(z) = 0
5. tanh(z) = +-1
6. coth(z) = +-1
Nice
Fundamentally are wrong. sin cos and tan are parts of triangles.
Due to which they are just some ratios of the sides.
So what someone taken out a series and find relationships.
It doesn't mean doing on that series means sin cos or tan is affected
I just found this channel and this video fries my brain like fries in a air fryer
I gonna try it in Calculator++.
Edit: it says "NaNNaNi"
RUSapache NaNi?!?!
@@Cloiss_ NaN is "not a number" in computer real (floating point) numbers. So it looks like calculator++ is trying to display the complex number a+ib as abi, possibly getting the number printing to display the '+' instead of the display logic (to avoid something like "2+-3i"), but because the value is "NaN" the +/- is not displayed?
"Nannani" sounds like something you'd shout in Japanese, like it means "wait up!" or something
tg^2 (x) + 1 = -1+1= 1/sin^2 (x), that's why 0=1/sin^2 (x), no such x exists.
I feel like if my maths teacher ever displeases me I should drop this on them
can it be lim[w->i] tan^-1 (w) ?
0:00 e^z=0 has no solution either
This is more of a no solution type where there would be no convergence so also no other set than the reals or complex would have a solution in it right
6:30 do you mean inverse tangent hyperbolic or inverse tangent? I see a natural log so I'm assuming you meant tanh inverse.
tan(z) = tanh(iz)/i
so naturally their inverse functions look very similar
@4:34 why didn't you use componendo-dividendo -_- ?
Great video.
Yo can u help me pls? I can't solve the integral of ((e^x)(x-1))/(x(x+e^x)) from 1 to 2.
how about tan(z)=i/2
Can we prove that tan(1)+tan(w)+tan(w^2)=tan(1). Tan(w).tan(w^2) where w is non real cube root of unity
I don’t know what the fuck is going on but I enjoy watching this solve these problems
ln(0) is -infinity so there is a answer(well yes its more of a limit)
Maybe one day you'll win a field medal :)
Complex numbers just confuse the hell out of me sometimes.
Great video
You haven't seen it entirely
Well, you just need to say: tanz=i, let be z=j solution for that equation and you create a whole new body of numbers
:O
3D complex numbers?
@@nuklearboysymbiote I have no idea
@@sebastianjakov4895 z would approach complex infinity
@@nuklearboysymbiote tanzential numbers?! 0_o
Where can I have this hoodie ? :D
Can't see the post
1234123412341234
Oh go to my community post and you will see the info for the contest
@@blackpenredpen I tried on youtube app and website using your link but it redirects me to your channel..
1234123412341234 ah I see. Then you will have to move to “community” then you will see the post.
Or you can check out the Teespring here for purchasing. Thank you. teespring.com/shop/the-defintion
The last one
I have a challenge for you: find analytically the complex solutions of the following, if possible:
ln (x) = e^x
Interesting. I shall have a go at this some time in the future
The zing sounds feel a bit out of place lol
Sin z = 2
2>1
-1
That’s for real solutions UwU
arctan(i) feels sad
Cosine feels unhappy coz he cannot join the game 😂. Thanks bprp and all the smart solutions in the comment!
If infinite is Undifineble then how we can put 1/infinity is exactly 0. I think it may be[.999999999..... /1.00000000000000......1] what u thinks
You are talking about the hyperreal numbers. Those have infinitessimals. ε is the smallest number larger than 0. In modern mathematics, we use limits instead. We say that as x approaches infinity, 1/x approaches 0, and, in the limit, equals zero.
This is written as lim(x->∞) 1/x = 0. Simple arithmetic using infinity is not possible, because 1+∞=∞=2+∞, so 1=2. Therefore, we always wrap infinity (and division by zero) into a limit, which can usually make it computable even when the actual expression makes little sense.
Originally, calculus had been developed with infinitessimals (by people like Isaac Newton), but it has been redefined using limits because of historical reasons.
fghsgh No, that is 100% inaccurate.
1. The real reason limits were made conventional in favor of infinitesimal quantities is because, back in the day, formal axiomatizations of mathematics did not exist, and as such, the understanding that the existence of infinite and infinitesimal quantities can be made consistent in first-order logic did not exist. In other words, it had nothing to do with them being confusing, it had to with not having the necessary formal, theoretical tools to set infinitesimal numbers on a rigorous footing. In mathematics, rigor takes superiority over intuitive understanding, so regardless of how less confusing the notion of infinitesimal quantities was, there just was not enough rigor to justify their usage.
2. You completely misrepresent how hyperreal numbers work. First, you claimed that the quantity ε/2 being smaller than ε is an impossibility. This is false. Numbers of the form a + bε, such that ε = 1/ω, where a and b are real numbers, are well-defined and the set has a well-ordering. It also has a different total-ordering which matches that of the real numbers. Also, ωε = 1, not any number. 0/0 can be any number, but that is not the same thing. If you extend the field to a wheel, however, then 0/0 is a well-defined quantity separate from any other.
fghsgh The definition of ε is simply 1/ω, nothing else. Nothing about it contradicts the existence of ε/2.
@@angelmendez-rivera351 I updated my answer. Thanks. Infinitessimals is not something high schoolers usually learn about.
Wolfram alpha says it is i*inf :)
Lol I checked and saw that too.
Oh, so this is just a masked version of ln(z) = 0!
arctan I.
Well 1=2 truly has no solutions
Tan(z)=i has no solution
So, we can scale lim x->i 😆
You forget the hypercomplex world hehehe
I’m in 6th grade
I’m dead after watching this
lets just make a new complex number called "j" for the solution to this
inb4 quaternion solutions smh
TAN(Z)=2
I just invented a new type of number, which I call ¥, such that tan(¥)=i.
You can thank me later, weak brained mathematicians 😎👌
I think tan(z)= i has no solution.
Ok… would this have something to do with the fact that cos(π/2) = 0? The invalid inputs exp(πi/2) and exp(-πi/2) remind me of that.
NukeML Not really. It has to do with the fact that log(0) is undefined, since the equation can be rearranged to be e^(iz) = 0, which has no solutions. Alternatively, it means arctan(i) and arctan(-i) are undefined.
First
👍👍👍
Ok, but does tan(z)=i have a hypercomplex solution?
I have sympathy for you.....
"I cannot be won!"
get a real mic.
I showed it by replacing tan with sin/cos and then by sin/√(1-sin^2)
Then, by squaring both sides i got sin^2/(1-sin^2)=-1
To be honest, I visualised the unit circle and figured that you can't have tan(z) = i because to get to i, you have to take an angle of pi/2 and then you have i divided by 0. I hope what I mean here is clear.
How about cos(x)=x?
I think this is interesting
Great video give going:)
Complex equation - no pun intended lol.
*sees inverse tangent
*pukes
A bit more detail on inverse circular/hyperbolic trig functions in the complex domain would be nice because it's not that obvious (fun fact, I contributed the example for arcsin(z) on Wikipedia).
Maybe it has something to do with the fact that to get +-i in the complex plane you need a 90 or 270 degree turn, which are invalid arguments for the tangent function
+-2i does have a solution though?
I integrated 1/(1+x^2) using partial fractions hence got imaginary definition of inverse tangent. Plugged in 1 and got pi/4 so C = 0, then plugged in pi/2 and got an answer of pi/2i. Whats going on?
I dare you to do the derangement of i
Edit: I didn’t find any solution in google so I did it and wanna see if i did it well
Hi BlackpenRedpen. I teach physics at university and I advised the students to watch your channel if they want to learn how to Integrate functions really, really quickly and skip lectures, in which they teach how to integrate... They were amazed ;) like I'am !!
Complex numbers are sneakier then we think
i am a math student and your solutions are really interesting...i love you...keep it up...
I just saw your video on imaginary triangle
made me think back to this video
tan(z) = i = i/1
create triangle with opp=i, adj = 1, hyp = sqrt(i^2 + 1^2) = sqrt(-1+1) = 0
this means cos(z) = 1/0......
sin(z) = i/0...
division by zero....
(probably not rigorous at all but cool nonetheless)
this basically goes back to the method you pinned with starting with cos^2(z) + sin^2(z) = 1
so tan(z) = i
implies tan^2(z) = i^2
tan^2(z) = -1
1+tan^2(z) = 0
sec^2(z) = 0..... (uh oh)....
1/cos^2(z) = 0 .....
1 = 0????
can use similar method for -i... since squaring a "negative" is "positive". i^2 = (-i)^2 = -1
[i can see that someone else in the comments (in reply of pinned comment has basically used the same method regarding tan^2(z) + 1 = sec^2(z)]
I thought any entire function had at most one value missing from the image though?
Oh, right, ok, so tan(z) can be infinity,
and the actual thing is that an entire function can be missing at most 2 values from the Riemann sphere, right?
And for tan(z) these are i,-i ?
Cool!
Interesting
Excellent presentation. Thanks. DrRahul Rohtak Haryana India
Can u do a video on application of derivatives and integrals? Like finding the value of 1.996^5 using derivatives 😅
ua-cam.com/video/bEylbdVM54A/v-deo.html
I was thinking D since y=tanx takes all real values except for npi/2, n is odd for which it's infinity, so there's no space for it to equal i
You discovered R3. I call it j hehe
Sin (pi/x) x = pi
Then, what's the tangent doing instead?
How does it transform the complex plane so that no complex value is mapped to ±i?
(Are those two the only exceptions?)
I think yes
نريد ترجمه
tan(z)=i
I know i is some special number thing in the imaginary world but is z just a pronumeral because looking at the sine graph the range is between -1 and 1 so how could you get 2?
Z implies a complex number, and the range of sin() being between 0 and 1 only applies to the real numbers.
Every equation should have a solution. If the complex world has no solutions for it, then maybe a greater arena of maths may have a solution. Maybe we could use quarternions.
it's crazy that when you think about it. only 3 has a solution and 25% doesnt. such a big number