which equation has no solutions: sin(z)=2, sin(z)=i, tan(z)=2, tan(z)=i?
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- เผยแพร่เมื่อ 26 ก.ย. 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?
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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!
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See the post
th-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
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
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@@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.
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.
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
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 !!
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
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?
you could just integrate dt/(1+t²)
1/(1+t²)=1/(t+i) - 1/(t-i) =int= ln((x+i)/(x-i))
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
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Hey bro I love your maths trick.Can you provide quick solution to sin100.sin120.sin140.sin180
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).
How about cos(x)=x?
I think this is interesting
Great video give going:)
Complex equation - no pun intended lol.
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.
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
Excellent presentation. Thanks. DrRahul Rohtak Haryana India
@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}.
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What is iverse tangent aproaching when z aproaches i
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
sometimes my professor says this equation(some equation) has no solution without trying to calculate it, and I have no idea why...
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).
*sees inverse tangent
*pukes
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
Complex numbers are sneakier then we think
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
Can u do a video on application of derivatives and integrals? Like finding the value of 1.996^5 using derivatives 😅
th-cam.com/video/bEylbdVM54A/w-d-xo.html
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.
Great vid- could you show by graphing as well?
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
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
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
Log of zero cannot be written as -infinity?
Nope, just like how you can't say 1/0 = +-infinity. The equation has no solution. e^b=0, is what we are saying. No value of b will work, though large negative values will, what you can say is the limit of e^b as b is negative infinity is 0, or rearranged the limit of ln(0) is negative infinity. However a limit is not a value. Hope that made sense.
@4:34 why didn't you use componendo-dividendo -_- ?
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
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.
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.
Teacher:Can we integrate x^x
Me:yes, bprp had video on that.
-_-
Lol. I will do a legit way for that later. By power series
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
Does tan(z)=i have solution in quaternions or octanions?
I believe it has No Solutions, at all.
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 :)
Great video
You haven't seen it entirely
Complex numbers just confuse the hell out of me sometimes.
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
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Can't see the post
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1234123412341234 ah I see. Then you will have to move to “community” then you will see the post.
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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
I have a question for you
How can I send it to you ????
arctan(i) feels sad
Sin z = 2
2>1
-1
That’s for real solutions UwU
Cosine feels unhappy coz he cannot join the game 😂. Thanks bprp and all the smart solutions in the comment!
So does this mean tan(z)=c is solvable for all complex c not equal to plus or minus i?
Yes!! You can just simply plug in any c you want in my formula for arc tan
Sweet
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.
Interesting
tan(z)=i
Does this mean
cos Z = i
is also undefined ?
its not, cos(z)=i principal is ±[π/2 + i log(√2 -1)]
@@Fokalopoka Thank you.
Wolfram alpha says it is i*inf :)
Lol I checked and saw that too.
arctan I.
نريد ترجمه
Oh, so this is just a masked version of ln(z) = 0!
Tan(z)=i has no solution
Well 1=2 truly has no solutions
Now I know angles can be imaginary...... GODDAMMIT MATH LEMME LIVE...
mod(x)=-1
So, we can scale lim x->i 😆
You forget the hypercomplex world hehehe
I’m in 6th grade
I’m dead after watching this
You discovered R3. I call it j hehe
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?