how to solve sin(x)=i?

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blackpenredpen

blackpenredpen

5 жыл бұрын

Learn how to solve this complex impossible-looking trig equation sin(x)=i. Of course, we need to use Euler's formula and the complex definition of sine.
sin(sin(z))=1 • Math for fun, sin(sin(...
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Пікірлер: 368
@blackpenredpen
@blackpenredpen 10 ай бұрын
sin(z)=2, kzfaq.info/get/bejne/aamPi6eQyKnJlX0.html sin(sin(z))=1 kzfaq.info/get/bejne/bsqgZpiLzbvWj6M.html
@sergiolozavillarroel3784
@sergiolozavillarroel3784 5 жыл бұрын
I solved this very easily: sin(z)=i z=arcsin(i) Easy isn't it?
@davidawakim5473
@davidawakim5473 5 жыл бұрын
The point is to figure out what arcsin(i) is through algebraic manipulation.
@crosisbh1451
@crosisbh1451 5 жыл бұрын
Did you doing in your head... without prompting‽‽
@seangrand3885
@seangrand3885 5 жыл бұрын
CrosisBH what ._.
@Abdega
@Abdega 5 жыл бұрын
Future mathematician right here ☝️
@R1ckr011
@R1ckr011 5 жыл бұрын
@@crosisbh1451 he is horribly UnFun in the explanation process
@PaddedShaman
@PaddedShaman 5 жыл бұрын
i don't need to be on the bottom if i don't want to
@poppo20202020
@poppo20202020 5 жыл бұрын
That's what she said!
@cdeyng
@cdeyng 5 жыл бұрын
HAHA, that was witty though. XD
@janv.8538
@janv.8538 5 жыл бұрын
_top comment_
@userBBB
@userBBB 5 жыл бұрын
when did he say this?
@dinamosflams
@dinamosflams 4 жыл бұрын
That's what Lilith said ( ͡° ͜ʖ ͡°)
@pelegmichael5489
@pelegmichael5489 5 жыл бұрын
8:00 "pi is an integer". I am personally offended.
@Albkiller22
@Albkiller22 5 жыл бұрын
But he wrote only n is an integer so was not wrong
@PaddedShaman
@PaddedShaman 5 жыл бұрын
π ∈ ℤ
@jabir5768
@jabir5768 5 жыл бұрын
well pi is an integer isnt it?
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Sarcastic Name I feel I have to make a public apology now Bc of that. : )
@kingbeauregard
@kingbeauregard 5 жыл бұрын
We are young Heartache to heartache We stand No promises No demands Pi is an integer Whoo
@VJZ-YT
@VJZ-YT 5 жыл бұрын
you are the only person in this world that makes math look fun. well done
@blackpenredpen
@blackpenredpen 5 жыл бұрын
VJZ thanks!!!
@branthebrave
@branthebrave 5 жыл бұрын
Definitely not the only one
@VJZ-YT
@VJZ-YT 5 жыл бұрын
@@branthebrave who else?
@branthebrave
@branthebrave 5 жыл бұрын
@@VJZ-YT Definitely a lot o teachers out there at any level that do that. Math always looks fun to me, so that doesn't really matter. You're probably asking for other youtube channels, so really any of the popular ones sometimes do like numberphile, 3blue, mathologer, standupmaths, but it depends what you call "fun," like maybe you mean really interesting. Think twice does that.
@hamez1324
@hamez1324 5 жыл бұрын
sin inverse both sides -> z= sin^-1 (i) ez
@joshuamason2227
@joshuamason2227 5 жыл бұрын
@@simpletn r/whoosh
@alexting827
@alexting827 4 жыл бұрын
BAD NOTATION XDD
@Mexa2105
@Mexa2105 5 жыл бұрын
I get impressed when you use the euler's identity by using as well the logaritms rules good video man
@dremr2038
@dremr2038 2 жыл бұрын
Same , he used the perfect technique to teach that concept
@cwo12cw
@cwo12cw 5 жыл бұрын
One answer is the negative imaginary natural log of the silver ratio. *how. Cool. Is. Thaat.*
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Clemente Wacquez : )))))
@alansmithee419
@alansmithee419 5 жыл бұрын
Maths does weird things with seemingly unrelated areas sometimes.
@98danielray
@98danielray 5 жыл бұрын
has to do with the quadratic equation that appeared
@nazishahmad1337
@nazishahmad1337 5 жыл бұрын
Silver ratio what's that I've heard of golden ratio only
@shoobadoo123
@shoobadoo123 5 жыл бұрын
alan smithee *math
@iansamir18
@iansamir18 3 жыл бұрын
Easy solution: sin(z) = i, so cos(z) = sqrt(2) by sin^2 + cos^2 = 1. Therefore, e^(iz) = cos z + i sin z = sqrt(2) + i(i) = sqrt(2) - 1, and z = -i ln(sqrt2 - 1) as desired.
@user-gp5zr9wb4z
@user-gp5zr9wb4z 5 ай бұрын
cos(z) = ±√2
@m_riatik
@m_riatik 5 жыл бұрын
i discovered you from the sin(z) = 2 video. i remember a lot of people were fighting in the comments because you said "conplex axis"
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Muriatik yea lol
@themeeman
@themeeman 5 жыл бұрын
F
@chatherinehu3804
@chatherinehu3804 5 жыл бұрын
Hahahah I think they pay the wrong attention
@andrecruzmarquez645
@andrecruzmarquez645 5 жыл бұрын
"You got to do more work to please people" ...
@dhay3982
@dhay3982 5 жыл бұрын
Now do the formula Sin(z)=a+bi
@shre6619
@shre6619 5 жыл бұрын
Z is just sin^-1(a+ ib)
@antimatter2376
@antimatter2376 5 жыл бұрын
@@shre6619 And what is the inverse sin of a complex number?
@thanoskalamaris3671
@thanoskalamaris3671 5 жыл бұрын
@Seife Zwei you take the formulas of sin-1(I) and switch i with a+bi
@antimatter2376
@antimatter2376 5 жыл бұрын
@@thanoskalamaris3671 That's not how math works
@themanagement69
@themanagement69 5 жыл бұрын
You can write any real number in a+bi form.
@david-yt4oo
@david-yt4oo 5 жыл бұрын
4:10 you always make really interesting videos, and some .real. good puns
@pedrocastilho6789
@pedrocastilho6789 5 жыл бұрын
You can also do by saying that e^iz=cos(z)+isin(z) Since cos^2+sin^2=1 Cos^2+(i)^2=1 Cos^2-1=1 Cos^2=2 Cos(x)=+-sqrt(2) Using that you have that e^iz=+-sqrt(2)-1 The rest is the same as the video :)
@enisheadpay
@enisheadpay 5 жыл бұрын
If you want a single formula you could rewrite the final answer as arcsin(i)=-i*ln(sqrt(2)+(-1)^(k+1))+k*pi for integer values of k.
@hazza6915
@hazza6915 5 жыл бұрын
For logarithm to be bijective you need to specify which theta you are taking and which half line you are removing also
@enclave2k1
@enclave2k1 2 жыл бұрын
" _i_ don't have to be on the bottom if _i_ don't want to" Brilliant pun and very useful.
@stigastondogg730
@stigastondogg730 4 жыл бұрын
Love this dudes passion for math!
@noahp4589
@noahp4589 3 жыл бұрын
i think a clever way to do it without knowing the complex form of sin would be by sin(z)=i sin²(z)=-1 1-cos²(z)=-1 cos²(z)=2 cos²(z)=plus or minus sqroot 2 then by reemplazing in euler's formula e^iz=cos(z)+isin(z) e^iz=plus or minus sqroot 2 +i² i really enjoyed the video thanks for being an awesome teacheeeer
@chatherinehu3804
@chatherinehu3804 5 жыл бұрын
I love your way of making maths fun , hoping to be the same person like you .
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Thank you!!
@krishabm1
@krishabm1 5 жыл бұрын
None of your videos are possible without e.... XD
@blackpenredpen
@blackpenredpen 5 жыл бұрын
: )
@gelatinaworld
@gelatinaworld 5 жыл бұрын
You need a high iq to get the E
@nuklearboysymbiote
@nuklearboysymbiote 4 жыл бұрын
@@gelatinaworld but im a silly man with a small
@JivanPal
@JivanPal 4 жыл бұрын
@@infernocaptures8739, kzfaq.info/get/bejne/mMCZqaaomMimZqs.html
@MrKhan-dw9vh
@MrKhan-dw9vh 5 жыл бұрын
I am missing "Blackpenredpen Yay!"
@Matthew-tu2jq
@Matthew-tu2jq 5 жыл бұрын
This is awesome i love the content you make ❤️
@prollysine
@prollysine 4 жыл бұрын
Szerintem ez totál elméleti, talán csak elméleti matek szempontból érdekes, de sok apró lépés eszembe jutott. Nagyon jól tanítasz, minden részletet bemutatsz gratulálok !
@quitecomplex6441
@quitecomplex6441 5 жыл бұрын
I just stumbled along this problem the other day and I solved it. I came on here to check my answers. Very cool problem indeed.
@24_santanurath56
@24_santanurath56 2 жыл бұрын
seriously i have fun by this video teaching style,This video is amazing
@admancr2823
@admancr2823 Жыл бұрын
I am absolutely passionated about complex numbers... It is just completely different from anything I have learned for 19 years of my life, sometimes is just crazy. Yet it is so useful that we use these numbers in electricity, quantum mechanics, Riemann's hypothesis which is the biggest mystery in Maths. Just amazing. Thank you for your work sir.
@dr.rahulgupta7573
@dr.rahulgupta7573 2 жыл бұрын
Excellent presentation ! Vow !!
@Iamnotyou29
@Iamnotyou29 3 жыл бұрын
I get fun to look your math problems. Thnx sir🙂🙂
@davidawakim5473
@davidawakim5473 5 жыл бұрын
This video was great :D
@user-td6pl6wk6s
@user-td6pl6wk6s 3 жыл бұрын
Thank you so much
@MrBeen992
@MrBeen992 4 жыл бұрын
8:04 "You write down where n is Z so people know you are cool" LOL
@AndDiracisHisProphet
@AndDiracisHisProphet 5 жыл бұрын
8:05 pi is an integer? Almost as good as three is smaller than two^^ Also, Sin(z)=2 was cooler, imho.
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Lol!!! Yup I still remember that one too!! Btw, the time mark should be 8:00
@AndDiracisHisProphet
@AndDiracisHisProphet 5 жыл бұрын
@@blackpenredpen Oh. I BPRP'uped the time stamp. Do you remember which video that was?
@blackpenredpen
@blackpenredpen 5 жыл бұрын
AndDiracisHisProphet log_2(3) vs log_3(5)
@AndDiracisHisProphet
@AndDiracisHisProphet 5 жыл бұрын
@@blackpenredpen such a trauma, that you still remember^^
@blackpenredpen
@blackpenredpen 5 жыл бұрын
AndDiracisHisProphet lol!! So did you!
@rezamohammadyousefi3317
@rezamohammadyousefi3317 2 жыл бұрын
Wow its very useful for me...tanku for that
@alanwolf313
@alanwolf313 5 жыл бұрын
Hello RPBP i really like your videos and I need some help. I was doing some math for fun the other day and tried to solve the integral of the xth root of x (or xˆ(1/x)) dx. How should I tackle this problem?
@backyard282
@backyard282 4 жыл бұрын
10:15 You can't call that z=arcsin (i), because arcsin is a function so it can't have infinitely many values, it gives a "principled" angle, while the set of all solutions to sin z = i include arcsin + 2pi*n and pi - arcsin + 2pi*n. The same way how the square root function gives you the principal root, and the set of roots are +/- the square root function.
@rainbow-cl4rk
@rainbow-cl4rk 5 жыл бұрын
I have question: e^(iz)=+-sqrt(2)-1 =cos(z)+isin(z) But sin(z)=i =cos(z)+ii =cos(s)-1 Cos(z)-1=+-sqrt(2)-1 Cos(z)=+-sqrt(2) Arccos(cos(z))=arccos(+-sqrt(2)) Z=arccos(+-sqrt(2)) It is correct?
@thesame7423
@thesame7423 2 жыл бұрын
Yeah but you still have to do more work for the arccos, cuz it's domain is only [-1;1] and sqrt(2)>1/-sqrt(2)
@yugeshkeluskar
@yugeshkeluskar 5 жыл бұрын
Can you generalize it for sin(?)=a+bi
@shoopinc
@shoopinc 5 жыл бұрын
Yeah, ill give it a try
@shoopinc
@shoopinc 5 жыл бұрын
@@Tom-qz8xw I've worked it down to a formula where you can input a and b and have it pop out the answer. But I must have made an algebra mistake somewhere because its slightly wrong. For the case in the video it gives me a value where I take the sin and it gives 1.5*i rather than i. So I'll do the derivation again and fix it.
@sergiolozavillarroel3784
@sergiolozavillarroel3784 5 жыл бұрын
@@shoopinc Done yet?
@RendeiRotMG
@RendeiRotMG 3 жыл бұрын
if you still need it. It's sin(z)=pi/2-i*ln(z±sqrt(z^2-1))
@soumyachandrakar9100
@soumyachandrakar9100 5 жыл бұрын
Would you mind doing a video on Fermat's Little Theorem?
@tylertorsiello8450
@tylertorsiello8450 3 жыл бұрын
this guy is my spirit animal
@manuelepedicillo864
@manuelepedicillo864 5 жыл бұрын
I want number theory videos 😭😭
@jakubfrei3757
@jakubfrei3757 5 жыл бұрын
Excatly
@gian2kk
@gian2kk 5 жыл бұрын
Escatly
@sergiolozavillarroel3784
@sergiolozavillarroel3784 5 жыл бұрын
Persicely
@srpenguinbr
@srpenguinbr 5 жыл бұрын
with complex integers!
@professorpoke
@professorpoke 3 жыл бұрын
You should watch Micheal Penn on KZfaq, for number theory problems.
@michael161
@michael161 Жыл бұрын
Math is so beautiful and helpful in our life.❤️❤️❤️
@af8811
@af8811 5 жыл бұрын
The important lesson from this, is : "Keep people in peace by not to do logarithm of negative numbers. Just don't do that and even touch it" (Prof. Steve) 😆😆😆😆😆👍👍👍👍👍
@blackpenredpen
@blackpenredpen 5 жыл бұрын
: ) #SteveIsMyStageName
@af8811
@af8811 5 жыл бұрын
Please don't be mad Professor ☺☺. I was joking. He he he he... Cause math is fun, isn't it Professor ??? 👍😊👍
@blackpenredpen
@blackpenredpen 5 жыл бұрын
@@af8811 Oh no, I wasn't mad at all. I just wanted to put that harsh tag whenever people comment "steve" : )
@af8811
@af8811 5 жыл бұрын
@@blackpenredpen :') i thought it's your real name, Professor.
@factsheet4930
@factsheet4930 5 жыл бұрын
My professor told me that it is possible to solve for z in the equation |z|=-1 Wolfram Alpha couldn't do it, can you make a video about it, if it is possible?
@jessehammer123
@jessehammer123 5 жыл бұрын
Fact Sheet I think your professor is wrong. In the real number set, there’s obviously no number that fulfills this. In the complex world, all complex numbers have positive magnitude. In the quaternion world, all quaternions have positive magnitude. Et cetera, I think. But I’m just a sophomore in high school, so what do I know?
@zuccx99
@zuccx99 5 жыл бұрын
It's impossible because it's undefined just like 1/x when x is 0.
@factsheet4930
@factsheet4930 5 жыл бұрын
I mean probably not as the distance definition but as square root of x to the power of 2 And yes Wolfram Alpha will say there is no solution but it also says that for x^(1/3)=-2, while clearly - 8 is the solution.
@98danielray
@98danielray 5 жыл бұрын
depends on how norm is defined. is this the usual norm?
@8bit_pineapple
@8bit_pineapple 5 жыл бұрын
Okay, so for the real numbers |x| is just defined as |x| = { x if x ≥ 0 {-x if x < 0 i.e. throw away the minus if there is one. For other numbers like the: Complex, Quaternions, Octonions, etc The ||z|| function is the distance from 0 to z. As such, you won't find an example where ||z|| < 0, in any of these sets of numbers. Distance functions map to values ≥0 as part of their definition. But supposing ... You had your own set of numbers "😂", Then you define a function f : 😂 → ℝ , where f(z) = -2 for some z∈😂 Everyone would think you're an ass if you wrote "Let |z| = f(z) when z∈😂" By all means you could... it's your mathematics... But... my recommendation would be for you to extend |z| with a new function "😊" And just put: 😊(z) = { |z| if z∈ ℝⁿ { f(z) if z∈😂 That way everyone will be happy with your notation.
@DrDirtyHarry
@DrDirtyHarry 5 жыл бұрын
The two general solutions look very similar. Is there a correspondence on the complex plane?
@gregoriousmaths266
@gregoriousmaths266 4 жыл бұрын
this is ez for me now, but it wouldnt be if it werent for ur vids thank u so much
@No_hope_No_fear
@No_hope_No_fear 5 жыл бұрын
Bprp: "Okay, thanks for watching" *almost dies laughing
@blackpenredpen
@blackpenredpen 5 жыл бұрын
LOLLLL
@user-co6rg9jt9x
@user-co6rg9jt9x 5 жыл бұрын
I love this "This like that"
@magnuschanduru6173
@magnuschanduru6173 5 жыл бұрын
Nice way of teaching..
@abdellh8079
@abdellh8079 3 жыл бұрын
Actually it is interesting , great job , keep going ,
@DrQuatsch
@DrQuatsch 5 жыл бұрын
I would like it more if you had taken sin(z) = i/2. -1+sqrt(2) and 1+sqrt(2) are not as nice as the golden ratio in your answer.
@muse0622
@muse0622 5 жыл бұрын
I Love Math. Blackpen Redpen YAY
@ryanguenthner823
@ryanguenthner823 3 жыл бұрын
"We have to do more work to appease people." Man, I fucking felt that. Great video.
@TheFinalRevelation1
@TheFinalRevelation1 5 жыл бұрын
Is that a mic ?
@1976kanthi
@1976kanthi 3 жыл бұрын
No it’s a thermal detonator
@anthonywong1781
@anthonywong1781 5 жыл бұрын
Are you sure you can just invert sin without specifying the domain and range for this question? The formal definition for the inverse of sin is for every x in [-π/2, π/2] , y in [-1,1] , arcsin(y) = x if and only if y = sin(x) though.
@madaaz6333
@madaaz6333 5 жыл бұрын
There may be a problem in this case. According to Wikipedia (complex logarithm) the property Log (z1 z2) = Log (z1) + Log (z2) is not generally valid when there is a negative number. What do you think about it?
@Reallycoolguy1369
@Reallycoolguy1369 2 жыл бұрын
Before watching the video, I tried this problem, and when I got to that step, I used the polar coordinate definition of the complex number (z=a+bi, z= r*cos[theta] + i*r*sin[theta]), then applied euler's formula (z=r*e^i[theta]), then took the natural log (ln(z)=ln(r) + i*[theta]). In this case r is |-1-sqrt(2)| and since this is a negative real number, on the complex plane the angle theta would be pi. So you end up with i*z=ln(1+sqrt(2))+i*pi, and then it's the same steps as the video. This doesnt require the product property of logarithms and I got the same answer, so I think we are good here. BPRP shows exactly what I'm describing in the sin(z)=2 video.
@tristancam7219
@tristancam7219 5 жыл бұрын
We have -1+sqrt(2) = 1/(1+sqrt(2)) therefore using the fact it is a quotient: -i*ln(-1+sqrt(2)) = (-i)*ln(1) - (-i)*ln(1+sqrt(2)) = i*ln(1+sqrt(2)). Can we now write down a shorter solution in only one expression ?
@bob53135
@bob53135 5 жыл бұрын
I don't think so, but we could have found the second solution easily, as if z is a solution to sin(z)=v, then (π-z) is also a solution…
@user-qb5gw7tc9e
@user-qb5gw7tc9e 4 жыл бұрын
z = (-1)^n * i * ln(√2 + 1) + πn cases : n = 2m and n = 2m -1
@Bicho04830
@Bicho04830 5 жыл бұрын
Yep but note that (1+√2)=(-1+√2)^(-1) (reader exercise) So ln(1+√2)=-ln(-1+√2), and therefore we can wtite it as: z=nπ +((-1)^n)ln(-1+√2)
@gilber78
@gilber78 3 жыл бұрын
wouldn't the full answer just by -i*ln(1+sqrt(2)) + pi*n since both answers are the same just offeset by pi and they both have the 2pi factor?
@sgrass471
@sgrass471 5 жыл бұрын
for the second answer wouldnt the pi and the 2*pi*m term add together giving us pi*(1+2m)? or in other words pi*q where q is an odd number? by the way love your videos
@francis6888
@francis6888 5 жыл бұрын
"2 screw" Gotta love subtitles.
@viletomedoze5036
@viletomedoze5036 5 жыл бұрын
Best part of the video " i don't need to be at the bottom if i don't want to"
@renardtahar4432
@renardtahar4432 5 жыл бұрын
vous etes formidable!
@maxchatterji5866
@maxchatterji5866 5 жыл бұрын
Hey BPRP, I have an Oxford maths interview in a week. Are there any cool maths tricks I should know about which would blow the interviewer away?
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Hmmm, I am not sure about math tricks, especially they should be the finest math people in the world. If I have to do it myself, I will definitely show them how to do math with blackpen and redpen in one hand. Best luck to you!!!! Keep me updated. I would love to hear how it goes!
@lilysowden4035
@lilysowden4035 5 жыл бұрын
I have a computer science interview next week as well!
@trueriver1950
@trueriver1950 5 жыл бұрын
Prove there are no boring positive integers. 0 1 is not boring because it is the identity for multiplication 2 is not boring because it is the smallest prime 3 is not boring because it is the closest prime to the previous 4 is not boring because it is the smallest composite 3 5 and 7 are not boring because they form the smallest value series of primes in arithmetic progression 6 is not boring because it is the first number with distinct prime factors This proof IS becoming boring so can we generalise it? Reductio ad absurdum If any numbers were boring one of them would be smaller than all the other boring numbers, and therefore would be interesting simply for that. Therefore the smallest boring number is NOT boring: which is absurd. Therefore there are no boring positive integers. QED
@omerhamzabilgin8963
@omerhamzabilgin8963 5 жыл бұрын
Good video :)
@andresidl
@andresidl 2 жыл бұрын
“i don’t like to be on the bottom” hahaha I see what you did there
@ridefast0
@ridefast0 5 жыл бұрын
Hi - I am probably wrong, but in your final answers you could start with (Z+2.pi.n) on the left hand side, so wouldn't it transfer across as -2.pi.n on the right hand side? I suppose it might depend on the odd and even symmetry of the sin() and cos() functions?
@NotBroihon
@NotBroihon 4 жыл бұрын
Yes, you are right notation wise. But it doesn't make any difference since n (and m) can be any integer (negative and positive). So the solution is still correct.
@l3igl2eaper
@l3igl2eaper 5 жыл бұрын
When are you going to teach Linear Algebra!?
@griffgruff1
@griffgruff1 3 жыл бұрын
Alternative solution: Let z = a+ib sin(a+ib) = sin(a).cosh(b)+cos(a).(i.sinh(b)) = i Equating real and imaginary parts gives a=0 and sinh(b)=1 So final answer is: a=0, b=0.88137
@ismaeljuniormoupe8881
@ismaeljuniormoupe8881 4 жыл бұрын
please the integral of e^cosx
@quantumcity6679
@quantumcity6679 5 жыл бұрын
thank's for watching...... 😁😘
@alansmithee419
@alansmithee419 5 жыл бұрын
Does anyone know of an app I could get for an android phone that plots complex number equations on graphs?
@roc6596
@roc6596 4 жыл бұрын
I learned complex numbers for high school, still though, didn't see any of this e number and all, is it only for calculus at a university? I'm from Brazil so I don't know if it's just cause here we don't have this for HS curriculum
@vikasdeep6393
@vikasdeep6393 5 жыл бұрын
Sir can you define log 0
@nullanon5716
@nullanon5716 5 жыл бұрын
If we plugged in the first solution into the z of the second solution, wouldn’t that make pi*(2m+1) = 0?
@haninyabroud7810
@haninyabroud7810 5 жыл бұрын
Thx i ♡ maths
@apotheosys1
@apotheosys1 4 жыл бұрын
Make a video showing that there is no z such that tan(z) = i
@blackpenredpen
@blackpenredpen 4 жыл бұрын
RealComplex Wait there isn’t?!
@blackpenredpen
@blackpenredpen 4 жыл бұрын
Wow very cool!!!! I just worked out. Thanks!!!
@WarpRulez
@WarpRulez 5 жыл бұрын
You missed a golden opportunity to mention that "e to the i pi equals -1" is the famous Euler's identity.
@godseye8785
@godseye8785 5 жыл бұрын
pretty sure he thinks most his viewers know that tho lol
@createyourownfuture3840
@createyourownfuture3840 2 жыл бұрын
All of his veteran viewers know that.
@General12th
@General12th 2 жыл бұрын
He also missed a golden opportunity to mention that 1 + 1 = 2, which is probably the most famous equation of all!
@edrodriguez5116
@edrodriguez5116 5 жыл бұрын
gotta do calc 2 again.
@blackpenredpen
@blackpenredpen 5 жыл бұрын
Yup!
@VKHSD
@VKHSD 10 ай бұрын
when he said "this guy" at around 5:00 he had the most perfect american accent
@samharper5881
@samharper5881 5 жыл бұрын
Please do a video about 1/(2+3/(4+5/(6+7/(8+9/(10+11/(12+...))
@ExzeneriteXll3492
@ExzeneriteXll3492 3 ай бұрын
By writing 2π, it is aproximating aproximated tau or 6.28
@raphaelh6791
@raphaelh6791 Жыл бұрын
Can you juste right -iln(-1+V2) + n pi ?
@afafsalem739
@afafsalem739 5 жыл бұрын
Well well
@sophiaabigai_l
@sophiaabigai_l Жыл бұрын
7:59 "you have to denote that pi is an integer" wait what
@MrBeen992
@MrBeen992 4 жыл бұрын
So the sin inverse of a complex number is also multivalued ?
@darysparta9676
@darysparta9676 3 жыл бұрын
4:10 when she wants to be on top for once
@AhmedMahmoud-yj6yx
@AhmedMahmoud-yj6yx 5 жыл бұрын
Find this limit with out using l'hobital's rule lim ln((x+sqrt(a^2+x^2)) /a) /x as x approches infinity
@AhmedMahmoud-yj6yx
@AhmedMahmoud-yj6yx 5 жыл бұрын
I said without using l'hobital's rule because it is the easiest way i want it the hard way 😂😂
@user-qi3mk4nr1g
@user-qi3mk4nr1g 5 жыл бұрын
Is arcsin(i) is same with the answer on this video??
@EduardoHerrera-fr6bd
@EduardoHerrera-fr6bd 5 жыл бұрын
Finally, bc is because!
@fujoridev
@fujoridev 3 жыл бұрын
1:45 Ёкарный, я думал он по-русски сейчас зашпарит!
@fariqjamil5484
@fariqjamil5484 10 ай бұрын
But that's the hypotenuse of the sine equation
@andrej6582
@andrej6582 3 жыл бұрын
Хорошо что язык математики и без переводчика понятен)
@user-de8nb8fn6s
@user-de8nb8fn6s Жыл бұрын
Постоянно этим восхищаюсь!
@sebastiantabara2325
@sebastiantabara2325 3 жыл бұрын
When u did ln-1 shouldnt u have also put a plus 2pin so in the answer u should have at the end smth like 2pi(m+n)?
@CaradhrasAiguo49
@CaradhrasAiguo49 5 жыл бұрын
10:09 Technically cannot do that (set arcsin(z) = ... + 2*pi*m) as it would no longer be a function
@JakeWaas
@JakeWaas 5 жыл бұрын
tfw implied graph cut
@angelmendez-rivera351
@angelmendez-rivera351 5 жыл бұрын
No, you can do that, just exactly in the same way you can write +/- when evaluating square roots. It is multi-valued function. If you want to be strict about it, the arcsin is a map from C to P(C), and the element of P(C) by which it is valued is unique, so it is a function.
@angelmendez-rivera351
@angelmendez-rivera351 5 жыл бұрын
it's me It is a function. What you have failed to understand is that arcsin is not map from C to C, but rather a map from C to P(C). The output in P(C) is unique, though it is a set.
@borg972
@borg972 5 жыл бұрын
I follow all the steps and everything's fine but I still can't understand how a function can be both periodic and unbounded at the same time. please help so I can make sense of this complex world..
@lenguyenvietcuong5379
@lenguyenvietcuong5379 Жыл бұрын
8:00 "π is an integer"
@Shouray9891
@Shouray9891 4 жыл бұрын
How you write sinz =[ e^iz-e^(-iz) ] / 2i
@manishkumarsingh3082
@manishkumarsingh3082 5 жыл бұрын
So good^_^
@GaryFerrao
@GaryFerrao 10 ай бұрын
8:00 know π is an integer. (quoted verbatim, but out of context 😂)
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