Chebyshev polynomials are designated as T because early transliterations from the Cyrillic rendered "Chebyshev" as "Tchebicheff" - similar to how "czar" was once (and sometimes still is) spelled "tsar".
@@pierreabbat6157 So maybe we should designate them as Ч(x). Cyrillic needs more representation in the symbols used in mathematics. I've heard that sha is used as a variable name, but that's the only one I know about so far.
@@pierreabbat6157 Interesting how there is no sounded letter in the original spelling where the y is in the transliterated spelling. Just the soft sign. I never expected that. I would've expected Чебйшёв.
This is probably a big coincidence, but you released a video on this topic EXACTLY after the end of one of the Russian Olympiads held by the KGB for schoolchildren, where there was a problem on EXACTLY THIS topic. I just wrote this Olympiad and remember this problem. We uncovered you, you are an overseas KGB agent...
Ordinary generating function for Chebyshov polynomial is G(x,t)=(1-xt)/(1-2xt+t^2) Exponential generating function for Chebyshov polynomial is e^{xt}cos(sqrt(1-x^2)t) Exponential generating function gives us formula with derivatives T_{n}(x) = limit (d^n/dt^n e^{xt}cos(sqrt(1-x^2)t), t = 0)
I’m not sure if it’s always true, but I noticed in the examples that you did, that when n is odd, we only see odd powers of cosine, and when n is even, we only see even powers. Its not much, but just anothher interesting fact I saw.
Maybe you can record video about Chebyshev polynomials Michael Penn derived formula with complex numbers but he stopped too early Formula for Chebyshev polynomials as a sum of monomials can be found at least three ways 1. By orthogonalisation with Inner product = Int(p(x)q(x)*1/sqrt(1-x^2),x=-1..1) 2. By solving recurrence relation a) define exponential generating function b) solve ordinary differential equation E''(t)-2xE'(t)+E(t) = 0 E(0) = 1 E'(0) = x Suggested method is Laplace transform c) expand E(x,t) = exp(xt)cos(sqrt(1-x^2)t) with Leibniz product rule (It looks like binomial expansion but instead of powers we have derivatives) d) we can go further and expand (x^2-1)^k with binomial expansion e) use fact that binomial(n,k) = binomial(n,n-k) and reindex the inner sum f) change order of summation g) calculate the sum sum(binomial(n,2*k)*binomial(k,m),k=m..floor(n/2)) I have problems with 2 g) 3. By solving ordinary differential equation (linear , homogeneous second order but not with constant coefficients) a) Let y(t) = cos(nt) b) differentiate y(t) twice to get y''(t) = -n^2cos(nt) y''(t) = -n^2y(t) y''(t) + n^2y(t) = 0 c) lets change of independent variable t = arccos(x) to get (1-x^2)y''(x) - xy'(x) + n^2y(x) = 0 d) lets solve this equation with power series y(x) = sum(a_{k}x^{k},k=0..infinity) e) we get two step recursion which can be easily expressed as a product but warning if we want to expand this product to factorial and later to binomial coefficient we may get factorial of negative number or division by zero Once we have Chebyshev T polynomials we can easily get Chebyshev U polynomials because U_{n}(x) = 1/(n+1) d/dx T_{n+1}(x) These three approaches works for all orthogonal polynomials
I think one small test we can do is to try to divide the equation so constant term is between -1 and 1 and check to see if sum of rest of the coeffecients is one or not by the fact which can be proved just by putting theta =0
T_n/2=(cos(θ)+i*sin(θ))^n+(cos(θ)-i*sin(θ))^n The first part is a circle function, the second part removes the sin part of that circle function, making it a real number because it's a conjugate. This is a Lucas Number thing. All properties Lucas Number have, this will also have. I'm sure the sum is easy to calculate.
You can prove the sum of coefficients just by induction Let S(n) be the sum of coefficients of T(n). We know S(0)=1 and S(1)=1. Then we can say S(n) = 2*S(n-1)-S(n-2) = 2*1-1 = 1
@@shreyan1362 Yes, that is the result. 16x^4+8x^3-16x^2-8x+1=0 can be solved using quartic methods, but I don't think there is a particularly simple substitution (certainly not simply in terms of x^2, as suggested). en.m.wikipedia.org/wiki/Quartic_function www.wolframalpha.com/input/?i=solve+16x%5E4%2B8x%5E3-16x%5E2-8x%2B1%3D0
We can use the effective Piezas methods for resolving quintic that are solvable by radicals. Then the solutions are the form a+(z_1)^(1/5)+(z_2)^(1/5)+(z_3)^(1/5)+(z_4)^(1/5). Where z_1,z_2,z_3,z_4 are roots of quartic polynomial. This quartic polynomial is solvable only using square root. And (z_i)^(1/5) is one root of unity. We can a few complicated ordering thats roots. (If the polynomial has one real solutions, then z_1,z_2,z_3,z_4 are real,the a+(z_1)^(1/5)+...+(z_5)^(1/5)). (If the polynomial has 5 real solutions is imposible express the roots in terms of real radicals (standard)).
One perhaps clearer way of writing the proof of this recurrence is exp(i(n+1)x) = exp(inx)exp(ix) + exp(inx)exp(−ix) − exp(inx)exp(−ix) = exp(inx)(2 cos x) − exp(i(n−1)x), from which the cosine ("real") and sine ("imaginary") parts can be separated.
I also noticed that the sum of the absolute values of the coefficients of T_(n+2) equals 2 times the sum of the abs values of the coefficients of T_(n+1) plus the sum of the abs values of the coefficients of T_n. I'm working on trying to prove it but I think you have to use a power of i times T_n(i) for the general term to get a positive integer. In terms of T_n(x), you are taking x=arccos(i), which is a multivalued function, but its principal value is π/2 - i ln(1+√2) ! 😅
I am thinking if you could transfer the knowledge of the C-P onto the harmonic series because it looks Like cos(x)=×, cos(2x)= x^2 So that cos(2a)+cos(2b)+cos(2c)+...cos(2z) = a^2+b^2+c^2..+z^2 Or: cos(a) + cos(2a)+cos(3a)...+cos(inf×a) = a+a^2+a^3...+a^inf And maybe isn't that similar to the way Euler solved the "Basel Problem"?
In France, we write it as Tchebychev so I guess it kinda depends on where you're from (maybe in Russian, it does start with the equivalent to the t of the Latin Alphabet)
kind of! the first letter in the Russian name (Чебышёв) is the closest thing you can get to a French t+ch consonant cluster, and the difference in orthography is just ... because of the difference in how ch is used.
@@blackpenredpen In French, CH sounds like an English SH, like you see in Michigan or Chevrolet. They use the T in front of ch, to make it sound like the CH in Charles. An example is the country of Tchad, that we call Chad in English.
Yes; there are two families of Chebyshev polynomials. Incidentally, there are numerous ways to write his name in the Latin alphabet, so be aware when you go looking. I think my favorite is Čebyšev.
If you have the cos(nx) formula, then sin(nx) = d/dx -cos(nx)/n works for all n except 0. Interestingly, the recursion remains the same sin((n+1)x) = 2cos(x)sin(nx) - sin((n-1)x) except starting values are different sin(0x) = 0 sin(1x) = sin(x) sin(2x) = 2cos(x)sin(x) - 0 sin(3x) = 2cos(x)(2cos(x)sin(x)) - sin(x) = -4sin^3(x) + 3sin(x) etc. See e.g. trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesRecursiveFormula.htm and related pages on that site.
Huh, I was looking into these on my own (without knowing the name) and was analyzing the coefficients to see what patterns I could find to create a general formula. Didn’t come up with anything. May go back and compute a few more terms.
Can you see the phi number into some solutions? In fact you may verify in Wolfram Alpha (for example) and you will find that local maxima of the function represented by polynomial function not equalled to zero are (1/(2 phi)) and (- phi/2), and the local minima (- (1/(2 phi))) and (phi/2).
I thinkt it is called T_n because Chebyshev was a russian, so you need a transliteration vom cyrillic to the latin alphabet. And a different common transliteration is Tschebyschow. But that is just a guess. Edit: should have looked into the comments.
Is there a general formular for the nth term of T? I mean can find the nth term without knowing the (n-1) and (n-2) th terms?(ex: cos 10x without knowing cos 9x and cos 8x)
First of all here we have yo (e with diaeresis, although diaeresis is usually omitted it is still there) and sh is hard so we say this name Chebyshov (more precisely Chiebyshov) Moreover accent is on yo (Accent is flowing and knowing where is accent is important for correct reading) You read it wrong and other take example from you and also read it wrong even if they where taught this language How I get formula for Chebyshov polynomials I started with cosine of sum then i wrote recursive relation for it cos((n+1)t)=cos(nx)cos(t) - sin(nt)sin(t) cos((n-1)t)=cos(nx)cos(t) + sin(nt)sin(t) cos((n+1)t)+cos((n-1)t) = 2cos(t)cos(nt) Let cos(t) = x and cos(nt)=T_{n}(x) T_{n+1}(x)+T_{n-1}(x)=2xT_{n}(x) T_{n+1}(x) = 2xT_{n}(x) - T_{n-1}(x) T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x) so we have recursion T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x) T_{0}(x) = 1 T_{1}(x) = x Here I used exponential generating function E(x,t) = sum(T_{n}(x)t^n,n=0..infinity) After plugging it into the recursive relation i have got second order linear ODE with constant coefficients (in fact initial value problem) This differential equation was easy to solve with Laplace transform Once i have got exponential generating function it appeared that it is easy to calculate nth derivative using Leibniz's product rule
I was wondering where the coefficient 4 in 4cos³θ came from, so I figured: cis(3θ)=cos³θ+3icos²θsinθ-3cosθsin²θ-isin³θ =cos³θ+3i(1-sin²θ)sinθ-3cosθ(1-cos²θ)-isin³θ.
In 5:00 you added cosnthetacos theta twice then you sdould have substracted it twice so that original value don't change but you added sinnthetasintheta which is wrong
Not entirely related, but I haven't been able to find anything online that shows me how to obtain the principal quintic form in terms of the general quintic equation's coefficients. Is it possible to even do this?
Chebyshev polynomials are designated as T because early transliterations from the Cyrillic rendered "Chebyshev" as "Tchebicheff" - similar to how "czar" was once (and sometimes still is) spelled "tsar".
Oooh I see. Thanks.
It should actually be "Chebyshóv" (Чебышёв) but, for some reason, people forget to yoficate it.
@@pierreabbat6157 the reason is that russians often write the letter е instead of ё. This letters are pronounced differently
@@pierreabbat6157 So maybe we should designate them as Ч(x). Cyrillic needs more representation in the symbols used in mathematics. I've heard that sha is used as a variable name, but that's the only one I know about so far.
@@pierreabbat6157 Interesting how there is no sounded letter in the original spelling where the y is in the transliterated spelling. Just the soft sign. I never expected that. I would've expected Чебйшёв.
If you read the subtitles, these are actually "chubby chef" polynomials.
😂
Close enough
@sco phone We believe in you and support you, subtitles!
I failed my calc exam because I was sitting between two identical twins ,so it was hard to differentiate
I didn’t fit in well at school, so I couldn’t integrate. But I was able to differentiate though.
I also experience an infinite series of similar pain. It makes me want to go to the L’Hôpital.
I didn't like my school, so I did the ole' la place transformation.
@@mgancarzjr Moving to a new country was tough for me. I decided to integrate by parts to slowly ease myself in
@@HeyKevinYT don't worry every pain has a limit... It's good that you went to L' Hopital.
Sheer excellence!
I would love to see you do more STEP questions!
Fun vid. Interesting .... Solving is satisfying. Great vid. ...Gotta get my trig book off the shelf.
This Is The Best Quintic Equation Ever!
There is a type of microstrip filter used in RF known as a Chebyshev filter as its topology is derived from Chebyshev polynomials.
I love the mic just as much as the math
This is probably a big coincidence, but you released a video on this topic EXACTLY after the end of one of the Russian Olympiads held by the KGB for schoolchildren, where there was a problem on EXACTLY THIS topic. I just wrote this Olympiad and remember this problem. We uncovered you, you are an overseas KGB agent...
I think he's might be a K Gamma 6 agent.
Ordinary generating function for Chebyshov polynomial is
G(x,t)=(1-xt)/(1-2xt+t^2)
Exponential generating function for Chebyshov polynomial is
e^{xt}cos(sqrt(1-x^2)t)
Exponential generating function gives us formula with derivatives
T_{n}(x) = limit (d^n/dt^n e^{xt}cos(sqrt(1-x^2)t), t = 0)
I’m not sure if it’s always true, but I noticed in the examples that you did, that when n is odd, we only see odd powers of cosine, and when n is even, we only see even powers. Its not much, but just anothher interesting fact I saw.
You are right and we can use the recursive formula to prove it.
Interesting to note the Chebyshev looks similar to the Fourier series!
Best videos ever
Maybe you can record video about Chebyshev polynomials
Michael Penn derived formula with complex numbers but he stopped too early
Formula for Chebyshev polynomials as a sum of monomials can be found at least three ways
1. By orthogonalisation with Inner product = Int(p(x)q(x)*1/sqrt(1-x^2),x=-1..1)
2. By solving recurrence relation
a) define exponential generating function
b) solve ordinary differential equation
E''(t)-2xE'(t)+E(t) = 0
E(0) = 1
E'(0) = x
Suggested method is Laplace transform
c) expand E(x,t) = exp(xt)cos(sqrt(1-x^2)t) with Leibniz product rule
(It looks like binomial expansion but instead of powers we have derivatives)
d) we can go further and expand (x^2-1)^k with binomial expansion
e) use fact that binomial(n,k) = binomial(n,n-k) and reindex the inner sum
f) change order of summation
g) calculate the sum
sum(binomial(n,2*k)*binomial(k,m),k=m..floor(n/2))
I have problems with 2 g)
3. By solving ordinary differential equation (linear , homogeneous second order but not with constant coefficients)
a) Let y(t) = cos(nt)
b) differentiate y(t) twice to get
y''(t) = -n^2cos(nt)
y''(t) = -n^2y(t)
y''(t) + n^2y(t) = 0
c) lets change of independent variable t = arccos(x) to get
(1-x^2)y''(x) - xy'(x) + n^2y(x) = 0
d) lets solve this equation with power series
y(x) = sum(a_{k}x^{k},k=0..infinity)
e) we get two step recursion which can be easily expressed as a product
but warning if we want to expand this product to factorial and later to binomial coefficient
we may get factorial of negative number or division by zero
Once we have Chebyshev T polynomials we can easily get Chebyshev U polynomials because
U_{n}(x) = 1/(n+1) d/dx T_{n+1}(x)
These three approaches works for all orthogonal polynomials
I think one small test we can do is to try to divide the equation so constant term is between -1 and 1 and check to see if sum of rest of the coeffecients is one or not by the fact which can be proved just by putting theta =0
The value between x and cos12° in the beginning was so complex that i thought how approch the value😂😂 but this problem solving was so great
Great video. Thank you
You teach math better than my teacher
T_n/2=(cos(θ)+i*sin(θ))^n+(cos(θ)-i*sin(θ))^n
The first part is a circle function, the second part removes the sin part of that circle function, making it a real number because it's a conjugate. This is a Lucas Number thing. All properties Lucas Number have, this will also have. I'm sure the sum is easy to calculate.
You can prove the sum of coefficients just by induction
Let S(n) be the sum of coefficients of T(n). We know S(0)=1 and S(1)=1. Then we can say S(n) = 2*S(n-1)-S(n-2) = 2*1-1 = 1
Wow
I really like his explaination. "How could we continue this video." A valid math proof 😂
Really like this🔥🔥
After finding that 1/2 is a solution of the equation, we have a biquadratic equation easily solvable with X=x^2
Can you explain?
16 x^4 + 8 x^3 - 16 x^2 - 8 x + 1
@@shreyan1362 Yes, that is the result.
16x^4+8x^3-16x^2-8x+1=0
can be solved using quartic methods, but I don't think there is a particularly simple substitution (certainly not simply in terms of x^2, as suggested).
en.m.wikipedia.org/wiki/Quartic_function
www.wolframalpha.com/input/?i=solve+16x%5E4%2B8x%5E3-16x%5E2-8x%2B1%3D0
After making your substitution x = cos(theta) and getting your polynomial in a nice form, your coefficients still sum up to 1. Kinda cool.
I want the whole world to see this video at 0:44 because here i have to memorise 200 TRIG FORMULAS ICHDJFJSNFJSBCNJD
We can use the effective Piezas methods for resolving quintic that are solvable by radicals. Then the solutions are the form a+(z_1)^(1/5)+(z_2)^(1/5)+(z_3)^(1/5)+(z_4)^(1/5). Where z_1,z_2,z_3,z_4 are roots of quartic polynomial. This quartic polynomial is solvable only using square root. And (z_i)^(1/5) is one root of unity. We can a few complicated ordering thats roots. (If the polynomial has one real solutions, then z_1,z_2,z_3,z_4 are real,the a+(z_1)^(1/5)+...+(z_5)^(1/5)). (If the polynomial has 5 real solutions is imposible express the roots in terms of real radicals (standard)).
The method of Piezas is a copy of the method of Watson and Young. A better aproach is the method of Malfatti or McClintock
The sum off all the coefficients is 1 because if you give to tita the value of 0 , you get a1+a2....+an = cos(n0)=1
One perhaps clearer way of writing the proof of this recurrence is
exp(i(n+1)x)
= exp(inx)exp(ix) + exp(inx)exp(−ix) − exp(inx)exp(−ix)
= exp(inx)(2 cos x) − exp(i(n−1)x),
from which the cosine ("real") and sine ("imaginary") parts can be separated.
I also noticed that the sum of the absolute values of the coefficients of T_(n+2) equals 2 times the sum of the abs values of the coefficients of T_(n+1) plus the sum of the abs values of the coefficients of T_n.
I'm working on trying to prove it but I think you have to use a power of i times T_n(i) for the general term to get a positive integer. In terms of T_n(x), you are taking x=arccos(i), which is a multivalued function, but its principal value is π/2 - i ln(1+√2) ! 😅
Video request: Find the maclaurin series of f(x)=(2-x^2)^1/3
Just a fun fact: 12 = 36/3 and cos 36 degrees (pi/5) is algebraic (either sin 18 or cos 18 has a nice form)
we can prove that sum of coefficient will be 1 like below:
cos(ntheta)=costheta {f(costheta)} if we check limit theta tends to zero both side ...
Chebyshev polynomialls are called Tn, Un, Vn, Wn, that's why not just Cn.
I am thinking if you could transfer the knowledge of the C-P onto the harmonic series because it looks Like cos(x)=×, cos(2x)= x^2
So that cos(2a)+cos(2b)+cos(2c)+...cos(2z) = a^2+b^2+c^2..+z^2
Or:
cos(a) + cos(2a)+cos(3a)...+cos(inf×a)
= a+a^2+a^3...+a^inf
And maybe isn't that similar to the way Euler solved the "Basel Problem"?
0:47 we all need that shirt !!
Teachers only : )
Most needed during exams
BPRP is the Quinticsential math professor.
It was so neat .tx
GF for cos(na) is
G(x)=(1 - cos(a)*x)/(x^2 - 2cos(a)*x+1)
with your idea x^5-20x^3+80x-32=0 → x=1/2 √3 √(10+2√5) +1/2 (√5-1)
tanh(3arctanh x) may look pointy, but it’s smooth on its entire domain.
Where can I get that trig shirt? Did not see it in the store. And it's upside down, nice!
In France, we write it as Tchebychev so I guess it kinda depends on where you're from (maybe in Russian, it does start with the equivalent to the t of the Latin Alphabet)
kind of! the first letter in the Russian name (Чебышёв) is the closest thing you can get to a French t+ch consonant cluster, and the difference in orthography is just ... because of the difference in how ch is used.
Ah I see!
@@blackpenredpen In French, CH sounds like an English SH, like you see in Michigan or Chevrolet. They use the T in front of ch, to make it sound like the CH in Charles. An example is the country of Tchad, that we call Chad in English.
Chebyshev Polynomial makes TRIG fun imo.
They do!!
I would have derived it using powers of cos + i sin
Can you find a similar formula for sin(n * teta)?
Yes; there are two families of Chebyshev polynomials. Incidentally, there are numerous ways to write his name in the Latin alphabet, so be aware when you go looking. I think my favorite is Čebyšev.
@@tomkerruish2982 West of Russia
If you have the cos(nx) formula, then
sin(nx) = d/dx -cos(nx)/n
works for all n except 0.
Interestingly, the recursion remains the same
sin((n+1)x) = 2cos(x)sin(nx) - sin((n-1)x)
except starting values are different
sin(0x) = 0
sin(1x) = sin(x)
sin(2x) = 2cos(x)sin(x) - 0
sin(3x) = 2cos(x)(2cos(x)sin(x)) - sin(x) = -4sin^3(x) + 3sin(x)
etc.
See e.g. trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesRecursiveFormula.htm and related pages on that site.
1:57 blackpenredpen...bluepen?
Huh, I was looking into these on my own (without knowing the name) and was analyzing the coefficients to see what patterns I could find to create a general formula. Didn’t come up with anything. May go back and compute a few more terms.
The name of the polynomial refer to my professor : chubby chef. Lol
BTW great one as always❤️
Absolutely beautiful
Can you see the phi number into some solutions?
In fact you may verify in Wolfram Alpha (for example) and you will find that local maxima of the function represented by polynomial function not equalled to zero are (1/(2 phi)) and (- phi/2), and the local minima (- (1/(2 phi))) and (phi/2).
Yes. This is what Grant of 3Blue1Brown calls "phi's little brother".
Yay!!
0:46 wait that's illegal!😂
Dear @blackpenredpen
Can you suggest some nice way to memorise trig identities? I have soooo many of them!! At least like 50
Agree
How about sin(nθ)?
It is (1/COS C) [ Only legends can understand 😉 ]
Sec see
@@jagula no man...... Some thing BOLD
@@sapy4124 i got it but i dont want to write it here xD
@@jagula modulus 0 :- l0l
I thinkt it is called T_n because Chebyshev was a russian, so you need a transliteration vom cyrillic to the latin alphabet. And a different common transliteration is Tschebyschow. But that is just a guess.
Edit: should have looked into the comments.
This transliteration looks like german
@@holyshit922 There are different transliterations. I'm not an exoert, but I assume the german transliteraion is Tschebyscheff
@@AndDiracisHisProphet Maybe, but why in most transliterations there is e when in Cyrillic there is e with diaeresis Чебышёв
@@holyshit922 I dunno. I asked Wikipedia
Excellent presentation. Wow !! DrRahul Rohtak India
Chevy Chase polynomials? : )
Is there a general formular for the nth term of T? I mean can find the nth term without knowing the (n-1) and (n-2) th terms?(ex: cos 10x without knowing cos 9x and cos 8x)
"Champion-Polynomial"
First of all here we have yo (e with diaeresis, although diaeresis is usually omitted it is still there) and sh is hard so we say this name Chebyshov (more precisely Chiebyshov)
Moreover accent is on yo (Accent is flowing and knowing where is accent is important for correct reading)
You read it wrong and other take example from you and also read it wrong even if they where taught this language
How I get formula for Chebyshov polynomials
I started with cosine of sum then i wrote recursive relation for it
cos((n+1)t)=cos(nx)cos(t) - sin(nt)sin(t)
cos((n-1)t)=cos(nx)cos(t) + sin(nt)sin(t)
cos((n+1)t)+cos((n-1)t) = 2cos(t)cos(nt)
Let cos(t) = x and cos(nt)=T_{n}(x)
T_{n+1}(x)+T_{n-1}(x)=2xT_{n}(x)
T_{n+1}(x) = 2xT_{n}(x) - T_{n-1}(x)
T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x)
so we have recursion
T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x)
T_{0}(x) = 1
T_{1}(x) = x
Here I used exponential generating function
E(x,t) = sum(T_{n}(x)t^n,n=0..infinity)
After plugging it into the recursive relation i have got second order linear ODE with constant coefficients
(in fact initial value problem)
This differential equation was easy to solve with Laplace transform
Once i have got exponential generating function it appeared that it is easy to calculate nth derivative using Leibniz's product rule
0:54 and what is formula for kth coefficient of Chebyshov polynomial nth degree ?
Apparently the intuition of T_n is that you add cos (n+1)t and cos (n-1)t to get 2 cos nt cos t
That moment at 20:10 remembers me of CrazyRussianHacker when he forgots tp cut theese parts 😂
Great job
Cool video ! Thank you
sum must be 1. suppose theta = 0, cos theta = 1, which forces the other side to have coef summed to 1
I was wondering where the coefficient 4 in 4cos³θ came from, so I figured:
cis(3θ)=cos³θ+3icos²θsinθ-3cosθsin²θ-isin³θ
=cos³θ+3i(1-sin²θ)sinθ-3cosθ(1-cos²θ)-isin³θ.
What is cis
WE WANT A PART2 OF THIS!!!!
woooo yes
The reason they use T for Chebyshev polynomials is because of the French Tchebychev.
In 5:00 you added cosnthetacos theta twice then you sdould have substracted it twice so that original value don't change but you added sinnthetasintheta which is wrong
I catch the part in which 1 is 1
Hi, you made a video about arc lenght= area can you do it on a 3d shape where surface area = volume
Just finished the the 100 integrals video....why did you change your look.?
To look like an ancient philosopher mathematician of course ;)
Not entirely related, but I haven't been able to find anything online that shows me how to obtain the principal quintic form in terms of the general quintic equation's coefficients. Is it possible to even do this?
Yeah, it is possible to do it, using a Tschirnhaus transformation of second order and apply the resultant.
Don't have time ... I have an Algebra & Calculus Exam tomorrow , But I'll make sure I watch this vid sometime ..
Thanks.
1:17 plot armor!😂
part 2 plsssssssssssssssssssssssssssssssssssssssss :)
Sir plz next video how to find 6th derivative of X.e^X by using leibnitz rule.
What is that object, That always you hold it in your right hand?????
*left. It is a mic.
@@andreffrosa yes, but more importantly, it's a pokeball!
Please change your pfp back to normal, not that the new one is bad, it is just that the old one is so iconic
Toppppppppp
#polynomial
Just use the quintic formula that 100%-ly exists .... 🌚
No moon day in the last?.... I guess?
20:03 The original "to the" : Am I a joke to you?
That's what I thought exponents were called, before I knew the term. I called them "ta-the's", because it is read aloud as "2 to the 4th".
My favourite comment: “One is one” 😁😁😁
That could be hard to be found being Chebyshev...
#QuinticEquation
Or da sab badiya😂😂
Super
@Cookie Monster good one🤣🤣
@@nandeesh2ninad Good e^(2*pi*i)!
@@nandeesh2ninad Or should I say?
गूड ए^(२*पी*ई)
BlackPenRedPen: **Makes a new video**
Me: #Yay
Pls help me integrate x^2 /{ (1-x^4) * sqrt ( 1+x^4) }
This is my iit jee integral
Even wolfram alpha failed lol
Pls help
Pls
No
Don't think there's an elementary function to the integral
Use integral calculator. Fellow jee aspirant here
Oh wow great question man
I seems like i have seen this integral somewhere🤔
Nice beard though
bro please
I also noticed the first coefficient of each identity is always equal to 2^(n-1)
@bprp @13:11 can u prove that 1 is 1 🤪
I want that t shirt 😂
Juste use the formulas when you have a 4th degree equation 🤪
The beard, oh dear.