Kon hai jiska video dekhne se phale hi hogaya tha 🌝 . To master the true art of advanced-level problem solving join our course, jeesimplified.com/set-of-60
4:19 Here we can also convert z^(2n-1)/2n-1 into a integral and switch the integration and summation. It would form a GP and then we can directly integrate
@@aryannamboodiri8974What works, works. Neither you, nor me, have studied complex analysis to formally analyse the convergence of the series. Just because something works for real numbers doesn't mean it works for complex numbers.
My solution development: General term- Re{e^2nix/(2n-1)(2n+1) 1/(2n-1)(2n+1)= (1/2)([1/2n-1] - [1/2n+1]) {Splitting the fractions} , now we have two summations we can solve them individually by two functions f(x)=Summation e^2nx We can easily find the f(x) as this is a GP so f(x) is known here , and now we multiply e^x on both sides (e^x)f(x)=Summation e^(2n+1)x (You get the idea here , we have the power 2n+1 on e, so we can integrate on both sides and we will get the coefficient of power in the denominator) integral (e^x f(x))= Summation (e^(2n+1)x)÷2n+1 , this is exactly what we needed, we can put x= ix in this function and we have the value for the first summation) Similarly, we can divide the function by e^x on both sides and find the value of 2n-1 wala summation by integration on both sides 👍 P.S. Using integration is a very common idea in binomial as well
10:29 glt hai bhai complex me aise log nhi le skte, i ko to e^i×pi/2, e^i×5pi/2,.. in general e^i(4n+1)pi/2 se bhi represent kr skte hai sbme log ki alag value aa jaayegi
12:00 bilkul sahi kaha bhai... aapka comment section me kuch zyada hi log batata hai ki atleast 15-20 minutes lagne wale sawaal 1-2 minutes me ho gaya... aur ab people are starting to say 7-10 minutes to sound a bit more believable 💀
bhai kuch log to ye bhi bolte hai ki i just stated 11th and complete in hardly 2 mins bhaiya ki chat se hi 2024 jee adv air 1 , inmo, rmo ke air 1 ayenge
Bhaiya just for solving , me x=0 dalke solve karta and opt me 0 dalke dekh leta , if I am lucky 3 opt eliminate ho jate , but learning ke liye bhaut op question tha
@@bluesky5794 hey hey I am not even able to score good in mains Math's but I like to explore Math's and just want satisfactory score in mains and advanced 😊.
Bhaiya trigonometry me to multiple types of complex formulas hote h jo ki in general bohot se books me bhi nhi diye rhte h. Aur waise formula se Q Advanced me bhi are h. Un formulas ko kaise janenge?
Bhaiya ho gaya !! maja aagaya ekdum !! Aise sawal indefinite integration mein bhi hote hai cosx aur sinx ko hatake real ya imaginary part of exponential function lelete hai meine waise sawal kaafi kr rkhe shayad isliye 1st baar mein strike ho gaya
@@themadtitan728 nhi jee mein to nhi aya but hamari coaching mein karaye the ek do aur mereko thoda calculus mein interest hai to mein karta rehta hu idhar udhar se uthake MATHS505 name se ek channel hai uspe aate hai aise badiya integration ke videos but wo jee advance relevent bhi nhi to i wont recommend ki tum abhi uske videos dekho
@@themadtitan728 nhi yaar pehle to kabhi nhi aya but hame coaching mein karaye the aur mereko thoda calculus mein interest hai isliye mein krta rehta hu idhar udhar se utha kar
Hello there, Thank you for the great video ! I used the Fourier series to solve this Cos(2nx) can be written cos((2n-1)x +x) and costly((2n+1)x - x) and proceeded forward In the end I got 1/2 - sinx * Sigma sin((2n-1)x)/(2n-1) The Sigma part is Fourier series of square wave Which is π/4 if 0
Denominator can be broken into (2r-1)(2r+1) then numerator has cos(2rx) submission which can be easily broken down into terms of 2r-1 and 2r+1 Then we have Sin series But integration and limit together can make it compex
I was done with the conceptual part in around 15 mins, then for calulation it took me 30 mins. many calculation errors i made. solved it many times to approch the correct answer
I am rarely able to do these questions but after taking simple hints I understand them very quickly And please keep sharing all these wonderful questions with us even after this 75 hard thing
Let the sum is f(x) Differentiate two times, you'll get f"(x) + f(x) = 1/2 Solution of this differential equation is: f(x) = 1/2 + Acosx + Bsinx For 0 < x < π Now put x= π/2 f(π/2)= 1/2 - π/4 f'(π/2) = 0 You'll get B = -π/4 , A = 0 Solved My Intro: JEE Advanced 2021 qualifier. Edit: Those who are saying ki class 12th mein 2nd degree equation nahin aati. Ye baat mai bhi jaanta hu, lekin class 11th mein agar SHM padha ho toh yaad Karo har jagah issi tarah ki equation use hoti hai 🗿
niche -isinx/2 hai so 1/i=-i so - sign cancle hojayega..... aur wo ln(i)=i (pi/2 + 2n pi) hi hoga but bhaiya ne simplest term li hai, general term nahi li.
you are actually completely right , the complex logarithm function is actually a multivalued function which repeats over 2πki and with this logic the answer should've been multivalued which certainly seemed weird and bothers me as well , however weirdly whenever you put a complex no. let's say i , into the Taylor series expansion of a function it tends to give the principal argument as the answer if it's centred somewhere at the unit circle , however if you analytically continue the function outside the radius of convergence and make one rotation about the pole of the function (at x=0) we get the answer as 2π + π/2 and so on , however I'm not really very sure as complex numbers are weird sometimes so you should probably ask your math teacher and tell me what his solution is 👍
Not meant to brag, but done in around 13 minutes being an 11th entering student : My approach: 1. Convert cos(2nx)= 1/2 (e^i2nx + e^-i2nx). 2. Separate the series into two series (one of positive powers and other of negative powers) . 3.Solve the first series by substituting a=e^ix. After few manipulations in first series, we will get : [(a²-1)/4a].[ln{(1+a)/(1-a) }]+0.5 4. Similarly in the second series, repeat the process by substituting b=e^-x. 5.Add both the parts and simplify to obtain the final answer.
I tried to differentiate and integrate the general term since kuch dino pehela aap na ekk series ko aisa solve kiya tha but kuch nehi Hua 😂. But really a monster question ❤.
Brother, this series is not equal to 1/2-(pi/4)sin(x) always. It is a periodic series with period pi. We can check this by x = 3pi/2, S = -(1/1.3) + (1/3.5) - (1/5.7) = (1/2)(-1+1/3+1/3-1/5-1/5+1/7+....) = (1/2)-(1-1/3+1/5-1/7+...) = 1/2-arctan(1) = (1/2)-(pi/4) from formula, x = 3pi/2 gives S = (1/2)+(pi/4), so the series is equal only in {(0,pi) +2k(pi)} S = 1/2+(pi/4)*(sinx) in (pi,2pi) + 2kpi; S = (1/2 )- (pi/4)(|sinx|) Not sure if it was MCQ with multiple options or only one option is given in JEE Adv
However, the other result can be achieved by seeing that- log(icot(pi/4)) = log(i) = i*pi/2 and log(i*cot(3*pi/4) = log(-i) = -i*pi/2; log(z) = log|z| + i arg(z) => arg(i*cot(x/2)) = pi/2*(|cotx/2|/cotx/2) = pi/2*(|sinx|/sinx) so Im(log(icotx/2) ) = pi/2*(|sinx|/sinx) S = (1/2)-(pi/4)*(|sinx|)
i was not able to think in this q mtlb the first three introductory steps were taken by me lkin ye sochna ki ye q complex ka h , it was mind boggling .
bhai complex toh soch liya tha but fir expansions dhyan mai nhi aayi itne questions karre hai bhout try ke baad trigo kaa savaal naa hone par yhi hoo sakta hai ki voh complex se hoo rha ho
dekh bro pehle nth term likh then dono ko alag alag likh phir just like infinite gp terms kat jaenge bas 1st term bachega jo n=1 dalke aa jaega aur last term k liye limit solve kr lena 1 page solution
6:37 at this point the expression of other term is written wrong It should be 1/2ln(1+z/1-z) -z. U didn't consider the half wala factor Edited: I didn't see further(sorry)
Bhaiya baaki mujhko pta nhi mainey yhi ekk kiya ki x ko zero dalkar aur 1/2 common nikallkar infinite V/N bana aur usmey last ke terms bahut badey honey ke karad kaat diye phir kya 1/2 answer baki , phir bhi ekk pura pie by 4 term rh gya , dusreyy x ke liye alg alg answer hoga ... that's my fault ❤❤❤ Hii likhdo mereko bhi kabhi kabhi akele akele editor rakhhey video bana rhi bss 😅😅
Bhaiya kash aap mera soln dekh pate, mujhe 2cos2x+1 mila hain finally , kyo wrong hain pta hi nhi chl rha maine hi complex numbers use kiye the maine cos 2rx ko exponential main expand kiya fir , telescopic series banayi jis me chaar alag alag summation bn rhi hain is form main sum ( 1 to infinity )= t^r/r is sum -ln(1-t) nikalti hain pr maine substitute kiya toh answer galat aa raha hain.
Mainee eshaa ekk question already kahii dekha tha but usmee sirf ln(1+x) ka paternn observerd krnaa thaa.... I think that's why i was able to solve it ✌🏼
i didn't knew how to solve it so just calculated what will be the value of this equation when x will be 2mpie and got 1/2 [when will i be able to solve advance level questions😔😔😔]
Kon hai jiska video dekhne se phale hi hogaya tha 🌝
.
To master the true art of advanced-level problem solving join our course, jeesimplified.com/set-of-60
Abhi sojata hu solution kal dekhunga🌝🌝
ho gya tha 🤣🤣
ye sab substitutions basic hain
Done bhiya 11-12 minute lag gye
Bhaiya aise 10 sawaal to mere allen ke test me bhare pade the (btw i m from star batch or tms in allen so this was easy obviously)
Yeah solved it. I have interest in calculas stuff. So I got it.
you couldve given me 5 days to solve this and i still wouldnt even think about complex numbers 💀
True
Point hai
Tru
But in 6th day u wud?
Indeed..6th day ban jata
4:19 Here we can also convert z^(2n-1)/2n-1 into a integral and switch the integration and summation.
It would form a GP and then we can directly integrate
No
@@puneetparashar2935 Dont answer No without having proper knowledge.
Nope the series does not converge for |z|=1
@@aryannamboodiri8974What works, works. Neither you, nor me, have studied complex analysis to formally analyse the convergence of the series. Just because something works for real numbers doesn't mean it works for complex numbers.
but that is the whole point of taylor series tbh
Bhaiya ye to Mera Sawaall hai thanks..❤
Bahut ache dost question kaha ka hai please bata do
@@YuvrajGupta-m2o NIT ka hai bhai
@@SidemenClips_77 tu surathkal me hai kya
Log expansion and complex substitution solved it when were preparing for IIT JEE -2010 . Brings back memories. Great content bro
Do you got IIT?
Bro could u tell the expansions?
My solution development:
General term- Re{e^2nix/(2n-1)(2n+1)
1/(2n-1)(2n+1)= (1/2)([1/2n-1] - [1/2n+1]) {Splitting the fractions} , now we have two summations we can solve them individually by two functions
f(x)=Summation e^2nx
We can easily find the f(x) as this is a GP so f(x) is known here , and now we multiply e^x on both sides
(e^x)f(x)=Summation e^(2n+1)x (You get the idea here , we have the power 2n+1 on e, so we can integrate on both sides and we will get the coefficient of power in the denominator)
integral (e^x f(x))= Summation (e^(2n+1)x)÷2n+1 , this is exactly what we needed, we can put x= ix in this function and we have the value for the first summation)
Similarly, we can divide the function by e^x on both sides and find the value of 2n-1 wala summation by integration on both sides 👍
P.S. Using integration is a very common idea in binomial as well
Pratham Bhai aapka comment pe pakka reply karega
Bhai tu Mumbai suburban area me rehta ha kya ?? Just asking becuase yaha pe ek lakshay name ka toppr h
Bhaiya how to develop thinking like you pl tell. I am in 11th jee 2025
nice solution!! but i think it will become more lengthy
Woh ioqm ke liye prepare kiya hai toh us lvl ki thinking hoti hi hai@@ninja0o0557
10:29 glt hai bhai complex me aise log nhi le skte, i ko to e^i×pi/2, e^i×5pi/2,.. in general e^i(4n+1)pi/2 se bhi represent kr skte hai sbme log ki alag value aa jaayegi
search branch cut for multivalued functions.
12:00 bilkul sahi kaha bhai... aapka comment section me kuch zyada hi log batata hai ki atleast 15-20 minutes lagne wale sawaal 1-2 minutes me ho gaya... aur ab people are starting to say 7-10 minutes to sound a bit more believable 💀
bhai kuch log to ye bhi bolte hai ki i just stated 11th and complete in hardly 2 mins
bhaiya ki chat se hi 2024 jee adv air 1 , inmo, rmo ke air 1 ayenge
Chup chap se x ko kuch mano 3-4 term tak plus kro aur options me eliminate krlo😂😂
Put x as 0
@@AnaySarda Ha mera MAtlab wahi Tha 😅
Aur agar options na ho toh 😅skip the question 😂
@@decentaditya1382 options must be there as the ans is in terms of x so it cant be a numerical qn
@@AnaySarda and what if the questions comes in this form
S = - π sin(x)/a + 1/b than find a+b
lo karlo x=0 put 🙂
Khilwa denge ... Bro's creativity is increasing day by day ( as content creater)😂😂
Why i got it in my recommendations, ab toh 3rd yr engineering ka khtm hone wala hai😂😂 kaafi convenient days the😢
Bhaiya just for solving , me x=0 dalke solve karta and opt me 0 dalke dekh leta , if I am lucky 3 opt eliminate ho jate , but learning ke liye bhaut op question tha
bro this is a really good question mujhe bilkul approach pata hi nahi tha thanks man
Already enjoyed this problem months back on Mathsmerizing channel, its solution covered my full two pages 😅
What's your expected marks in jee adv brother...
@@bluesky5794 hey hey I am not even able to score good in mains Math's but I like to explore Math's and just want satisfactory score in mains and advanced 😊.
Bhaiya trigonometry me to multiple types of complex formulas hote h jo ki in general bohot se books me bhi nhi diye rhte h. Aur waise formula se Q Advanced me bhi are h. Un formulas ko kaise janenge?
4:03 but z ko aise kaise summation ke bahar le liye?
Because there is no n in z
z = cosx +isinx
No summation index
Thanks Bhaiya firse waisa type ka question lane ke liye jisme sequence and complex ka involve ho.learned a lot..
bhai 25-30 mins lag gaye but mazaa hi aa gaya bhai saab mast question... khud se solve karne ki alag hi feel aate hai
Bhaiya ho gaya !! maja aagaya ekdum !!
Aise sawal indefinite integration mein bhi hote hai cosx aur sinx ko hatake real ya imaginary part of exponential function lelete hai meine waise sawal kaafi kr rkhe shayad isliye 1st baar mein strike ho gaya
Bhai aisa sawal kabhi aya bhi hai jee mai? Jee advance, mains ke pyq aur module bhi laga li mujhe toh aisa question kahi nhi mila
Tu pyq karta hai kya bass? @@themadtitan728
@@themadtitan728 nhi jee mein to nhi aya but hamari coaching mein karaye the ek do aur mereko thoda calculus mein interest hai to mein karta rehta hu idhar udhar se uthake MATHS505 name se ek channel hai uspe aate hai aise badiya integration ke videos but wo jee advance relevent bhi nhi to i wont recommend ki tum abhi uske videos dekho
@@themadtitan728 nhi yaar pehle to kabhi nhi aya but hame coaching mein karaye the aur mereko thoda calculus mein interest hai isliye mein krta rehta hu idhar udhar se utha kar
@@adityaagarwal2504 more power to you. me feeling nostalgia recalling the lockdown quora grind
We have solved such questions in NIT Rourkela in Waves and Optics subject..
It uses the cis method.. quite interesting problem❤❤
Seeing the problem,i thought about of series,trigo,but complex numbers solution is mind-blowing
Hello there,
Thank you for the great video !
I used the Fourier series to solve this
Cos(2nx) can be written cos((2n-1)x +x) and costly((2n+1)x - x) and proceeded forward
In the end I got 1/2 - sinx * Sigma sin((2n-1)x)/(2n-1)
The Sigma part is Fourier series of square wave
Which is π/4 if 0
i used both fourier and z transform to solve this
bhai yeh hota kya hai??? JEE me hai kya ye?
@@nohateplease777 Nhi, 2nd sem 1st yr, ka problem hai
Denominator can be broken into (2r-1)(2r+1) then numerator has cos(2rx) submission which can be easily broken down into terms of 2r-1 and 2r+1
Then we have Sin series
But integration and limit together can make it compex
Hi
Which app you use for making video recording and notetaking in your videos
I was done with the conceptual part in around 15 mins, then for calulation it took me 30 mins.
many calculation errors i made.
solved it many times to approch the correct answer
i have one doubt if the series or you can see say f(x) is even function before summing but it becomes neither odd nor even after summing the series
Ln1+ x wala tagda socha hai complex number tak toh socha bhi tha par this sum was ausum
bhaiya once u said complex involved ,,uske baad log socha easy tha ,,but ya the initial thought was OP
btw happy teachers day bhaiya . i really actually learn a lot from you . your channel is adding value in my jee journey, thank u 😁😁😁😁
😄thanks a tonne dhanika
Bhaiyaa
Trigonometry se nahi hoga kya
Kuch manupulaⁿ vagera karke
???
nope.
It's actually a Fourier series you can look that up if you want to
Didn't thought of complex at all😅
I am rarely able to do these questions but after taking simple hints I understand them very quickly
And please keep sharing all these wonderful questions with us even after this 75 hard thing
Let the sum is f(x)
Differentiate two times, you'll get
f"(x) + f(x) = 1/2
Solution of this differential equation is:
f(x) = 1/2 + Acosx + Bsinx
For 0 < x < π
Now put x= π/2
f(π/2)= 1/2 - π/4
f'(π/2) = 0
You'll get
B = -π/4 ,
A = 0
Solved
My Intro:
JEE Advanced 2021 qualifier.
Edit: Those who are saying ki class 12th mein 2nd degree equation nahin aati.
Ye baat mai bhi jaanta hu, lekin class 11th mein agar SHM padha ho toh yaad Karo har jagah issi tarah ki equation use hoti hai 🗿
Samaj ni aya pr dekh ke accha laga❤😂
Wow! But we have to also remember the solution for this type of diff eqn to use this method.
@@OkayO7-jc6hh
SHM nahin padha kya ab tak 🗿
I'm batman
Bhai aapne toh question ki chaamod di 💀
9:38 .... There should be
(-i cot(x/2) ) .......but also ln(i) is not only iπ/2 there should be addition of 2nπ
niche -isinx/2 hai so 1/i=-i so - sign cancle hojayega..... aur wo ln(i)=i (pi/2 + 2n pi) hi hoga but bhaiya ne simplest term li hai, general term nahi li.
Thanks for helping......but answer ya correct nahi hoga na ....iska multiple answers possible hoga na
@@SumitKumar-lq1lr imo, yes you are right
you are actually completely right , the complex logarithm function is actually a multivalued function which repeats over 2πki and with this logic the answer should've been multivalued which certainly seemed weird and bothers me as well , however weirdly whenever you put a complex no. let's say i , into the Taylor series expansion of a function it tends to give the principal argument as the answer if it's centred somewhere at the unit circle , however if you analytically continue the function outside the radius of convergence and make one rotation about the pole of the function (at x=0) we get the answer as 2π + π/2 and so on , however I'm not really very sure as complex numbers are weird sometimes so you should probably ask your math teacher and tell me what his solution is 👍
Bhaiya complex nhi soch paya but lnx soch liya tha thank you for preparing this question
Will try these question one month before my jee adv 2025
I also have a question of this level (maybe higher) I want to send it to you... Where can i do it?
Send here plz.
@@Mathematician6124 here?where?...I am confused!
@@DKAIN_404, here
can anyone explain how bhaiya took IZI = 1 at @2:16
Pretty easy question honestly but had a lot of fun. thanks for uploading this video
Benstokes ye easy tha kya
@@NAYANKUMAR-mx9mb depending on the person, yeah this was easy. at least easier than many other monsters i personally have dealt with
which jee was this from
Similar qs -
Evaluate cos(π/3) + 1/2 cos(2π/3) + 1/3 cos(3π/3) + .... ∞
ln(sin(x/2))?
@@Udipi-cg8sv 0
acha question tha.... loved it... complex number wala part toh guess kar liya par ln(1+x) ka expansion yaad nahi tha 😬
Bhaiya par ye question konsi book ka material
How can I send some of my doubts and questions
10:36 ln(i) is not simply i.pi/2 but rather 2ni.pi/2... so there will be infinitely many solutions
Badhiya question tha bhaiya 👏🏻
Bhaiyaaa 40min se soch rha tha kuch aisa dimaag me ek baar ke liye strike kiya (complex wali baat)but fir m kuch aur hi concept lagane lag gya
bhaiya pls upload on yt again
BHAI KAISE SOCHE KI KIS QUESTION ME KYA LAGA SAKTE HAI
bhaiyya yaha pe jab dono summation split hote hai tab dono HP ke form me hai- usse seedha hp summation formula laga ke simplify kar sakthe hai na?
Not meant to brag, but done in around 13 minutes being an 11th entering student :
My approach:
1. Convert cos(2nx)= 1/2 (e^i2nx + e^-i2nx).
2. Separate the series into two series (one of positive powers and other of negative powers) .
3.Solve the first series by substituting a=e^ix.
After few manipulations in first series, we will get :
[(a²-1)/4a].[ln{(1+a)/(1-a) }]+0.5
4. Similarly in the second series, repeat the process by substituting
b=e^-x.
5.Add both the parts and simplify to obtain the final answer.
Bruhh I'm also entering 11th but not able to get a shit out of it.
@@NeetusGallery5 No need to worry... 😊
I am also entering 11th pcm and dont know shit about it 😭😶
foundation wala hai tu.... btw vwry good being able to solve this level of questions at this age
@@pratikIIT24 Self study(Class 11+12) + Coaching(Class 10 at IIT foundation level)
How to send question?
2 min mai hogya
easy tha
bas cool ban ne ke liye bola 😅
btw love this que
This can also be done by fourier analysis.
Dimag garm karrha ye question, kabhi solve nhi krunga aise question mai realistically.
Bhaiya Q kaise share krenge apko ?
Sir sach mein bahut acha question and solution tha
Is there any alternative solution to this ?
I tried to differentiate and integrate the general term since kuch dino pehela aap na ekk series ko aisa solve kiya tha but kuch nehi
Hua 😂. But really a monster question ❤.
Brother, this series is not equal to 1/2-(pi/4)sin(x) always.
It is a periodic series with period pi.
We can check this by x = 3pi/2, S = -(1/1.3) + (1/3.5) - (1/5.7) = (1/2)(-1+1/3+1/3-1/5-1/5+1/7+....) = (1/2)-(1-1/3+1/5-1/7+...) = 1/2-arctan(1) = (1/2)-(pi/4)
from formula, x = 3pi/2 gives S = (1/2)+(pi/4), so the series is equal only in {(0,pi) +2k(pi)}
S = 1/2+(pi/4)*(sinx) in (pi,2pi) + 2kpi;
S = (1/2 )- (pi/4)(|sinx|)
Not sure if it was MCQ with multiple options or only one option is given in JEE Adv
However, the other result can be achieved by seeing that-
log(icot(pi/4)) = log(i) = i*pi/2 and
log(i*cot(3*pi/4) = log(-i) = -i*pi/2;
log(z) = log|z| + i arg(z) => arg(i*cot(x/2)) = pi/2*(|cotx/2|/cotx/2) = pi/2*(|sinx|/sinx)
so Im(log(icotx/2) ) = pi/2*(|sinx|/sinx)
S = (1/2)-(pi/4)*(|sinx|)
This problem also came in csir net with little change, asking about it's uniform convergence .
Bhai mai toh cos ke expansion se soch ra tha
Ur Ans is wrong my dude😊
Sir isko appne jyada tough bana Diya isko aasani se solve kar sakte hai
i was not able to think in this q mtlb the first three introductory steps were taken by me lkin ye sochna ki ye q complex ka h , it was mind boggling .
too easy agar sahi concept pata hein toh.
Is problem ko infinte series se 2min pe ho jayega by using partial sum of sequence.
😂
Bhaiya try karta hu to nhi banta hai pr solution turant samjh aa jati hai😢
does ln(i) = i*pie/2 is correct because it can also be any integer solution of it
Like i5π/2
bhai complex toh soch liya tha but fir expansions dhyan mai nhi aayi
itne questions karre hai bhout try ke baad trigo kaa savaal naa hone par yhi hoo sakta hai ki voh complex se hoo rha ho
Sir
Ish ka ggrafically mean showes
How to send you ques?
Bas ln wali cheez ka idea nahi laga tha baki vo hint leke hogya tha
Hey bro!
Our teacher once said that expansions are valid only for real numbers. Is that true?
Bhaiya isi time pe vdo dalna hmeshs... raat me jb pdhte time neend aati h to vdo kholke ques pr matha lagane lgta hu and all set neend gayab
Somehow this video appeared .I subscribed to this channel right after watching this video
Lol the moment i saw this i thought about complex numbers
dekh bro pehle nth term likh
then dono ko alag alag likh
phir just like infinite gp terms kat jaenge
bas 1st term bachega jo n=1 dalke aa jaega
aur last term k liye limit solve kr lena
1 page solution
i figured out taking the real part of the complex number myself but uske baad clueless betha raha and nahi huan
Legends leave this question unattempted 😂.
Jee level nahi hai Olympiad level hai 😢😢😢
Ha actually
Giving JEE Adv. At exam hall is same as solving olympiad questions at home.
Giving JEE Adv. At exam hall is same as solving olympiad questions at home.
@@notsodope834 nah not true itna bhi jyada pressure nahi hota relevance ek cheej hoti hai sir
Not actually. But ya , the easier ones of the first stage of Olympiad. IOQM 2 markers
Mind bogling question
Damn. Really creative question. I didn't know the expansion of logarithmic series so I couldn't solve beyond that part but the question was great.
Dropper this time took 3 mins to understand and voila🎉
6:37 at this point the expression of other term is written wrong
It should be 1/2ln(1+z/1-z) -z. U didn't consider the half wala factor
Edited: I didn't see further(sorry)
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Bhaiya baaki mujhko pta nhi mainey yhi ekk kiya ki x ko zero dalkar aur 1/2 common nikallkar infinite V/N bana aur usmey last ke terms bahut badey honey ke karad kaat diye phir kya 1/2 answer baki , phir bhi ekk pura pie by 4 term rh gya , dusreyy x ke liye alg alg answer hoga ... that's my fault ❤❤❤
Hii likhdo mereko bhi kabhi kabhi akele akele editor rakhhey video bana rhi bss 😅😅
Bhaiya kash aap mera soln dekh pate, mujhe 2cos2x+1 mila hain finally , kyo wrong hain pta hi nhi chl rha maine hi complex numbers use kiye the
maine cos 2rx ko exponential main expand kiya fir , telescopic series banayi jis me chaar alag alag summation bn rhi hain is form main
sum ( 1 to infinity )= t^r/r
is sum -ln(1-t) nikalti hain pr maine substitute kiya toh answer galat aa raha hain.
Though of complex and even did it till taking real part but never thought of log series . 😢
Mainee eshaa ekk question already kahii dekha tha but usmee sirf ln(1+x) ka paternn observerd krnaa thaa.... I think that's why i was able to solve it ✌🏼
i didn't knew how to solve it so just calculated what will be the value of this equation when x will be 2mpie and got 1/2
[when will i be able to solve advance level questions😔😔😔]
JEE advanced have fairly easy questions than this.. work your way up to there by practising good questions
@@manishsingaria3895 thanks for your kind words bro 😄
DW, this is ISI UGB/olympiad level question, JEE A has easier questions
Bhaiya, I emailed you a question please check it once
The Question, why on earth the complex numbers will strike one's mind....why...is it practice or is it more neural connections...
ek achha que tha sequences ka
1+2x+6x^2+20x^3+66x^4+212x^5.......
Sawal to aasan hi tha par hua nahi. same concept questions cengage me bhi phada tha.
ln(1+x) kaa expansion nhi strike Kiya 😢 chalo nvm
me ln(1+x) aur e^x ke expansion ka phle soch betha to usi par dhyaan gya, complex par socha nhi
Maja aa gaya thank u bhaiya
Why don't u just substitute x as a known value of cos theta and substitute the same
Maybe it's a dumb approach but that's what I thought
10 min karne ke baad nhi hua tab jaake pata chala that it's 1x3,1x5 etc and not 1.3,1.5. Hp bana ke try kar raha tha
Meri Ques dekh ke hee phat gyi