This kind of video, where you show your thought process and consider which route to go and even hit a dead end is very very nice as it teaches how to tackle the problem instead of simply presenting a deus ex machina solution.
I like how you say "this guy" making numbers look like living things that make your life easier and many times Hard. This is a great integral you solved I loved it.
The integral - after using the fact that the integrand is even, using the triple angle formula can be written as 1/8 Im{ integral (e^i3x - 3e^ix)/x^3 dx } where the integral runs from -infinty to infinity. We can analytically continue into the complex plane, separate the two integrals, run a contour along an infinite semi-circle in upper half plane, and a small-semi circle in upper half plate, circulating the singlarities at z = 0 of the function. Using Jordan's lemma to determine that the integral around the large semi-circle is 0, and using Cauchy residue (no poles within contour) means that the integral is equivalent to integrating in the complex plane the above integral around an infinitely small semi-circle, centred at z = 0. The result is 1/8 Im(0.5*2*i*pi*residue at z = 0). The residue, of the above integrand at z = 0 is -3 (which can be quickly checked by expanding the exponential numerator to quadratic term. Plugging this in gives the answer...no Feynman...
Yes I did thought of the same way whenever the limits are till infinity with some power of x in denominator I always first try to use the residue theorem
If you Laplace transform this integral you'll see why the value for the 3rd power is different from the first two powers. Essentially you're doing a convolution, which amounts to taking a moving average over a sliding window of a rectangular function. For the first 2 powers, the window isn't wide enough to affect the value of the average over the moving window, but for the 3rd power, eventually we are averaging zero contributions from outside the rectangle which brings the moving average down. 3blue1brown did an AWESOME video into this: m.ruclips.net/video/851U557j6HE/видео.html
You have 60hz hum coming from your microphone JSYK. Try to turn your microphone up more without clipping over 0dbFS, or look and see if any wires are crossing over a power wire from your interface.
Because sin^3(x)/x^3 is an even func you can write the integral to be 1/2 of the same integral over the real line. Then after you take the d/dt and use the sin^2(x)=1-cos^2(x) you get an odd function over a simatric interval, so it's 0. So you don't need to take the second d/dt you did
This is excellent and the video led me to Math 505's generalized version of sin^n(x)/x^n which looks like a great beast for you to work your magic on and possibly make understandable at around calc 2 level :). I struggled following the differentiation portion
@omograbi gives the same answer, but he could have continued with sin(3tx)/x, making the substitution a = 3tx, arriving at the integral of from 0 to infinity of (sina)/a which gives pi/2.
Both integrals will have the same exact value. Performing the substitution u=t*x for the first one will imply that 1/x=t/u and dx=(1/t)du, which makes the t's cancel out. Likewise, for the second one, if you let w=3*t*x, that will imply that 1/x=3*t/w and dx=(1/(3*t))dw, which makes the 3*t's cancel out.
Hello ! I hope you see my comment I saw this nice question so that I recommend it The question is : solve the system of equations a = exp (a) . cos (b) b = exp (a) . sin (b) It can be nicely solved by using Lambert W function after letting z = a + ib Hope you the best ... your loyal fan from Syria
Hey sir, Feynman's technique is mad cool but... Where should I set the parameter? Is there any "rule" to follow? I mean, you are supposed to put the parameter on a place in which after deriving, the integral is easier to solve, but it would be marvelous if you have a structured guide that tells you where to put it depending on the situation. It would be great a video like... "When and where to use Feynman's technique" Thanks sir.
10:16 I didn't get this step. Firstly how did he replaced the sin(3tx) with just sin(tx) and in the next step after substituting tx as u, he should get a 't' after integration which he missed as well. It should have been -3πt/8 + 9πt/8 considering his previous step of omitting 3 from the sin is right. Can anyone help me with this.
I didn't get that either, he switched from sin(3tx) to sin(tx), must be a mistake over there and maybe despite of that the result is the same by coincidence.
The first issue is definitely a mistake. With regard to the u=tx substitution this gives dx=du/t. When the substitution is performed it gives integral of sinu/(tx) du and as u=tx this gives integral of sinu/u du with the limits of integration being u=t×0=0 and u=t×infinity=infinity. This then is just the straightforward known integral with u instead of x with the same limits of integration giving pi/2.
There was no mistake, The integral from the type: integral from zero to infinity of sin(Ax)/x dx always equal to π/2, because: (Integral from zero to infinity of sin(tx)/x dx) let tx = u x = u/t dx = du/t and then: the integral of sin(u)/(u/t) × du/t equal to: t×sin(u)/u × du/t t and t cancel out (the integral from zero to infinity of sin(u)/u du)= π/2 and because of this: the integral from zero to infinity of sin(3tx)/x = the integral from zero to infinity of sin(tx)/x = π/2 Hope I helped you :)
I know it’s not the same integral, but I recommend checking out the use of complex contour integration to come up with a general formula for the integral of sin (x^n)/x^n on the same interval. ruclips.net/video/ovj71qp7C4k/видео.html
After you simplified I''(t), where you split the integral into two integrals, how did you get rid of the 3tx inside of the sin function? In the next step, there is just a tx and no 3tx, how can you do this? Can someone explain, please?
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!) That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2. If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2. Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer. I hope this helped!
10:22 i believe you miss the 3tx in the argument of the right most sin integrand. however a similar substitution of w = 3tx for the right integrand at this step gives the exact same conclusions.
I don't think we need to do the u-substitution separately for the sin(3tx) to see that it is also π/2. We can reason directly from the sin(tx)/x integral. Since it's result is π/2 for any t value that means even 3t is a value that works. So t=1,2,'3',4,5,'6',7... work which includes the multiples of 3 ie. t=3,6,9...
I lost something at 10:16. You got 9/4 int(0,inf) sin(tx)/x. But the previous step it was 3/4 sin(3tx)3x/x^2 ?? Shouldn't the argument to the sin function still be 3tx? How did that 3 disappear?
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!) That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2. If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2. Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer. I hope this helped!
@@boomgmr6403 he should have written it as sin3tx and shown that making the substitution u = 3tx would take the integral from 0 to infinity of (sinu)/u which gives pi/2.
I have found a pretty nice question that I will suggest Solve the system a = exp (a) . cos (b) b= exp (a) . sin (b) Can be solved easily using Lambert "W" function by computing ( a+ib ) And thanks
I think using the triple angle formula sin(3x)=3sin(x)-4(sin(x))^3 is easier. Unlike what other have said, there is no need to use contour integration.
Sorry for the naive question, but when solving for c, what if you let t equal any multiple of pi? Wouldn't that change what c is? Why is it possible to choose the "easiest" value?
Integrate by parts twice with D I sin^3(x) 1/x^3 then substitute u=3x finally i got 3/4Int(sin(x)/x,x=0..infinity) then i calculated Laplace transform L(sin(t)/t) and plugged in s = 0
When you do the u substitution when computing I''(t), you get constant 3pi/4, which is fine for all t>0, but how do you justify that at t=0 where the integral appears to give zero when you go to fix your constants of integration?
5:20 this is the main weakness in your approach ive noticed When you arrive at two viable techniques to modify the integration you force yourself to choose. Just do both!
Assume that identity is true. Then take log base b on both sides. Each side simplifies to 1. Logarithms are increasing and 1=1 so the identity must be true.
I am confused by the evaluation of I'(0) and I(0) at 12:14 and 13:26 respectively. Shouldn't (sin(x)/x)^n be equal to 1 at x=0 for positive integers n? Is there something I am missing? In order to have a nice value for I'(t), we need cos(tx) to be 0. In other words, we need to evaluate at tx= pi(k-1)/2, which changes the expression a lot and affects the next integration that finds I(t).
If t = 0 then the sin(tx) term is 0 and the whole expression goes to 0. Remember that I is a function over t not over x, so x = 0 doesn't need to be considered anywhere.
This is from the 100 integrals part 2. See the full video here ruclips.net/video/jQz1gQ24OHc/видео.html
I was wondering why you looked so tired, now I understand 😂
YES MU PRIME THINKS OUTSIDE THE BOX ALSO DO THE MERITAVIEN AND MIND YOUR DECISION TOO THEY ARE OUTSTANDINGLY DIFFERENT AND
DR PEYAM TOO
I Watch your video from India
We need a tutorial about where to use each pen
He has a RUclips short
Characteristic part
agree
hahaha I have a tutorial!
youtube.com/@mehmetturk6583
This kind of video, where you show your thought process and consider which route to go and even hit a dead end is very very nice as it teaches how to tackle the problem instead of simply presenting a deus ex machina solution.
@julw9138 who asked?
@@maalikserebryakovhuh?
Expressing sin³(𝑥) in terms of sin(3𝑥) and sin(𝑥) using the triple angle formula in the first place seems helps.
I like how you say "this guy" making numbers look like living things that make your life easier and many times Hard. This is a great integral you solved I loved it.
Looking at the clock and hearing it being synchronized with my own wall clock makes me feel like I am in the class :D Great integral!
The integral - after using the fact that the integrand is even, using the triple angle formula can be written as 1/8 Im{ integral (e^i3x - 3e^ix)/x^3 dx } where the integral runs from -infinty to infinity. We can analytically continue into the complex plane, separate the two integrals, run a contour along an infinite semi-circle in upper half plane, and a small-semi circle in upper half plate, circulating the singlarities at z = 0 of the function. Using Jordan's lemma to determine that the integral around the large semi-circle is 0, and using Cauchy residue (no poles within contour) means that the integral is equivalent to integrating in the complex plane the above integral around an infinitely small semi-circle, centred at z = 0. The result is 1/8 Im(0.5*2*i*pi*residue at z = 0). The residue, of the above integrand at z = 0 is -3 (which can be quickly checked by expanding the exponential numerator to quadratic term. Plugging this in gives the answer...no Feynman...
Indeed.
Yes I did thought of the same way whenever the limits are till infinity with some power of x in denominator I always first try to use the residue theorem
Except u can do this without any of that, u overcomplicate the simple
Well, yes, _obviously_ any Calc 1 student already knows how to do that, so why not challenge yourself to trying a different method?
If you Laplace transform this integral you'll see why the value for the 3rd power is different from the first two powers. Essentially you're doing a convolution, which amounts to taking a moving average over a sliding window of a rectangular function. For the first 2 powers, the window isn't wide enough to affect the value of the average over the moving window, but for the 3rd power, eventually we are averaging zero contributions from outside the rectangle which brings the moving average down.
3blue1brown did an AWESOME video into this: m.ruclips.net/video/851U557j6HE/видео.html
Reminds me of the Borwein integrals a bit
14:30 man i know that happiness and you have to experience it atleast once in a lifetime!
You have 60hz hum coming from your microphone JSYK. Try to turn your microphone up more without clipping over 0dbFS, or look and see if any wires are crossing over a power wire from your interface.
Because sin^3(x)/x^3 is an even func you can write the integral to be 1/2 of the same integral over the real line.
Then after you take the d/dt and use the sin^2(x)=1-cos^2(x) you get an odd function over a simatric interval, so it's 0. So you don't need to take the second d/dt you did
This is excellent and the video led me to Math 505's generalized version of sin^n(x)/x^n which looks like a great beast for you to work your magic on and possibly make understandable at around calc 2 level :). I struggled following the differentiation portion
10:14 shouldn't it be sin(3tx)/x?
Or it's anyway the same answer?
I would say it's a different answer.
@omograbi gives the same answer, but he could have continued with sin(3tx)/x, making the substitution a = 3tx, arriving at the integral of from 0 to infinity of (sina)/a which gives pi/2.
its the same answer, integral of sin(tx) / x from 0 to ∞ is π/2, so putting 3t instead of t will be the same
Yes, it‘s the same result, but it is confusing
Both integrals will have the same exact value. Performing the substitution u=t*x for the first one will imply that 1/x=t/u and dx=(1/t)du, which makes the t's cancel out. Likewise, for the second one, if you let w=3*t*x, that will imply that 1/x=3*t/w and dx=(1/(3*t))dw, which makes the 3*t's cancel out.
Hello ! I hope you see my comment
I saw this nice question so that I recommend it
The question is : solve the system of equations
a = exp (a) . cos (b)
b = exp (a) . sin (b)
It can be nicely solved by using Lambert W function after letting z = a + ib
Hope you the best ... your loyal fan from Syria
Email him!
The alternative way is Fourier transform, split it into (sinx/x)(sin²x/x²) then convolution time!
I used that technique for the integral of sin²x/(x²(1+x²)), seen in ruclips.net/video/S52DapoH17M/видео.html
Hey sir, Feynman's technique is mad cool but... Where should I set the parameter? Is there any "rule" to follow?
I mean, you are supposed to put the parameter on a place in which after deriving, the integral is easier to solve, but it would be marvelous if you have a structured guide that tells you where to put it depending on the situation. It would be great a video like... "When and where to use Feynman's technique"
Thanks sir.
10:16 I didn't get this step. Firstly how did he replaced the sin(3tx) with just sin(tx) and in the next step after substituting tx as u, he should get a 't' after integration which he missed as well. It should have been -3πt/8 + 9πt/8 considering his previous step of omitting 3 from the sin is right. Can anyone help me with this.
I didn't get that either, he switched from sin(3tx) to sin(tx), must be a mistake over there and maybe despite of that the result is the same by coincidence.
The first issue is definitely a mistake. With regard to the u=tx substitution this gives dx=du/t. When the substitution is performed it gives integral of sinu/(tx) du and as u=tx this gives integral of sinu/u du with the limits of integration being u=t×0=0 and u=t×infinity=infinity. This then is just the straightforward known integral with u instead of x with the same limits of integration giving pi/2.
There was no mistake,
The integral from the type: integral from zero to infinity of sin(Ax)/x dx always equal to π/2,
because:
(Integral from zero to infinity of
sin(tx)/x dx)
let tx = u
x = u/t
dx = du/t
and then:
the integral of sin(u)/(u/t) × du/t
equal to:
t×sin(u)/u × du/t
t and t cancel out
(the integral from zero to infinity of sin(u)/u du)= π/2
and because of this:
the integral from zero to infinity of sin(3tx)/x = the integral from zero to infinity of sin(tx)/x = π/2
Hope I helped you :)
@@amirbasson532 it was a mistake not making any reference to this fact as it was an assumption which has evidently caused confusion to several people.
@@amirbasson532 ohhhh that explains it, thanks!
I know it’s not the same integral, but I recommend checking out the use of complex contour integration to come up with a general formula for the integral of sin (x^n)/x^n on the same interval.
ruclips.net/video/ovj71qp7C4k/видео.html
Everytime I watch one of your videos, it feels like an emotional rollercoaster.
Another integral for the collection!
I just bought calculus clothing off your store. My weird maths science wardrobe is increasing. I am happier than I was 30 mins ago now.
After you simplified I''(t), where you split the integral into two integrals, how did you get rid of the 3tx inside of the sin function? In the next step, there is just a tx and no 3tx, how can you do this? Can someone explain, please?
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!)
That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2.
If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2.
Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer.
I hope this helped!
10:22 i believe you miss the 3tx in the argument of the right most sin integrand. however a similar substitution of w = 3tx for the right integrand at this step gives the exact same conclusions.
Maybe I missed it but at 10:16 how did the second term go from sin(3tx) to just sin(tx)?
Same doubt
it was a mistake but the integral is still pi/2
yeah It should be 3tx but as he already mentioned, the result is π/2 regardless of the input.
I don't think we need to do the u-substitution separately for the sin(3tx) to see that it is also π/2. We can reason directly from the sin(tx)/x integral. Since it's result is π/2 for any t value that means even 3t is a value that works. So t=1,2,'3',4,5,'6',7... work which includes the multiples of 3 ie. t=3,6,9...
@@noopcode but yeah I think he just missed writing the 3
I lost something at 10:16. You got 9/4 int(0,inf) sin(tx)/x. But the previous step it was 3/4 sin(3tx)3x/x^2 ?? Shouldn't the argument to the sin function still be 3tx? How did that 3 disappear?
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!)
That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2.
If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2.
Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer.
I hope this helped!
We can get the same result by a sampled function and sampled squared in frequency domain then taking area at freq = 0
I understood absolutely nothing of all this but still watched the whole video...
use infinity & beyond. works wonders
That was some Feynmania!
I like to think that he's just in somebody else's office at 12:30 in the morning doing integrals
At 11:54 you dont write sin3tx again, is that a mistake?
dont you then get integral sin3tx/x dx? Does that change something?
At 10:50 he says you can have any constant multiple and it will always be pi/2
Yes, but also no. The result is always pi/2 so it's not significant.
@@Gamedolf I see
@@boomgmr6403 he should have written it as sin3tx and shown that making the substitution u = 3tx would take the integral from 0 to infinity of (sinu)/u which gives pi/2.
Actually what is Feynman's trick ? I am a high school student and this isn't in my syllabus but I am eager to know
you can use de moivre angular expansion to write trig^n as a sum of linear trig
I have found a pretty nice question that I will suggest
Solve the system
a = exp (a) . cos (b)
b= exp (a) . sin (b)
Can be solved easily using Lambert "W" function by computing ( a+ib )
And thanks
To take out √-1 put e^iπ in the place of -1
Please do it !
I think using the triple angle formula sin(3x)=3sin(x)-4(sin(x))^3 is easier. Unlike what other have said, there is no need to use contour integration.
10:14
In the second integral it should have been Sin(3tx) in numerator and I think it would be difficult to get the final answer in this context.
Fantastic. Need to rehearse my trig identities!
Wouldn’t complex analysis work here?
Thank You for this video.
How did sin(3tx) became sin(tx)?
Are they equal?
I've the same question also ... that it makes my brain explode
I think it was a mistake that didn't matter to the answer, because the integral of sin(ax)/x dx is the same for any nonzero a
Hi bprp, thank you, the complicated calculation can be followed. Yes, Mr. Feynman could not only joke... I will study a lot...
Damn, that's hardcore - nice going!
9:32 the 2 appears like magic
Sorry for the naive question, but when solving for c, what if you let t equal any multiple of pi? Wouldn't that change what c is? Why is it possible to choose the "easiest" value?
At 10:20, in the sex nd integral, it should be sin(3t).
A fastest way is applying the formula sin^3(x)=1/4(3sinx-sin3x)
And using integrate by parts
At 6:20 why do we have cos(tx)-cos(2tx)?Where does the minus come from? And why do we have a plus afterwards?
at 10.19 sin(3tx) becomes sin(tx) after the separation of the 2 integrals. Is this correct?
bro u are insane pls make more viedos like this, i realy like it
Integrate by parts twice with
D I
sin^3(x) 1/x^3
then substitute u=3x
finally i got
3/4Int(sin(x)/x,x=0..infinity)
then i calculated Laplace transform
L(sin(t)/t) and plugged in s = 0
There's a mistake in the 2nd derivative of I(t) in the 2nd line you got sin(3tx) in the 2nd term. All respect to u it's a hard integral .
I didn't understand what u did but i would have used product rule of integration using ILATE😅
Greetings from the BIG SKY. Nothing like a bit of calc to end the day.
sin(3tx) right high corner , where did he go?
Bro u should also make a video doing this from contour integration using residue theorem and Jordan's lemma 😁😁
I solve these question in my paper . But your techniques are quite impressive
When you do the u substitution when computing I''(t), you get constant 3pi/4, which is fine for all t>0, but how do you justify that at t=0 where the integral appears to give zero when you go to fix your constants of integration?
5:20 this is the main weakness in your approach ive noticed
When you arrive at two viable techniques to modify the integration you force yourself to choose. Just do both!
I was lost for most of the video !! I have a long way to go.
Would this work? I(t)=(integral)sin^tx/x^t
10:13 If I'm not mistaken, it's supposed to be sin(3tx) but you wrote sin(tx)....please clarify?
what is the practical use of this equation?
10:13 wouldn't it be sin(3tx) instead of sin(tx) ??
Good catch but
ruclips.net/video/HmuYjxcRIXo/видео.html
So I think it doesn't matter.
Sin3xt/x
put 3xt=u
x=u/3t
3tdx=du
dx=du/3t
integral sin3xt/xdx=(sinu×du/3t)/u/3t
=integral sinu/udu =pie/2
we can even use sin3theta here
right?
On the right half of the board, the second step where you took negative common, shouldn't it be -(9/4) and so on?
No he split the expression at the +, so the left half is for -sin(tx)•x and the right half is for sin(3tx)•3x
After the 100-x series I am surprised you want to even SEE another integral ever again.
At 10:14 : error sin(3tx) not sin(tx)
10:14 why the sin (3tx) suddenly becomes sin(tx)
Good catch but
ruclips.net/video/HmuYjxcRIXo/видео.html
So I think it doesn't matter.
It is so good. I am really happy to see that solution. Thank
Sir, please explain why there exists two types of vector products...
There's three, actually
Wait what? Feynman invented math? I thought Feynman was a physicist . . .
I adore your videos i watched the half of 100 integrals)
Can you solve the integral of :
ln(sinx+cosx)/(cosx-sinx) dx
How can we write d/dx( sin^3x) = 3sin^2x cosx ?????? 😏
請問曹老師，封面的獎牌是？（數奧？還是⋯）
馬拉松獎牌 因為這是我一次做一百題數學的影片片段
Can you do the taylor series of sin and work from there?
PLEASE MAKE A VIDEO ON FORBENIUS METHOD OF SPECIAL FUNCTIONS
make a video on solving 100 Putnam Calc 2 Problems
Me trying to figure out is it 12 Am or Pm in the clock
Love from India
我想祝福你新年快樂happy Lunar New Year
Why have you stopped making frequent videos?
Very very easy question
what if d/dx((sinh(x*ln(x)))^(2x*cos(x)))
Is this integral generalizable for n?
i have never seen you that tired🥺
Muchas gracias
Borwein integrals ?
9:45 why did you differentiate the numerator here but didn't you differentiate the denominator of x^2?
Because x^2 is a constant in t world.
Bro what is your educational qualifications?
Hey blackpenredpen, I can't figure it out, but why is x^(1/log_b(x)) equal to b (the base of the logarithm)?
Assume that identity is true. Then take log base b on both sides. Each side simplifies to 1. Logarithms are increasing and 1=1 so the identity must be true.
@@spaghetti1383 That was a very clear explanation, thank you :)
Have you done PhD in maths?
How to become mathematician?
The second term in I'''(t) expression on RHS is not correct
10:12 how did he get rid of the 3 on sine's angle?
Good catch but
ruclips.net/video/HmuYjxcRIXo/видео.html
So I think it doesn't matter.
14:42 sin(3tX) ---》 sin(tX) ?!!! I have a problem here 😕
What about Nth power
I am confused by the evaluation of I'(0) and I(0) at 12:14 and 13:26 respectively. Shouldn't (sin(x)/x)^n be equal to 1 at x=0 for positive integers n? Is there something I am missing? In order to have a nice value for I'(t), we need cos(tx) to be 0. In other words, we need to evaluate at tx= pi(k-1)/2, which changes the expression a lot and affects the next integration that finds I(t).
If t = 0 then the sin(tx) term is 0 and the whole expression goes to 0.
Remember that I is a function over t not over x, so x = 0 doesn't need to be considered anywhere.
@@amtep Right, of course! Thank you!