The most interesting differential equation you have seen.
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- Опубликовано: 30 июл 2024
- Super FUNctional Differential Equation is here to save the day from the dastardly chalk. Chalkboard didn't know it at first but today would be like no other. Out of nowhere, chalk attacks chalkboard leaving a ton of marks. Like, it was a lot trust me. anyway, Super FUNctional Differential Equation saves the day with some sick moves. Only way to find out the rest is to watch this until the last name of the credits is but a shadow. Also, like the video, subscribe and join the Patreon.
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It never came to my mind that a div.equation could have a polynomial AND an exponential AND a trigonomic solution. Cool video
That's when we're lucky.... There are always those annoying non elementary functions that surface from the most innocent equations
@@yahhav345
Also note that the equation is nonlinear
Finite vs infinite polynomials
I mean, isn't that already possible in normal linear differential equations? All you need is Eigenvalues of 0,1,i etc. and maybe repeat them a few times if you want. The simplest example I can think of is y^''''-y^'''-y^''+y^'=0, which unless I've screwed up should have basis functions of X, Sin(x), Cos(x), and Exp(X)
I mean, trig functions are just exponential functions in a complex sort-of way
11:56 probably meant to divide by e^{2kx} so that the exponential factor is out of equation.
Yes yes correct 🤗
Production value improving, little basic mistakes on the blackboard remain.
I approve!!! 😂
Should be dividing by e^(2kx) in the second case
He did; he simply omitted the 2 when he wrote the reciprocal.
@@byronwatkins2565 More like forgot, but it's a minor mistake, true.
@@elsurexiste The important part is that he recovered from it.
@@byronwatkins2565 I'll leave dividing by the second e^kx as a hw exercise
Thank you! I knew I was missing something in his solution. I'm glad it was a mistake on his part lol Even if it does wash out in the end.
The trigonometric case is actually included in the exponential case if k is allowed to be imaginary, such as k = b*i. You would take the real and imaginary parts to get both real trigonometric solutions.
This one is a bit weird because the DE isn't linear, so you can't just separate a solution into a sum of real and imaginary parts and have each one work. I think you need to solve the two exponentials case such that the cross term on the right is 0, and find where that cancels out the imaginary parts.
@@iabervon It does work out because it's valid for all k, even if imaginary. If you substitute the exponential form into the differential equation the real part and imaginary parts of the equation must also be equal. You can then take the real and imaginary parts as separate equations. The LHS of each is f(2x), which is trivial to solve for f(x). The RHS is irrelevant as the work was already done when finding the solution to the exponential form. I also did the work from the entirely real approach afterwards and the results are the same.
@@iabervon Superposition does not work, you're right, but any solution
*1/k^3 * e^{k*x}, k ∈ ℂ*
can still be written in terms of *"sin(..) + i cos(..)",* using "Euler's Formula". You just cannot split real- and imaginary part to generate new solutions, since the differential equation is non-linear.
God damn i just post my comment observing the same thing and then right afterwards I see this lol.
@@carstenmeyer7786 Yeah, this is right. You can set e.g. k=bi to get f(x) = i/b^3 * exp(ibx) as a particular solution, which you can then express as f(x) = -sin(bx)/b^3 + i cos(bx)/b^3 by way of Euler's formula... but this is only a "trigonometric solution" in the sense that "it has sine and cosine in it"; really it's still the same underlying exponential solution. It doesn't work to just take the real & imaginary parts as separate solutions. You would have f(2x) = -sin(2bx)/b^3, f'(x) = -cos(bx)/b^2, and f''(x) = sin(bx)/b; multiply the latter two & get f'(x)f''(x) = -sin(bx)cos(bx)/b^3 = -sin(2bx)/(2b^3) ≠ -sin(2bx)/b^3.
Trigonometric solution: let f(x) = a sin(bx + c). (If the solution ends up being a cosine, then we will just have c = -π/2 so that's covered. Also that's a-times-sin, not asin i.e. arcsin.)
Then f(2x) = a sin(2bx + c), f'(x) = ab cos(bx + c), and f''(x) = -ab² sin(bx + c).
Multiplying the two derivatives, f'(x)f''(x) = -a²b³ sin(bx + c) cos(bx + c).
Apply the double-angle formula for sine, i.e. sin(2x) = 2sin(x)cos(x), to the product to convert it to f'(x)f''(x) = -a² b³ sin(2bx + 2c) / 2.
So now we must choose the parameters a, b, c so that a sin(2bx + c) = -a² b³ sin(2bx + 2c) / 2.
Inside the argument of sin(), we have c = 2c -> c = 0.
From the amplitude, we have a = -a² b³ / 2 which can be rearranged to a = -2/b³. This requires dividing by a and b, but we can safely assume a ≠ 0 and b ≠ 0, since either a = 0 OR b = 0 just leads to f(x) = 0 anyway.
Thus, any function of the form: *f(x) = -2sin(kx)/k³* will also satisfy the original equation!
What if the trigonometric solution is tan?
This makes me feel stupid as hell, I only spotted the case where k=1 😅
In the polynomial case, a=0 is valid, and results in f(x)=0, which is a correct solution. It appears to violate the n=3 finding, but this is based on deg(f'(x)) = deg(f(x))-1 which is not true if deg(f(x)) is zero. The correct value is deg(f'(x))=max(0, deg(f(x))-1) and if you plug this into the degree analysis of the original equation you also get the solution n=0.
Nice. And the power series corresponds with n=oo, which is also a solution, kind of.
Hmmm.... do u know degree of non-zero constant polynomial is 0 while degree of zero polynomial is not defined?
@@khoozu7802 The degree of zero polynomial is usually defined as negative infinity, to satisfy the product rule (deg(p*q)=deg p+deg q)
@@coc235that's right, not a real number
But is f(x)=0 a polynomial or an exponential result?
I'd love to see more content on more difficult subjects such as your research! Great vid!
This is my first time to see a differential equation solution without integration. Hats off!
you must be new around here
Some years ago I did A Level maths so I love watching your videos since watching your videos my math skills have dramatically improved and I can understand maths at a much higher level than before I love your videos
Michael Penn is fantastic. Love watching
I like your transitions. #FeedTheAlgorithm
Perfect and clear explanation,also an interesting d.e.;really an excellent Teacher,Thank you.
Consider f(x)=A·cos(αx)+B·sin(βx)+C with A,B,C,α,β∈ℝ for any x∈ℝ.
Then f(2x)=A·cos(2αx)+B·sin(2βx)+C
=A·(cos²(αx)-sin²(αx))+2B·cos(βx)sin(βx)+C
and f'(x)f''(x)=(-αA·sin(αx)+βB·cos(βx))(-α²A·cos(αx)-β²B·sin(βx))
=(-αA·sin(αx)+βB·cos(βx))(-α²A·cos(αx)-β²B·sin(βx))
=α³A²·sin(αx)cos(αx)-β³B²·cos(βx)sin(βx)+αAβ²B·sin(αx)sin(βx)-α²AβB·cos(αx)cos(βx) for any x∈ℝ.
If f(2x)=f'(x)f''(x) and A=C=0 and B≠0, then
2B·cos(βx)sin(βx)=-β³B²·cos(βx)sin(βx)
⇒ 2B=-β³B² ⇒ B=-2/β³
⇒ *f(x)=-2/β³·sin(βx) with β∈ℝ\{0} for any x∈ℝ.*
If f(2x)=f'(x)f''(x) and B=0 and C=A≠0, then
A·(cos²(αx)-sin²(αx))+A=α³A²·sin(αx)cos(αx)
⇒ 2=α³A·tan(αx)
⇒ *f(x)=A·(cos(αx)+1) with A,α∈ℝ\{0} for x=arctan(2/(α³A))/α.*
the first one is very similar to the one we got with exponents, but the next one is not valid because it's only for one suitable x. Or otherwise it's not trigonometric because frequency is dependent on x. And even if it was you would need to take that into account when differentiating
I don't know how u get 2=a^3Atan(ax) but I think u should separate it to two conditions, first is
a=b and second is a=/=b, then comparing the coefficient of each terms.
@@khoozu7802 _"I don't know how u get 2=a^3Atan(ax)..."_
A·(cos²(αx)-sin²(αx))+A=α³A²·sin(αx)cos(αx)
*⇒ 2cos²(αx)=α³A·sin(αx)cos(αx) with /A, because A≠0 and with 1-sin²(αx)=cos²(αx)*
⇒ 2=α³A·tan(αx) *with /cos(αx) with the assumption of cos(αx)≠0 and with sin(αx)/cos(αx)*
There you go.
_"I think u should separate it to two conditions, first is a=b and second is a=/=b, then comparing the coefficient of each terms."_
If you think so, then you can do that so for yourself. Let me know, how it went.
@@zsoltnagy5654 I got the same solutions with Colby for fun, u might have a view with his solutions where u can find in his comment.
@@zsoltnagy5654 I know there is something wrong with ur solution because to find the function f(x) for all x, we are gonna to use comparing the coefficient of each term , we are not solving for x=?
Haven't watched a video since 2021; oh my goodness, you've poured a lot into this!! I love your videos sir, thank you for being awesome
Interestingly, "pairs" of exponential functions (+/-k) can be combined to give f(x) = 2/k^3 Sinh(k x). Or simply using f(x)=a Sinh[kx] as in the video works to get this as well.
I took an integrable systems class as part of my MS work, and whole families of problems had solutions related to sech(x). I think these functions are often overlooked.
This equation is not linear, so you can't get a new solution by summing two different solutions.
@@VideoFusco I agree. Two arbitrary solutions of a nonlinear equation cannot be combined to get another solution. However, this is a special case that happens to work where 2 particular solutions out of an infinite set of solutions happen to combine to 1 particular solution out of another infinite set of solutions.
Wonderful video. I don't know a lot of english (i'm from latam), but I could understand this video. Thanks for the knowledge.
I like the colors of the chalks you are using. They are brilliant.
This is very cool. Something I have never been exposed to. The "degree operator" itself was a new concept. Never needed it before.
That's a cool equation you got there 🥶
Great video
11:47 I've known about trivial solutions for a very long time, but "boring solution" is a much better name 😆
20:49
Fantastic!!
The equation seems like something one could apply (generalize) classical lie-group theory to. It would certainly be neat to see what kind of symmetries (I guess an infinite parameter group) the equation admits and how these lead to the infinite set of solutions.
Interestingly, the determining equation turns out to be a normal PDE. However, the equation does not seem to admit any Lie-Point symmetries. Maybe/ Probably there are higher-order Lie-Bäcklund Symmetries that give rise to the family of solutions. But checking that by hand is no fun xD
The green at 12:00 should say 1/e^(2kx)
Well that was interesting. Also elegant.
Interested video👍
Math is truly one of the world's greatest spectator sports.
For the exponential solution, the result is correct, but you divided by e^(2kx), not by e^(kx). It eliminates all expontial terms though... (a typical Michael Penn mistake)
The derivative does not always decrease the decree, whitvh is (only) the case for the zeropolynomial. This makes f(x)=0 another valid polynomial solution
At 11:50, he acknowledges the 0 solution, but is ignoring it for being trivial.
He did name the f(x)=0 solution tho
derivative keeps same degree for any constant polynomial. still only matters if f'' is 0, which means f(2x)=0 so the only solution is the trivial one. sorry, pedantic math brain
@@leedanilek5191 some people say the zeropolynomial has degree -infinity or somthing similar
Most of your stuff is too high level for me, but Imma keep watching and watching until it isn't.
Is it only me having such a hard time with the chalk noises. I'm so tempted to throw my phone across the room.
😮😉😄
Idk what to say, it was a nice video
-2sin(x) is a trivial trigonometric solution 😊
“But we know what the derivative does to the degree: it decreases it by exactly one” was repeated twice
better colors, popup fact also felt nice, connection slids can be improved. loved as allways :) 10q
14:00 Oh Frobenius you were here all along, my guiding moonlight
Marvellous!
Now, it seems tempting trying to substitute the a_n by x^n, and explore the families of polynomials and it's roots!
hmmm i was thinking... if differential equations exist and *functional* differential equations exist, is there a way to map them into each other? maybe knowing about one can help us with the other
yay. I got the 3 basic solutions explored in the video. proud.
@11:28 Are you actually multiplying by 1/exp(2kx)?
At x=0 we get
f(0)=f'(0)f''(0) -> f''(0)=f(0)/f'(0)
If f'(0) is not zero, then this defines f''(0).
Taking the derivative of f(2x)=f'(x)f''(x) at x=0 we get
2f'(0)=f''(0)f''(0)+f'(0)f'''(0) -> f'''(0)=2-f''²(0)/f'(0)
If f'(0) is not zero, then this defines f'''(0).
Thus we may repeat, defining all higher derivatives at x=0, except for perhaps a family of cases where a 0/0 might throw a spanner in the works, but a cursory look suggests that that just covers the case f(x)=0 where all derivatives are zero, which is a solution.
This leads me to think that if f(0) and f'(0) are known, then generally all derivatives and thus the function itself is known (assuming it to be differentiable arbitrarily often, i.e. "well-behaved").
Exceptions may be some solutions that do not have x=0 in their domain.
For example, say f(0)=1 and f'(0)=-1, then from the above it follows that f''(0)=-1, f'''(0)=3, etc. such that f(x)=1-x-½x²+½x³+...
This is closely related to that power series approach, except that this shows -- imho -- that apart from a0 and a1 you're not free to choose much. The family of solutions has only "two degrees of freedom".
This is reminiscent of integrating a 2nd order ODE.
My favourite functional equation is f(x) = f((x+0)/2) + f((x+1)/2) on [0, 1]. You can guess the solution 2x-1 and it’s easy to show that this is the only continuously differentiable solution. So what other functions can be the solutions? Well, the only badly non-differentiable function I know is the Weierstraß function. Indeed, one can show that the other solutions are all of Weierstraß type (represented using the same power series with any function having the same periodicity conditions in place of sine)!
I think there are other solutions; c*(x -0.5) should be valid for any constant c
@@power-l5z Indeed, the solutions to this equation form a vector space. Neglecting to mention this was sloppy.
At 11:30 wouldn't it be dividing by e^(2kx) and not e^(kx) ?
Yes I noticed that as well
is there a way to graphically visualize the a/the solution/s?
At 11:30 aren't you supposed to divide everything by e^(2kx) instead of e^(kx)?
why under the exponential section, when you divided ce^2xk by e^kx you end up with just c , law of exponents says that you subtract when dividing so answer should be e^kx corect?
hey sir could plz suggest me a road map to study mayhemtics
in my country mathematics studies are poor
Differential equations really are the coolest bits of math IMO. That said I haven't really taken abstract or much proof based mathematics so I am not sure.
There's no differential equations or calculus at all in the IMO. Do you mean functional equations?
@@mixedfeelings9020 - I think he means "in my opinion"!
Oh dear lord I apologise xD
I wonder if there is a general solution like the equation y’=2y+x for example
15:21 look the owl
for series solutions, Integrate the equation, int(f(2*x))=1/2*(f'(x))^2+c and ... use series
Nice Bro
If you mix the potential of inertia in wave meaning and couple it with some abnormal polynomial equation
How to we know that thrid Solution results in a convergent power series
Wait does the exponential solution also extrapolate to complex numbers?
C = -i, k = ì? Meaning that you would end up with a sin(x) trigonometric function that way?
It does, but those are the wrong numbers. C=1/k^3 gives C=i for k=i.
Nice
Are there other solutions besides x^3 and e^x?
I misread this as f(2x)=f’+f’’ but I haven’t found a solution to that one. It’s not a polynomial at least, or exponential.
y = 4x^3/3 is not a particular case of the series solution. Its coefficients do not satisfy the recursive pattern.
Hey! Small note about the box transitions between solutions: that causes disorientation on large monitors to people who are susceptible to such things (like myself). It causes a sensation that's vaguely related to brief motion sickness due to my eyes saying I'm spinning while my semicircular canals say otherwise.
If you want my suggestion, I suggest brief fade transitions (blends) between the shot and the card. We want to avoid giving the sensation of motion where motion isn't expected.
Maybe I dozed off re-watching this, but was there ever a reference to the source of this problem? Yes, you can always just make one up, but most applications of this "niche subject" have the form of differential-delay equations, where the variable is translated, not re-scaled. So, is there an application?
Well, my lazy retiree's scholarship finally yielded some dividends. I at least discovered the "pantograph equation", which describes such mechanical linkage devices using a first-order ODE with a rescaling parameter λ, generally constrained to the open unit interval. So, a bit different from this example. Guess I'll have to review "reduction of order" whenever I get the time to play with this again. Also, ignoring the scaling parameter, the form of this equation screams Sturm-Louisville, so has anyone tried to obtain a Fourier Series solution yet? (Again, time permitting ...)
Has someone generalized this for f(a*x)=f'(b*x)*f''(c*x) with a,b,c real not zero? It would also be interesting to understand if that differential equation has a real-world phenomenon that is describes. Regarding the currently ongoing discussion about AGI: IMHO not a single AGI is any close to being able to develop these solutions nor could it even follow the explanations.
Wolfram Alpha doesn't so, either?
The trigonometric solution is quite simple. Take k=i in the exponential solution => f(x)= -sin x +icos x
In the polynomial case wouldn't if we included decimal powers change the value
Интегрируя обе части, можно представить уравнение в следующем виде:
F(2x)=[f'(x)]^2+C, где F(x)=integral[f(x)dx]
или
F(2x)=[F"(x)]^2+C
@11:32 divide by e^2kx
A complete general solution to this equation?
It seems clear that any solution will have to be defined at least on a semi-infinite interval,
either (-inf,-a] or [+a,+inf) for some positive a, so that f(x) can always be compared to f(2*x), f(4*x), etc.
This regress also means that all solutions will have to be C-infinity.
Non-analytic solutions on [a,inf) could exist in great variety, constructed using e.g. a C-infinity function on [1,2] which vanishes along with all its derivatives at both endpoints,
then extending it to [2,4), [4,8), etc.
Solutions analytic near the origin, but with an essential singularity at the origin, I cannot analyze yet,
but if there is no more than a pole at the origin, then ALL the real solutions will be one of these:
f(x)=0, f(x)=4/9*x^3, f(x)=+2*c^3*sinh(x/c), f(x)=-2*c^3*sin(x/c), f(x)=c^3*exp(x/c).
This can be seen by analyzing cases according to the integer N,
where N is the least exponent on x in the non-zero terms of the Taylor series or Laurent series for f(x).
Reasoning about the recurrence relations between the series coefficients quickly establishes
that the cases N3 are all impossible, leading to contradictions.
The case N=3 produces only the cubic solution.
The case N=1 produces only the sin and sinh solutions.
The case N=0 seems to generate a two-parameter family of solutions that includes the exponentials.
One parameter can be the scale factor, c, and it is convenient to take the other parameter to be b=f'''(0)-2, which is independent of c.
Numerical experimentation with Maple makes it look very likely that, after excluding the values of b generating the solutions already mentioned, all other values of b generate a meaningful-looking Taylor series with a radius of convergence that turns out to be zero.
But, this looks hard to prove in general.
Not too sure about these new transitions, would be less startling if it was just a drop down when showing "polynomial solutions"
exponential solution: Little mistake on the board. You divided by 1/exp(2kx)
f(x) = 0 is polynomial, exponential and trigonometric... and it fits equation, but I barely heard anything bout it in this video
It is definitely mentioned as a "boring solution" at 11:50, i.e. in the exponential case.
But if you divide ce^(2kx) by e^(kx), yo do not get c 🤔.
What's the general formula?
11:32 meant 1/e^(2kx)
Excellent, clear-cut explanation...thank you!
Particular solutions for the equation: y=(4/9)x^3, y=(2/k^3) sinh(kx), y=(-2/k^3) sin(kx), y=(1/k^3)e^(kx)
The solutions have common property: if y=f(x) is one of them, then so is y=(1/k^3)f(kx) for an arbitrary constant k.
* To verify for any function, let g(x)=(1/k^3)f(kx), then: g(2x)=(1/k^3)f(2kx), g'(x)=(1/k^2)f'(kx), g''(x)=(1/k)f''(kx).
* Plugging into the equation "f(2kx)=f'(kx)f''(kx)", we get "g(2x)=g'(x)x''(x)".
* Therefore, if f(x) is a solution, then g(x) also satisfies the equation.
So when we test some functions like "y=A·sin(x), B·sinh(x) or C·e^x" and we find "A=-2, B=2 and C=1", then we can extend them to the first line.
Interesting observation.
f(x) is not a function, x is the varible, f is the function. nice video
Trigonometric solution: f(x) = - sin x + i cos x
At 17:30 isn't this neglecting terms of x with exponents between n and 2n? Clearly the product produces these terms
Trigonometric solution:
Let f(x)=Acos(kx)+Bsin(kx)
f(2x)=Acos(2kx)+Bsin(2kx)
f'(x)=-Aksin(kx)+Bkcos(kx)
f"(x)=-Ak^(2)cos(kx)-Bk^(2)sin(kx)
multiplying theses and simplifying we get
f'(x)*f"(x)=k^(3)*[AB(sin^(2)(kx)-cos^(2)(kx))+(A^(2)-B^(2))(sin(kx)cos(kx)]
Using double angle trig identities we can write this as
=-ABk^(3)cos(2kx)+0.5*(A^(2)-B^(2))k^(3)sin(2kx)
Going back to the original equation we have
Acos(2kx)+Bsin(2kx)=-ABk^(3)cos(2kx)+0.5*(A^(2)-B^(2))k^(3)sin(2kx)
Setting the sine and cosine parts equal to each other we get
(1) A=-ABk^(3) and
(2) B=0.5*(A^(2)-B^(2))k^(3)
Dividing (1) by A on both sides we get
1=-Bk^(3) or B=-1/k^(3)
Substitute into (2) and rearrange
-1/k^(3)=0.5*(A^(2)-1/k^(6))k^(3)
-1/k^(3)=0.5*A^(2)k^(3)-0.5/k^(3)
-0.5/k^(3)=0.5A^(2)k^(3)
A^(2)=-1/k^(6)
Since k^(6) is always positive, A must be imaginary. In other words, there are no pure real sinusoidal solutions to the equation.
f(x)=-2*sin(x)
Your mistake was assuming you could divide by A. If A = 0, you get solutions with just sine and no cosines such as the -2sin(x) posted above. On the other hand, your solution with A imaginary actually leads to Penn's solution for exponentials, just with his k replaced with your ik.
@@burk314 I replace k with ik, then sin(kx) become sin(ikx), not sin(kx) anymore
@@khoozu7802 Yes, but sin(ikx) = i sinh(kx), so complex values turn it into exponentials.
The argument about degree is wrong, it only proves the max degree is 3. This allows that a could actually be zero, which is turn forces b, c, d zero, and thus f=0 is a legitimate polynomial solution.
Can you find a general solution to f:R→R f^(n)(x)=f(ax) that converges on all real numbers.
(The first part is the nth derivative of f(x))
I didn’t know that Sting did math.
I got f(x)=c*sin( cbrt(-2/c)*x) for any c.
1 sec after seeing this, one can directly notice that the exponential function is onenof the solutions xD
f(x) = -2sinx works
f(x) = n sin( cube root(-2/n) * x )
With n =\= 0
f(x)=-2/k^3*sin(kx) is a family of trigonometric solutions.
This comes naturally from the 1/k^3 e^kx family of solutions, letting k = ±i, and recognizing sin(x) = (e^ix - e^-ix)/2i
I'm not seeing how you can immediately get the sine solution from the exponential one even with that identity. The functional differential equation is in no way linear, so you can't just take linear combinations. We don't even get equivalent solutions with cosine despite cos(x) = (e^ix+e^-ix)/2.
@@burk314 good point, you're right
at least one
Infinitely Many Non-Linear Equations in Infinitely Many Unknows ~ IDIC (Vulcan philosophy) at 17:47.
Putting f(x) = A sin(x) we get A sin(2x) = 2 A sin(x) cos(x) = (A cos(x) ) . (- A sin(x) ), so A^2 + 2 A =0. Hence A = -2 is a non-trivial solution, so one solution is f(x) = -2 sin(x)
f(x)=-2/b^3*sin(bx)
I think there's a mistake at 8:18. Shouldn't it be 4b = 18ab ?
But we know a to be 4/9, so 18*4/9=8.
@@samwalko oh right
I saw the same thing. The second equation from like terms does not determine anything so long as b ≠ 0. If b = 0 we get solution given by MP. On the other hand, since the video has a second-order diffeq (even if it is nonlinear), there should be at least one free parameter. Let b be a free parameter, then the third coefficient in the expansion is:
6 a c + 4 b² = 8c / 3 + 4b² = 2 c, or c(b) = - 6 b²
d = 2 b c, or d(b) = - 12 b³. The polynomial solution f(x) ≠ 0 is:
f(x; b) = (4/9) x³ + b x² - 6 b² x - 12 b³.
@@paulkohl9267 Except for b nonzero, those are not solutions. If you work them out, f(2x) will have quadratic term 4bx^2, but f'(x)f''(x) will have quadratic term 8bx^2.
Am I the only one that saw the face drawn in on the chalkboard with the eraser marks?
[e^(ax) - e^(-ax)]/(a³)
It's a family of valid solution that I have checked out.
Subbing in (a -> ia), we get trig solution.
f(x) = -2sin(ax)/a³
This is the only trig solution I believe.
Only sine and cosine have repeating derivatives. Cosine has non zero value at x=0 while its derivative is 0.
So only sine passes the test.
Could someone explain why we don’t need parentheses on the m terms at 16:55?
It's because they are suffixes (subscripts).
@@Chris_5318 I meant that I don’t understand why we write n-m+1 instead of n-(m+1)=n-m-1. Why should we not bother changing the sign of the constant terms?
@@averagegamer9513 I see what you mean. Sadly (happily) I've drunk to much wine to check whether Michael did another of his required typos.
It looks like the series solution has a0 and a1 as free parameters.
Without watching video:
One approach: y = c x^3
y' y" = c^2 * 18 * x^3 = c * 8 * x^3
=> c = 8/18
Another approach: y = c exp(kx)
y' y" = c^2 k^3 exp(2kx) = c exp(2kx)
=> c = k^-3
Where does the terms @16:55 come from?
Collecting all terms with the same power.
You missed the Penn fact concerning Cauchy's formula that he pointed out a few seconds before that.
en.m.wikipedia.org/wiki/Cauchy_product
@@landsgevaer I saw the fact
I’m not sure how it applied to get all the terms that showed up
@@landsgevaer the fact had a convolution symbol in it
Not sure how that’s this
@@Happy_Abe Convolution is * . The circle tends to be a general operator, often function composition.
But anyway, see the link I posted above.
@@landsgevaer Thank you
Appreciate it