it's absolutely wild that middle school kids are watching this, but I agree that your presentation makes it easy to follow even when I didn't know the methods before
@@emanuellandeholm5657 I don't see how you would think this was about analytics after having watched Kamaal tell about how there's a commenter who shows these videos to middle schoolers
I have got so used to listen your"ookay cool" that many times I myself use "ookay cool" in same tune whether while solving questions in maths or escaping after disturbing my neighbours. And also I am going in my 10th grade now and I am watching your videos since I entered in the 9th grade. And you have taught me many techniques like Feyman's , substituting x by -x or 1/x, Laplace transform, gamma ad digamma, beta and a lot more. I still remember the happiness which i had got after solving a double integral using all the techniques I had learnt from you and after getting a 2page solution my answer matched with that of yours. I thank you for all the knowledge and fun you gave me Love from India
@@maths_505 Thanks a lot, it would be a pleasure. You are right, I am French, but the only reason why I choosed this French flag is to excuse my poor English 😅. I like your channel, I follow about 10 maths channels, including the most famous. I am retired, I was software analyst, but maths has always been a hobby. At the moment I am working about the real generalization of the factorial function, and its development into a pylonomial + rational fraction. Best regards.
A thing I do like about your videos is the direct and natural way you present them. Including swearing haha. But I do understand the issue about middle schoolers. Had these videos back in time would have been a great motivation! Regarding the video, interesting as always 🎉
I am an engineer , I enjoy the simplicity of your solutions in videos even for integrals , I think thry must teach this way even in college , we dont need the dominated convergence to conclude a limit of an all ready easy one
I got lazy(?) and just used a power series. The initial equation shows that all derivatives exist - so you know you can do this. Expand at x=0. Spot that the series you get has infinite radius of convergence. Then spot sin and cos in there 😂. Your solution is much more elegant!
This was interesting! Such a cool solution. By the way, I’m fine with less curses. Not moralizing, just saying, it can be distracting. I really like your “aside” discussions. Always interesting and genuine. Middle school kids, ya know, I think it’s better than good. One of the reasons so many people don’t love math is they don’t have a clue what it is or what you can use it for. They’ve literally in their whole life never seen it done even once… in their whole life! These middle schoolers will. Some of them will get inspired. And it’s because of you… reducing curses. That’s like a blessing, dude, for all those kids.
There is an easy way to solve this. The only functions which when added to its derivative at opposite parity that gives an even function x^2 + a are trigonometric plus a polynomial particular solution. Then you substitute a polynomial of degree 2 to determine the coefficients.
Cool problem! My first thought was representing f as the sum of an odd and an even function. Let's see how 505 tackles this. Edit: Yeah, you get the same development by simply differrentiating Still a real nice problem and all!
@@maths_505 As I understand you, you told that you differentiate f(-x) and receive -f‘(-x). But did you tell why? For me, I can‘t explain the minus sign - except: you have chosen. Can you explain that for me, please?
@@andreasworle752 the minus sign was just a consequence of the chain rule. If you're asking about my train of thought to actually arrive at that step, well it was just something I saw that could work and I worked my way up from there.
@@maths_505 Oh! Thank you very much! I thought too complex and had a very special assumption in my mind - and hadn‘t thought about the chain rule. Thank you for clarification!
Heh, I used to cuss up a storm in middle school. I wonder if it's not the kids ears that are the concern, but the "delicate sensibilities" of the parents, folks more likely to want to brag to their buddies about the cool math tutorials they keep their kids busy watching, but worried that their buddies will look down on them if they check them out too only to blush when they get hit by an f-bomb or three along the way.
I saw x^2+a was even and there was f(x) terms and f(-x) terms, so I split f(x) into odd and even, e(x)+o(x) that means f'(-x) ended up being -e'(x)+o'(x) This let me get a relationship between the odd and even parts of f(x) ending with f''(x)+f(x)=x^2+2x+a which I then solved and substituted back to get f(x)=A(sinx-cosx)+x^2+2x+a-2 Full development: f'(-x)+f(x)=x^2+a e(x)+o(x)=f(x) e(x) is even, o(x) is odd e(x)-o(x)=f(-x) e'(x)+o'(x)=f'(x) e'(x) is odd, o'(x) is even e'(x)-o'(x)=-f'(-x) -e'(x)+o'(x)=f'(-x) -e'(x)+o'(x)+e(x)+o(x)=x^2+a (-e'(x)+o(x))+(e(x)+o'(x))=x^2+a x^2+a is even o(x)-e'(x)=0 o(x)=e'(x) o'(x)=e''(x) o'(x)+e(x)=x^2+a e''(x)+e(x)=x^2+a o'(x)+e(x)=x^2+a o''(x)+e'(x)=2x o''(x)+o(x)=2x e''(x)+o''(x)+e(x)+o(x)=x^2+2x+a f''(x)+f(x)=x^2+2x+a f_h''(x)+f_h(x)=0 f_h(x)=Asinx+Bcosx f'_h(-x)=Bsinx+Acosx (A+B)sinx+(A+B)cosx=0 B=-A f_h(x)=A(sinx-cosx) f_p(x)=bx^2+cx+a+d f'_p(-x)=-2bx+c bx^2+(c-2b)x+a+c+d=x^2+a b=1 c-2b=0, c=2 c+d=0, d=-2 f(x)=A(sinx-cosx)+x^2+2x+a-2 e(x)=-Acosx+x^2+a-2 o(x)=Asinx+2x e'(-x)=-Asinx-2x o'(-x)=Acosx+2 e(x)+o'(-x)=x^2+a e'(-x)+o(x)=0
Bit of a clickbait thumbnail. There's nothing wrong with the solution at all, just requires an extra step because of the further differentiation we did to make it a 2nd order ODE
it's absolutely wild that middle school kids are watching this, but I agree that your presentation makes it easy to follow even when I didn't know the methods before
Thanks mate
@@emanuellandeholm5657 I don't see how you would think this was about analytics after having watched Kamaal tell about how there's a commenter who shows these videos to middle schoolers
I have got so used to listen your"ookay cool" that many times I myself use "ookay cool" in same tune whether while solving questions in maths or escaping after disturbing my neighbours.
And also I am going in my 10th grade now and I am watching your videos since I entered in the 9th grade. And you have taught me many techniques like Feyman's , substituting x by -x or 1/x, Laplace transform, gamma ad digamma, beta and a lot more.
I still remember the happiness which i had got after solving a double integral using all the techniques I had learnt from you and after getting a 2page solution my answer matched with that of yours.
I thank you for all the knowledge and fun you gave me
Love from India
But final answer is wrong, we need f(x) = c1(sinX-cosX) +x2+2x+a-2
Oh yeah I forgot that. It's cool, it's just a matter of rewriting it at the end since everything that proceeded it was correct.
Agree
from Morocco thank you for this clear complete explanations
Hi,
"Ok, cool" : 4:35 , 5:27 , 8:04 , 9:02 ,
"Terribly sorry about that" : 5:14 .
I'm assuming you're french and I hope I visit France soon cuz I really want to meet you bro 😂
@@maths_505 Thanks a lot, it would be a pleasure. You are right, I am French, but the only reason why I choosed this French flag is to excuse my poor English 😅.
I like your channel, I follow about 10 maths channels, including the most famous.
I am retired, I was software analyst, but maths has always been a hobby.
At the moment I am working about the real generalization of the factorial function, and its development into a pylonomial + rational fraction.
Best regards.
A thing I do like about your videos is the direct and natural way you present them. Including swearing haha. But I do understand the issue about middle schoolers. Had these videos back in time would have been a great motivation!
Regarding the video, interesting as always 🎉
Thanks mate
Another soliution is to divide f in parity even+odd and evn-odd will verify some equation and you can conclude
Very smart analysis. Thanks.
I am an engineer , I enjoy the simplicity of your solutions in videos even for integrals , I think thry must teach this way even in college , we dont need the dominated convergence to conclude a limit of an all ready easy one
I really love your solution development, no strict methods, just intuition and basic guidelines
The strict methods and formalism are incredibly important. I just skip that route to make it accessible to someone who hasn't taken a DEs class yet.
@@maths_505 what are some formalisms I should be aware of for this problem then? PS have you heard of operational calculus? Is it good?
Take a shot every time he says ”ooookay cool” 🙃
"which implies..."
Idolatria não faz bem
I got lazy(?) and just used a power series. The initial equation shows that all derivatives exist - so you know you can do this. Expand at x=0. Spot that the series you get has infinite radius of convergence. Then spot sin and cos in there 😂.
Your solution is much more elegant!
This was interesting! Such a cool solution. By the way, I’m fine with less curses. Not moralizing, just saying, it can be distracting. I really like your “aside” discussions. Always interesting and genuine. Middle school kids, ya know, I think it’s better than good. One of the reasons so many people don’t love math is they don’t have a clue what it is or what you can use it for. They’ve literally in their whole life never seen it done even once… in their whole life! These middle schoolers will. Some of them will get inspired. And it’s because of you… reducing curses. That’s like a blessing, dude, for all those kids.
Very nice.it'l be nice to see more problems like this in the channel.
There is an easy way to solve this.
The only functions which when added to its derivative at opposite parity that gives an even function x^2 + a are trigonometric plus a polynomial particular solution. Then you substitute a polynomial of degree 2 to determine the coefficients.
while f(x)=x
than f(-x) = -x and -f ` (-x) = 1
that's a countre example
in 2:00
Your presentation is so cool👍
Cool problem! My first thought was representing f as the sum of an odd and an even function. Let's see how 505 tackles this.
Edit: Yeah, you get the same development by simply differrentiating Still a real nice problem and all!
Gud
Left out some of the particular solution at the last equation
Yeah my bad😂
f(x)=x^2+2x-2+a
Thanks for this nice problem. But what is your argument for getting -f'(-x) by differentiating f(-x) in detail?
Thanks for your answer
My argument?
@@maths_505 As I understand you, you told that you differentiate f(-x) and receive -f‘(-x). But did you tell why? For me, I can‘t explain the minus sign - except: you have chosen. Can you explain that for me, please?
@@andreasworle752 the minus sign was just a consequence of the chain rule. If you're asking about my train of thought to actually arrive at that step, well it was just something I saw that could work and I worked my way up from there.
@@maths_505 Oh! Thank you very much! I thought too complex and had a very special assumption in my mind - and hadn‘t thought about the chain rule. Thank you for clarification!
@@andreasworle752 😂😂 it's okay my friend. Happens to me quite often.
Heh, I used to cuss up a storm in middle school. I wonder if it's not the kids ears that are the concern, but the "delicate sensibilities" of the parents, folks more likely to want to brag to their buddies about the cool math tutorials they keep their kids busy watching, but worried that their buddies will look down on them if they check them out too only to blush when they get hit by an f-bomb or three along the way.
Yeah but I think I do have some more responsibilities now. I've got over 50k subs now so I might as well be more careful.
I saw x^2+a was even and there was f(x) terms and f(-x) terms, so I split f(x) into odd and even, e(x)+o(x)
that means f'(-x) ended up being -e'(x)+o'(x)
This let me get a relationship between the odd and even parts of f(x)
ending with f''(x)+f(x)=x^2+2x+a
which I then solved and substituted back to get
f(x)=A(sinx-cosx)+x^2+2x+a-2
Full development:
f'(-x)+f(x)=x^2+a
e(x)+o(x)=f(x)
e(x) is even, o(x) is odd
e(x)-o(x)=f(-x)
e'(x)+o'(x)=f'(x)
e'(x) is odd, o'(x) is even
e'(x)-o'(x)=-f'(-x)
-e'(x)+o'(x)=f'(-x)
-e'(x)+o'(x)+e(x)+o(x)=x^2+a
(-e'(x)+o(x))+(e(x)+o'(x))=x^2+a
x^2+a is even
o(x)-e'(x)=0
o(x)=e'(x)
o'(x)=e''(x)
o'(x)+e(x)=x^2+a
e''(x)+e(x)=x^2+a
o'(x)+e(x)=x^2+a
o''(x)+e'(x)=2x
o''(x)+o(x)=2x
e''(x)+o''(x)+e(x)+o(x)=x^2+2x+a
f''(x)+f(x)=x^2+2x+a
f_h''(x)+f_h(x)=0
f_h(x)=Asinx+Bcosx
f'_h(-x)=Bsinx+Acosx
(A+B)sinx+(A+B)cosx=0
B=-A
f_h(x)=A(sinx-cosx)
f_p(x)=bx^2+cx+a+d
f'_p(-x)=-2bx+c
bx^2+(c-2b)x+a+c+d=x^2+a
b=1
c-2b=0, c=2
c+d=0, d=-2
f(x)=A(sinx-cosx)+x^2+2x+a-2
e(x)=-Acosx+x^2+a-2
o(x)=Asinx+2x
e'(-x)=-Asinx-2x
o'(-x)=Acosx+2
e(x)+o'(-x)=x^2+a
e'(-x)+o(x)=0
Or, without differentiating any equation, we can use matrix exponential.
Math is easy,english is hard 7:20
What is the domain of alpha?
All real numbers obviously
Bit of a clickbait thumbnail. There's nothing wrong with the solution at all, just requires an extra step because of the further differentiation we did to make it a 2nd order ODE