I liked the explanation of why it works. I know the technique, but I am always distrustful of it, like it is a trick or I am misremembering it. Hopefully I will be able to trust it more better now.
This video was shown to my pre-cal class by the teacher, this video helped me understand a fast way to do this type of equation, thank you for teaching this. (My teacher also is envious that you have blackboard)
10:37 still don't get it, when you make x zero aren't you still dividing by zero even though the x in the denominator on both sides have been cancelled out, why does it work? is it because undefined=undefined works as a valid equation?
To be sure, you couldn't put x=0 into the original equation. When we multiply everything by x, we have made a different equation, the only purpose of which is, it helps us solve the original equation. So while we can't do x=0 in the original equation, we can in the helper equation.
I guess it's a bit like limits, where you can take a derivative of 0/0 and instead of putting in the numbers, you solve it first and end up with something else
In my opinion this cover up has some advantages when roots of the denominator are real and distinct We need to remember which value we have already used If we allow using complex numbers then there is residue method which gives us partial fraction decomposition (Just like for inverse Laplace transform but without this exp(st) factor)
If we use it for integration then we can use polynomial long division and Ostrogradsky method of isolation rational part of integral and then we have integral with integrad which has only distinct roots of denominator
Strictly speaking, you can't just plug zero in the last step as it must be that x is different from 0. However you can take the limit of both sides of the equation as x goes to zero which gives you the same result as plugging in zero.
I guess its the same thing as multiplying the entire equation by the original denominator so that there is no fraction, then just let x equal all the poles until you find values for A,B,C etc
I liked the explanation of why it works. I know the technique, but I am always distrustful of it, like it is a trick or I am misremembering it. Hopefully I will be able to trust it more better now.
You can't cover up how good Prime Newtons is as a teacher! 🎉😊
This one wins 🏆 🙌 👌 👏 🤣 😂
Sure, he's hot too❤
Prime Newton always covers his head so fashionably!
This guy getting me interesting more.. I am 10th grader, I like learning/watching such advanced mathematics and this guy is a good place...
Mathematically brilliant teacher.
"Let's get into the video" is my new favorite phrase
This video was shown to my pre-cal class by the teacher, this video helped me understand a fast way to do this type of equation, thank you for teaching this. (My teacher also is envious that you have blackboard)
😮😮😮😮😮😮😮😮😮🎉🎉🎉🎉 you deserve it ❣️❣️❣️❣️
10:37 still don't get it, when you make x zero aren't you still dividing by zero even though the x in the denominator on both sides have been cancelled out, why does it work? is it because undefined=undefined works as a valid equation?
Same thing I was wondering
To be sure, you couldn't put x=0 into the original equation. When we multiply everything by x, we have made a different equation, the only purpose of which is, it helps us solve the original equation. So while we can't do x=0 in the original equation, we can in the helper equation.
I guess it's a bit like limits, where you can take a derivative of 0/0 and instead of putting in the numbers, you solve it first and end up with something else
you take the limit where x goes to 0
your explanations are so good . thanks sir
I'm learning this this semester so yippee
What brand of chalk do you use? Another great video. Great, full explanation.
In my opinion this cover up has some advantages when roots of the denominator are real and distinct
We need to remember which value we have already used
If we allow using complex numbers then there is residue method which gives us partial fraction decomposition
(Just like for inverse Laplace transform but without this exp(st) factor)
If we use it for integration then we can use polynomial long division and Ostrogradsky method of isolation rational part of integral
and then we have integral with integrad which has only distinct roots of denominator
Nicely done!
Thank you, Teacher!
İ love this video very much..
Wow 😳 this awesome
Thank you so much!
❤️🙏
Wonderful method!
Strictly speaking, you can't just plug zero in the last step as it must be that x is different from 0. However you can take the limit of both sides of the equation as x goes to zero which gives you the same result as plugging in zero.
Man with whole due of respect screw all my teachers they were idiots never did this
You look like an actor 😁
I guess its the same thing as multiplying the entire equation by the original denominator so that there is no fraction, then just let x equal all the poles until you find values for A,B,C etc
Oh wow!
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