@@BriTheMathGuy Carl Bender has a series of lectures on asymptotic and perturbative analytical methods of aproximation, posted to youtube under the titles of 'mathematical physics'. Anyways, in one lecture Carl drops the wonderful math 'pun' to the class: IFF infinity = 1/0, then "rotating" both sides 90 degrees results in 8 = -10 which simplifies to 0 = -18, then rotating back results in the desired "result" 0 = 1/infinity...
I did this purposely in the test for the problem "Find 2 irrational numbers whose sum is also irrational", my teacher ticked it correct, telling "I too watch Bri".
I remember having a book in math class with these fun bonus pages. It showed a drawn picture of a kid cancelling out the 9s of 19/95 and the teacher is watching in horror. :D
I wont Sir 8th grader here And btw this trick is there in the last chapter of our math book ... playing with numbers .. They have asked us to find more such numbers like 64/16 Could u please help me to get some ??
The differentiation is not a mistake. this is what differentiation does--- it turns each variation into an independent additive contribution. The bottom variation follows the power rule ,the top variation follows the exponential rule, and therefore adding them together is correct.
First of all, as an audience following you more than 3-4 years, i am glad that your audience is growing. What i want to say, however, is that the differentiation of x^x seems to be correct. As derivative is a linear operator, we can use superposition principle and it is distributive along different eigenfunctions. I will check it out as soon as i wake up.
The one about d/dx x^x isn't wrong. It's actually based in multi variable calculus. See it as d/dx( x^y(x) ) With y=x Use the chain rule for multi variable. d/dx = partial dx + partial dy * dy/dx. = partial dx + partial dy d/dx( x^y) + d/dy (x^y) = y*x^(y-1) + x^y ln(x) Then replace by x by y x*x^(x-1) + x^x ln(x)
The d/dx(x^x) one turns out to be correct, since the "incorrect procedure" is actually essentially correct in this case. This is exactly the process you would follow if you applied the multivariate chain rule to the problem. Regard x^x as the composite of the two variable function x^y with the map x|-->(x,x) and apply chain rule, and you get that it's the sum of the partials of x^y where you replace y by x.
pretty much the same with the log(1+2+3), you can do what he did when those are multiplied rather than added together, and the sum of 1, 2 and 3 is the same as their product so it's essentially correct
Your improper operation at the 4:30 is in fact proper when the arguments add to 90. Which it is in this example. Your final one is an illustration that 6 is a perfect number.
Actually, the final one is not, the fact that a number is a perfect number does not mean that the sum of its positive factors equals the product of its positive factors E.g. 1 + 2 + 4 + 7 + 14 = 28 1 × 2 × 4 × 7 × 14 = 784 = 28²
The proof of the last one is: log(1+2+3) = log(6) = log(2 . 3) = log(2) + log(3) The other side: log(1) + log(2) + log(3) log 1 is just 0 so it can be canceled And we have log(2) + log(3) Just put the left side and right side next to each other and tada! We proved the identity! :)
nice vid. Another alternative for the "find two irrationals whose sum is irrational" is to do root(2) + root(3) = root(2 + 3) = root(5) which is irrational : )
This would make a great April Fools Video. I once wanted to do something like this for all of the questions on a test, but since I wanted to get a perfect grade at the end of the year, saner heads prevailed.
At 2:23 this is actually a concept derived from multivariable calculus. It works if a function is composed of any number of functions of x like this: df/dx = ðf/ða * da/dx + ðf/ðb * db/dx + ðf/ðc * dc/dx + ... for example (although trivially): f = x^6 a = x b = x^2 c = x^3 f = abc df/dx = bc * 1 + ac * 2x + ab * 3x^2 = x^2 * x^3 + x * x^3 * 2x + x * x^2 * 3x^2 = x^5 + 2x^5 + 3x^5 = 6x^5 df/dx = 6x^5
0:40 Another way of doing this is to subtract each digit of each numerator and denominator from each other: 9/2 - 25/10 = (9 - 2 - 5)/(2 - 1 - 0) = 2/1 = 2
@@anshumanagrawal346 If you have, u and v being functions of x and f being a function of u and v, then the multivariable chain rule states: df/dx = ∂f/∂u du/dx + ∂f/∂v dv/dx (The actual version would have u1,u2,....un functions, but let's just go with the smaller case.) Now, call f(u,v)=u^v with u(x)=x,v(x)=x: df/dx = vu^(v-1)•1+u^v ln(v)•1 which is essentially what you write in the original statement - take one x constant, differentiate with the other and do the same for other x. This applies well and good to all such applications. Obviously there has to be a reason when a trick works everywhere. ;)
@@nikhilnagaria2672 Wow, thank you so much, I have wondered for a long time about this cause this same logic is followed by so many derivative rules that I highly suspected there was some such rule but I couldn't find it anywhere online as I didn't have the right keywords, I even asked this question in a comment on a video on this channel a while ago 😂
Like when I first learned this "trick" from my teacher that for differentiation of the functions of the type f^g, to take the derivative pretend one is a constant and the other is the variable, and then vice versa and add the results, after I proved this for any generic f and g using logarithmic differentiation, after I while I realised it also applies to the product rule, the division rule, and (somewhat trivially) to the sum rule as well
2:23 This could be a legit math tric: Lets say f(a, b) = a^b And a(x) = x and b(x) = x (partial f/partial a) = b * a^(b-1) (partial f/partial b) = a^b * ln(b) da/dx = 1 db/dx = 1 Now apply the multivariable chain rule: df/dx = (partial f/partial a) * da/dx + (partial f/partial b) * db/dx = b * a^(b-1) * 1 + a^b * ln(b) * 1 Now we substitute a=x and b=x to get: df/dx = x * x^(x-1) + x^x * ln(x) = x^x + x^x * ln(x) By (partial f/partial a) I mean the partial derivative.
The 3 × 9 one kind of works. You divided 81, or 9^2, by 3, so it's basically saying (a^2)/b = (a)(b), where a and b are real numbers. Simplifying, and if a is not 0, then a = b^2. So as long as this is true, it will work
Now the next challenge for the viewers is to find a general solution (for most of the examples) for values where the incorrect method will get the right answer
This is also somehow a mathematically correct derivation. Partial derivative rules can be applied to multiple instances of the same variable, not just different variables.( Think about it. If it works for every single possible function other than x, why would it not work for x too. Meta-calculus. ) Just pretend the other xs are (different) constants and add the results up. D x^2 = D xx = D _x + -> _ D x_ = -> _ = 2_ = 2x D xsinx = D x_ + -> _ D _sinx = -> _cosx = sinx + xcosx D x^(xsinx) = D x^ _ + -> _ x^( _ -1) D _ ^($sinx) + -> $log _ cosx _ ^($sinx) D _ ^(x$) = -> $log _ _ ^($x) = (xsinx)x^(xsinx-1) + xlogx cosx x^(xsinx) + sinx logx x^(xsinx) which is a monstrosity but does simplify to the correct solution of x^(xsinx)(sinx+logx)(sinx+xcosx)
The answer for the derivative of x^x follows from the chain rule for a function of two variables. If f(x,y)=x^x, then d/dx f(x,x) = f_x (x,x) +f_y(x,x) = x * x^*(x-1) + (ln x) x^x
Actually the power rule and the exponent rule are just specific cases for a general rule of derivatives of f(x)^g(x), where you “add the two rules” together. Because the derivative of a constant is 0, the power rule and the exponent rule will cancel out one half of the equation or the other, giving their respective rules
1:21 multiples of 9 for me was always the easiest for its reoccurring pattern: (10*n + 9 - n) with n is 0 to 9 1 x 9 = (10*0 + 9 - 0) = 9 2 x 9 = (10*1 + 9 - 1) = 18 . . . 10 x 9 = (10*9 + 9 - 9) = 90
5:43 this is a direct result of the fact that 1+2+3=1x2x3. The same is true of any collection of numbers which have equal sum and product. ln(1+1+1+3+3) = ln(1)+ln(1)+ln(1)+ln(3)+ln(3)
2:23 It’s actually CORRECT, because x^x can be written as f(u,v), where f(u,v)=u^v, u=x, v=x. df/dx = ∂f/∂u ∙ du/dx + ∂f/∂v ∙ dv/dx =x∙x^(x-1)∙1 + x^x∙ln(x)∙1 =x^x∙(ln(x)+1)
@@Firefly256 that's not there of course when you write it down. It just means "raised to the power of", it's quite common on UA-cam since its formula writing tools aren't the greatest here.
@@Firefly256Depending on if you write x² or x^2, you get 5 (in the former case) or -5 (in the latter case, since the ^ unfolds into the -), so those are both correct
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The only one I have seen before is the derivative of 1/x. But this is a great collection!
Glad you enjoyed them!
@@BriTheMathGuy Carl Bender has a series of lectures on asymptotic and perturbative analytical methods of aproximation, posted to youtube under the titles of 'mathematical physics'. Anyways, in one lecture Carl drops the wonderful math 'pun' to the class: IFF infinity = 1/0, then "rotating" both sides 90 degrees results in 8 = -10 which simplifies to 0 = -18, then rotating back results in the desired "result" 0 = 1/infinity...
I did this purposely in the test for the problem "Find 2 irrational numbers whose sum is also irrational", my teacher ticked it correct, telling "I too watch Bri".
its true, I was the test paper
hes totally correct, im the ink in his pen he wrote with can confirm
its very true,i was the table
I'm the principal of his school can confirm
I'm the air they breathed during 3 seconds
I remember having a book in math class with these fun bonus pages. It showed a drawn picture of a kid cancelling out the 9s of 19/95 and the teacher is watching in horror. :D
I was just taught to do it this way
its even correct, oh the horrors
Let him cook
This is why teachers want you to show your work
Please don't make these mistakes yourself.
Please.
I wont Sir
8th grader here
And btw this trick is there in the last chapter of our math book ... playing with numbers .. They have asked us to find more such numbers like 64/16 Could u please help me to get some ??
The differentiation is not a mistake. this is what differentiation does--- it turns each variation into an independent additive contribution. The bottom variation follows the power rule ,the top variation follows the exponential rule, and therefore adding them together is correct.
First of all, as an audience following you more than 3-4 years, i am glad that your audience is growing. What i want to say, however, is that the differentiation of x^x seems to be correct. As derivative is a linear operator, we can use superposition principle and it is distributive along different eigenfunctions. I will check it out as soon as i wake up.
@Bacon Hair This video is a joke, don't do any of the things shown in it. They're funny coincidences, but still technically wrong.
The one about d/dx x^x isn't wrong. It's actually based in multi variable calculus.
See it as d/dx( x^y(x) )
With y=x
Use the chain rule for multi variable.
d/dx = partial dx + partial dy * dy/dx.
= partial dx + partial dy
d/dx( x^y) + d/dy (x^y)
= y*x^(y-1) + x^y ln(x)
Then replace by x by y
x*x^(x-1) + x^x ln(x)
Thanks for the tips on problems I should AVOID when introducing a topic to my students!
You suck.
Mr west i got u
“A function can’t just be equal to zero.”
Banach spaces: “Foolish mortal. You know nothing of my powers!”
Funnily enough, there are null-adic (zero arguments) functions that return constants
Idk it would be as Lebesgue measurable as it can get
As an engineer, I don't see what's wrong at 2:12
pi = e = 3
The d/dx(x^x) one turns out to be correct, since the "incorrect procedure" is actually essentially correct in this case. This is exactly the process you would follow if you applied the multivariate chain rule to the problem. Regard x^x as the composite of the two variable function x^y with the map x|-->(x,x) and apply chain rule, and you get that it's the sum of the partials of x^y where you replace y by x.
pretty much the same with the log(1+2+3), you can do what he did when those are multiplied rather than added together, and the sum of 1, 2 and 3 is the same as their product so it's essentially correct
Your improper operation at the 4:30 is in fact proper when the arguments add to 90. Which it is in this example.
Your final one is an illustration that 6 is a perfect number.
Actually, the final one is not, the fact that a number is a perfect number does not mean that the sum of its positive factors equals the product of its positive factors
E.g.
1 + 2 + 4 + 7 + 14 = 28
1 × 2 × 4 × 7 × 14 = 784 = 28²
The proof of the last one is:
log(1+2+3) = log(6) = log(2 . 3)
= log(2) + log(3)
The other side:
log(1) + log(2) + log(3)
log 1 is just 0 so it can be canceled
And we have log(2) + log(3)
Just put the left side and right side next to each other and tada!
We proved the identity! :)
This case works because
1 + 2 + 3 = 1 × 2 × 3
@@W_Qimuel exactly
nice vid. Another alternative for the "find two irrationals whose sum is irrational" is to do root(2) + root(3) = root(2 + 3) = root(5) which is irrational : )
but this is not true.
@@asr2009 duh Mr. Obvious
Aha breaking the tower law of algebraic field extensions aren't we?
I would simply say √2 + √3 can't be rational, because it'd be a tad silly if it was, wouldn't it?
@@aravenbythenamealex Proof by "That'd be kinda weird"
This would make a great April Fools Video. I once wanted to do something like this for all of the questions on a test, but since I wanted to get a perfect grade at the end of the year, saner heads prevailed.
😂
At 2:23 this is actually a concept derived from multivariable calculus. It works if a function is composed of any number of functions of x like this:
df/dx = ðf/ða * da/dx + ðf/ðb * db/dx + ðf/ðc * dc/dx + ...
for example (although trivially):
f = x^6
a = x
b = x^2
c = x^3
f = abc
df/dx = bc * 1 + ac * 2x + ab * 3x^2 = x^2 * x^3 + x * x^3 * 2x + x * x^2 * 3x^2 = x^5 + 2x^5 + 3x^5 = 6x^5
df/dx = 6x^5
For the x^x example, it is almost correct, you can look at the function y^x and take the full derivative, then substitute y=x.
yeah that's what I want to say. It's actually another way of taking the two partial derivatives and apply the chain rule.
To be more rigorous, take a(x)^b(x), use multivariate chain rule to differentiate with respect to x. Then insert a(x)=b(x)=x.
Honestly I found the tricky answers more complivated than the real ones :D
Because they're not right lol
Please make part 2 this is too good!
0:40 Another way of doing this is to subtract each digit of each numerator and denominator from each other: 9/2 - 25/10 = (9 - 2 - 5)/(2 - 1 - 0) = 2/1 = 2
One of my students recommended your channel. Thank you 😊.
The subtracting fraction literally made me so happy! I can finally use common sense (although wrong) on a math problem and be right!
Oh lord that arccos being converted into si(x) and then converting to 6 got me good.
Awesome video! I was thinking of wrong ways to get the right answers a few days ago 😅
Glad you liked it!
The one at 4:46 is pretty easy of you recall cos(x)=sin(90°-x). Still, found that one hilarious
And this is why the ends don't justify the means.
“Not what I expected but that works too”
This math is making me cry.
professor: prove that *big complex formula* = *another big complex formula*
me: *multiplies both sides by 0, leaves*
Another funny example, To solve x^2 = 25 (positive solution) simply cancel out the 2s to get x = 5
I feel pain while watching this.
When your math teacher shoes this to you 💀
2:34, this is actually mathematically correct, and is just an example of the multivariate chain rule.
Fair enough :)
Really? I was just taught this as a trick
@@anshumanagrawal346 If you have, u and v being functions of x and f being a function of u and v, then the multivariable chain rule states:
df/dx = ∂f/∂u du/dx + ∂f/∂v dv/dx
(The actual version would have u1,u2,....un functions, but let's just go with the smaller case.)
Now, call f(u,v)=u^v with u(x)=x,v(x)=x:
df/dx = vu^(v-1)•1+u^v ln(v)•1
which is essentially what you write in the original statement - take one x constant, differentiate with the other and do the same for other x. This applies well and good to all such applications. Obviously there has to be a reason when a trick works everywhere. ;)
@@nikhilnagaria2672 Wow, thank you so much, I have wondered for a long time about this cause this same logic is followed by so many derivative rules that I highly suspected there was some such rule but I couldn't find it anywhere online as I didn't have the right keywords, I even asked this question in a comment on a video on this channel a while ago 😂
Like when I first learned this "trick" from my teacher that for differentiation of the functions of the type f^g, to take the derivative pretend one is a constant and the other is the variable, and then vice versa and add the results, after I proved this for any generic f and g using logarithmic differentiation, after I while I realised it also applies to the product rule, the division rule, and (somewhat trivially) to the sum rule as well
I just found this video after my math exam earlier, the algorithm hates me.
5:00
this is the best solution ive ever seen
I’ve met the masterpiece before it go viral.
In the 64/16 example, you can actually further cancel out the 1s to get ∠
genius
18/2
8+1=9
This started my entire grade 4 math class trying to use the digits in every problem to find the answer
Now my brain hurts.
This is extremely cursed
5:00 had me laughing
Same 😆
I plugged it into a calculator just to make sure.
That is the video what I want. Thanks!
Glad to hear it!
2:23
This could be a legit math tric:
Lets say f(a, b) = a^b
And a(x) = x and b(x) = x
(partial f/partial a) = b * a^(b-1)
(partial f/partial b) = a^b * ln(b)
da/dx = 1
db/dx = 1
Now apply the multivariable chain rule:
df/dx
= (partial f/partial a) * da/dx + (partial f/partial b) * db/dx
= b * a^(b-1) * 1 + a^b * ln(b) * 1
Now we substitute a=x and b=x to get:
df/dx
= x * x^(x-1) + x^x * ln(x)
= x^x + x^x * ln(x)
By (partial f/partial a) I mean the partial derivative.
These are very funny. Thanks Bri!
0:06 no, 4's vertical part cancels with 1, leaving something like rotated V, and V is 5.
"Use the fraction bar as a negative" hahahaha
mother of newton
"Cancel the n from (sin x)÷n = six = 6"
The 3 × 9 one kind of works. You divided 81, or 9^2, by 3, so it's basically saying (a^2)/b = (a)(b), where a and b are real numbers. Simplifying, and if a is not 0, then a = b^2. So as long as this is true, it will work
The way I thought of that one was 3 x 9 = 3 x 3^2 = 3^3 and 81/3 is just 3^4/3 = 3^3 = 27, but I definitely didn't think of general cases as well
Congrats on 100K subscribers Bri! 😊
Thanks so much 😊
This is painful to watch, but also damn entertaining xD
The 3*9 part was hilarious 😂🤣🤣🤣
It's not every day that a math video makes me laugh out loud, but this one did. Congrats.
I really shouldn't be watching this a day before the math test😂
Now the next challenge for the viewers is to find a general solution (for most of the examples) for values where the incorrect method will get the right answer
dude this is just mindblowing you are an awesome mathematician 🥶🥶🥶🥶🥶
3:06 This is actually a great way to remember the derivative of x^x :D
This is also somehow a mathematically correct derivation.
Partial derivative rules can be applied to multiple instances of the same variable, not just different variables.( Think about it. If it works for every single possible function other than x, why would it not work for x too. Meta-calculus. )
Just pretend the other xs are (different) constants and add the results up.
D x^2 =
D xx =
D _x + -> _
D x_ = -> _
= 2_ = 2x
D xsinx =
D x_ + -> _
D _sinx = -> _cosx
= sinx + xcosx
D x^(xsinx) =
D x^ _ + -> _ x^( _ -1)
D _ ^($sinx) + -> $log _ cosx _ ^($sinx)
D _ ^(x$) = -> $log _ _ ^($x)
= (xsinx)x^(xsinx-1) + xlogx cosx x^(xsinx) + sinx logx x^(xsinx)
which is a monstrosity but does simplify to the correct solution of
x^(xsinx)(sinx+logx)(sinx+xcosx)
The answer for the derivative of x^x follows from the chain rule for a function of two variables. If f(x,y)=x^x, then d/dx f(x,x) = f_x (x,x) +f_y(x,x) = x * x^*(x-1) + (ln x) x^x
That's what I thought, isn't it just the chain rule?
@@DennisComella That's right. It's the chain rule for multi-variable functions.
I came here not understanding how to do math correctly. Now I don't even understand how to do it incorrectly.
This man is so good at math that he can get the right answer by getting the wrong answer
I NEED MORE
another one that works in only one case involves square roots :
✔(2 2/3) = 2✔(2/3)
Actually the power rule and the exponent rule are just specific cases for a general rule of derivatives of f(x)^g(x), where you “add the two rules” together. Because the derivative of a constant is 0, the power rule and the exponent rule will cancel out one half of the equation or the other, giving their respective rules
Some of these could be useful as mnemonics, if taught correctly
My teacher went mad after watching this
4:34 I like how you totally didn't have to cancel out the 90s, because it's still something over itself, you just did it to piss us off
the radical long division one works for all x * x^2 because you’re really solving for x^3 which is x^4/x. and the easiest way to get to x^4 is x^2^2
Always make enough mistakes so that they cancel each other out
🤣
There we have, a principle of statistics..lol
1:21
multiples of 9 for me was always the easiest for its reoccurring pattern:
(10*n + 9 - n) with n is 0 to 9
1 x 9 = (10*0 + 9 - 0) = 9
2 x 9 = (10*1 + 9 - 1) = 18
.
.
.
10 x 9 = (10*9 + 9 - 9) = 90
2:40 actually isn't so off. It's basically partial derivatives.
This was funny because it’s true.
I miss seeing you in your videos, one of the best looking MathTubers around.
Keep up the great work.
I appreciate that! Thanks so much for watching!
Haha 😹 ur absolutely right 😌
He does a great job making higher level math very approachable for all. We could use someone like him for the lower levels.
5:43 this is a direct result of the fact that 1+2+3=1x2x3. The same is true of any collection of numbers which have equal sum and product. ln(1+1+1+3+3) = ln(1)+ln(1)+ln(1)+ln(3)+ln(3)
Beautiful! Absolutely beautiful! (Also I made my teacher cry)
These are like math puns
2:23 It’s actually CORRECT, because x^x can be written as f(u,v), where f(u,v)=u^v, u=x, v=x.
df/dx = ∂f/∂u ∙ du/dx + ∂f/∂v ∙ dv/dx
=x∙x^(x-1)∙1 + x^x∙ln(x)∙1
=x^x∙(ln(x)+1)
To get d/(dx) (2x)
Cancel out d to get 2x/x
Cancel out x to get 2
I AM LITERALLY CRYING RIGHT NOW. PLEASE MAKE IT STOP.
I'm not a teacher, I'm a student, yet I got very upset at this video!! Great work, I won't be looking at this one again though.
😂
This makes my head hurt so much I started laughing
I showed EVERY SINGLE PROOF to my math teacher (she was shocked as hell).
1:44, this is interesting and can be proven algebraically
Another favourite of mine is x^2=25. You just cancel out the 2s
What about the ^
@@Firefly256 that's not there of course when you write it down. It just means "raised to the power of", it's quite common on UA-cam since its formula writing tools aren't the greatest here.
@@vit.budina ², copy it
It is incomplete, because it is missing the x = -5 solution then.
@@Firefly256Depending on if you write x² or x^2, you get 5 (in the former case) or -5 (in the latter case, since the ^ unfolds into the -), so those are both correct
2:40 I really doubt someone with this level of mathematics trying to calculate derivatives
Task failed successfully
This is a nightmare for every mathematician.
and at 5:03, it was so funny that you just cancel out the n's and ended up with the word six🤣🤣🤣🤣🤣🤣
As someone who's gone through Calculus 3, watch this video is rage inducing.
This video made me think irrationally 😂
i was one of those people who allways used wrong rules and got the right awnser
2:59 i did this in my maths class today although it was wrong but my teacher corrected me
I couldn't stop my laughtr for entire video as my laughter also watches your yt channel except it is shy.
Yeahhhh 100k! Congrats dude
Thanks so much!!
Great way to remember what the derivative of x^x looks like
"If you write things that equal between equal signs, I will mark you correct."
This is the math i love
im showing this to my math teacher and you can't stop me
the sin x over n being six was the most clever, it literally relies on spelling, and in an alternate universe this method wouldn't work
My fav is cancelling out the “n” in sin
Still I am horrible in math.
I hate it very much 😊
Crongrats for your 100K subs. Keep up the great work 👍
Thank you so much 😀
time to use these methods to defy the laws of maths in class
So what I’m hearing is that if I don’t know how to do a problem there’s a slight chance I can logic my way through them
I will store this vid for when I know the terms that I don't understand now.