*I hope you did enjoy today's video :) Even though the results of the video were rather trivial, it did still not take away from the surprise!* Other than that, here are all of today's relevant links Interested in more surprising mathematics and various other STEM topics? Why not try out Brilliant then? =D brilliant.org/FlammableMaths Life is Complex Merch! :D teespring.com/life-is-complex Better Quotient Rule: ruclips.net/video/oCuQIumw8zc/видео.html
@@integralboi2900 Yes, and this implies f'/f = 1/(1 - x) or, f = 0, the former of which implies that h = g, where g piecewise equals -ln(1 - x) + A when x < 1, -ln(x - 1) + B when x > 1, and h piecewise equals ln(-f) + C when f < 0, ln(f) + D when f > 0. If f < 0 and x < 1, then ln(-f) + C = -ln(1 - x) + A implies -f·exp(C) = 1/(1 - x)·exp(A), which implies f = K/(1 - x) with K < 0 and x < 1. If f < 0 and x > 1, then ln(-f) + C = -ln(x - 1) + B, which implies f = K/(x - 1) with K < 0 and x > 1. If f > 0 and x < 1, then by analogous argument, f = K/(1 - x) with K > 0 and x < 1. If f > 0 and x > 1, then by analogous argument, f = K/(x - 1), with K > 0 and x > 1. Combining all the possible cases for the most general solution, if f' = (x·f)', then f = K/(1 - x) with arbitrary real K for all nonzero x, which actually just forces the solution in the video.
Proof( done before watching the video) Well, the difference between the functions (1/1-x) and (x/1-x) is 1, a constant. So, on differentiating both sides and remembering that differentiation is additive, we get that the difference between the derivatives is equal to the derivative of 1, i.e, 0. So, the derivatives of the functions are equal. Quod erat demonstrandum. Anyhow, cool video!:)
One very trivial math fact that nevertheless surprised me: roll two dice with possibly differing number of sides (but with normal consecutive numbering starting with 1 -- this excludes for example, tens ten dice or the doubling cubes used in backgammon) The odds both dice show the same number depends only the number of sides on the die with the greater number of sides. The number of sides on the other die simply doesn't matter.
I need a proof or an explanation, i mean imagine two dice: 4 side and a 6 side. If i throw the 4 side first i have 1/6 chance to match, but if I throw the 6 first then I have 1/4 chance to match, did I miss something?
@@EvidLekan If you roll the six-sided first and it comes up 5 or 6, what chance does the four-sided have to match? Anyway, consider both dice rolled together. There are 4×6=24 possible rolls. Only four of those have both dice the same. That is 4/24 chance which reduces to 1/6.
If I roll the dice with the greater number of sides first (let’s call that m and the smaller number of sides is n), then there’s a n/m chance that I roll a number that exists on the dice with n sides. The probability that the n-sided dice matches the number on the m-sided dice, is 1/n. So there’s a 1/m chance that they match. If I roll the n-sided dice first, there’s a 1/m chance that the m-sided dice matched the number rolled on the n-sided dice. Adding those probabilities together, we get a 2/m chance of them matching. Is this correct?
I’ve shown students the same issue when integrating certain functions using different methods. For example, try integrating x/(x + 1)² by substitution and by parts.
Something similar happens with certain trig integrals. If you have an integral that can be solved in multiple different ways, there's a chance the results will look completely different, but are actually only off by a constant value.
Geometric series? Just say 1 = (1-x)/(1-x). Thus x/(1-x) +1 = (x+1-x)/(1-x) = 1/(1-x). Thus, the two functions differ by a constant (1) and must have the same derivative. I remember something similar happening once when using two different methods of integration I got the solutions ln|2x| + c and ln|x| +c and thinking something went wrong before realizing that I could use log rules to show that ln|2x| and ln|x| differ by a constant.
The chalkboard's counterweight is probably too heavy. Add some weight to the chalkboard, or if it is possible, remove some weight from the counterweight, then it'll stop moving.
just so you know two equation can result in the same derivative. it demonstrate this idea pretty well through antiderivative of that +C part of antiderivative. I dont see it is so hard to get tho. +1 is basically moving that whole graph up 1 unit vertically. it does not affect the shape of the graph, therefore same derivative
x/(1-x) = 1/(1-x) - 1 so the two functions are 1/(1-x) and 1/(1-x) - 1 since the other one just differs by a constant their derivative will be the same, right? Although the info used in the video is still pretty useful
Can't agree more with simple stuff being the most surprising. There are so many complicated ideas I'm perfectly fine with but it's always the simple stuff which slows me down
Ah yes I remember coming across this problem when I learnt logs too. Something like because the log(ax) = log(a)+log(x) and differentiating drops the log(a) because it’s a constant right?
I think it's a matter of actually subtracting them to find out if they differ by a constant, rather than looking at how the expressions seem to relate to each other.
Reminds me of having to come to terms with the fact that d/dx ln(2x) is still 1/x... I didn't like it but the derivative rule said it was so, but then I thought about it some more. ln(ax) = ln(a) + ln(x). d/dx ln(a) = 0, so it checks out.
The most easy way to see this, is the following: Shift the two functions by 1 into negative x-direction, then you get -1/x and -(x+1)/x= -1/x - 1. Since the second one is just shifted by 1 in negative y-direction, the derivatives are equal.
that means that f(x)=1/(1-x) holds the functional identity (x*f(x))'=f'(x). You can generalize this identity with any function g(x) such that (g(x)*f(x))'=f'(x)and get f(x)=1/(1-g(x)). That way you get some more suprising identities such ass (x^2/(1-x^2))'=(1/(1-x^2))' or even choose crazier fucntions g(x). Anyways this can be explained by adding 1 and substracting 1 the same way you did in the video, but I still find it interesting! very nice video
1/(1-x) - C = (1 - C(1-x))/(1-x) = (1 - C + Cx)/(1-x) Shifting 1/(1-x) down by 1 gives x/(1-x) Shifting 1/(1-x) down by 2 gives (-1 + 2x)/(1-x) Shifting 1/(1-x) up by 1 gives (2 - x)/(1-x) These are all vertical translations and thus preserve the derivative. When you had to use log tables at school because slide rules were not allowed, you multiplied two values by finding the logs of the two values, adding them and finding the antilog of the result. Therefore, it should not come as a surprise that [log(ax)]' = [log(x)]' Turns out there is another way to multiply two numbers using cosine tables: cos A cos B = (cos (A-B) + cos (A+B))/2 - you just need to scale down cos A and cos B first.
I mean two derivatives are the same if the functions are off only by a constant, in this case the constant is -1. Adding -(1-x)/(1-x) to the first equation gives the second, verifying this.
x/(1 - x) = (x - 1)/(1 - x) + 1/(1 - x) = -1 + 1/(1 - x). Therefore, if x |-> f(x) = x/(1 - x), x |-> g(x) = 1/(1 - x), x |-> h(x) = -1, then f = h + g, and since D[h] = 0 almost everywhere (in fact, in this particular case, actually everywhere), D[f] = D[g]. The fact that (log')(a·x) = (log')(x) is relevant in measure theory, because this idea gives rise to the idea of the Haar measure, which has the characteristic of being scale invariant.
But if you subtract the two functions you get 1? I dunno seems really trivial! ... Any rational fucntion y = (ax + b)/(cx + d )can, by division be written as a y axis translation of a function of the form e/cx + d...and thus will have the same gradient.
And by extension, given any polynomials f(x), g(x) with degree n, there exists a polynomial h(x) of degree at most n-1 such that the derivative of f(x)/g(x) and h(x)/g(x) are the same.
I was wondering about the value of x where all of this make sense and the domain of the fuction log with product property...for example the geometric series formula works for certain value of x and you can only use the product property of the log function if the numbers are positive, or am i wrong or missing something?
cotangent squared and cosecant squared are clearly different functions and *surprise* they have the same derivatives. :) Fun fact, the same is true for the hyperbolic versions of them. Now about the Riemann zeta function...
Lol i thought today the same thing that sec²x and tan²x have same derivatives(slopes), then i realized after few seconds, it's obvious 😂🤣🤣, what a coincidence and i saw this video now. Sry for being late, but honestly it was timed correct😂
I thought about it, and then it became clear to me: 1/(1-x)=(1-x)/(1-x) + x/(1-x)=1 + x/(1-x) So they are clearly having the same derrivative only using simple transformations and no Taylor Expansions
Pre-watch, from the thumbnail: That's really neat!! Never would have dreamed this up, but once you write it down & stare at it for a bit, it becomes obvious. Here's how. Claim: f'(x) = g'(x), where f(x) = 1/(1-x) and g(x) = x/(1-x) Well that claim is equivalent to f'(x) - g'(x) = 0; (f(x) - g(x))' = 0 But f(x) - g(x) = 1/(1-x) - x/(1-x) = (1-x)/(1-x) = 1 And differentiating both sides gives the result. f'(x) - g'(x) = 0 PS. Like your shirt says, life is antisymmetric. Fred
Lol I actually encountered something like this myself! I was dealing with trigonometric functions and two 'different' functions seemed to have the same derivative. I later realized that the two functions just differed by half.
All you have to remember is to add (or subtract) your constant of integration in front (or back) of your function and you can make any number of the derivatives of the functions equal to each other. Happy derivable derivations day. Have a Flammy flam-a-boi-le math day.
Thank you for this video. It is Not trivial but perhaps elementary. How do you integrate 1/x when you do not yet know about the number e? Historically that is called the quadrature (squaring) of the hyperbola.
Reminds me of the doubly infinite series: ...+x^3+x^2+x+1+x^-1+x^-2+x^-3+... = (1+x+x^2+...) + (x^-1+x^-2+...) = 1/(1-x) + x^-1/(1-x^-1) = 1/(1-x) + 1/(x-1) = 1/(1-x) - 1/(1-x) = 0 So all integer powers of any number add to zero!? I believe attributed to Euler (who else). Not too difficult to spot the error, but funny.
Just curious, how did you happen to come across this problem? I had to do this exact problem a few days ago, and did a double take when I saw the thumbnail.
I never would have come to the conclusion just by eyeballing the expression that x/(1-x) = 1/(1-x) - 1 but it is. Here's proof: x/(1-x) = x/(-x + 1) = x/-(x -1) = ((x-1)+1)/-(x-1) = -(1 + 1/(x-1)) = - 1 + 1/-(x-1) = -1 + 1/(1 + (-x)) = 1/(1-x) -1
Although it was trivial, and easy to see that the trick fir the proof was to add -1+1 in the numerator, thisbwas still a cool video. Trivial is truly not the opposite to interesting
If i was confronted with whether two simple functions have the same derivative, my first thought might have been to, i don't know, compute both derivatives.
I dont get the remember part at the beginning. Why (f/g)' = f'/g' ? This is obviously not true since this would mean that the derivative of x = x/1 would be 1/0 and not 1 ?! Or am i missing something?
Often identities involving derivatives are only true when all functions involved are continuous and can be resolved. As you've produced a divide by 0 error the expression may no longer hold. This function y=1/0 doesn't have any conditions under which it is true - it can't be resolved (you can't graph it). The reason for this is that when these identities are found, it is assumed that derivatives act like fractions, which if you want to be mathematically rigorous is not the case - specifically, it's not the case when not all of the functions involved are continuous or can be resolved. There may be other cases where it isn't true but I don't know what they are if there are any. This is my best guess based on my current understanding, it may not be applicable to this specific identity, or may even be completely wrong. Alternatively, I have found several situations where you can end up cancelling terms in a problem and getting something like 1=1/0 or 2=1/0 etc, 0 just doesn't behave well in maths. This may simply be another case of 0 doing weird things that don't make sense. essentially it just means that whatever you did isn't valid - not because you did it wrong, but simply because doing it lead to a divide by 0 error.
The part about the geometric series confused me greatly. If you're approaching zero maybe they get close but not if you're approaching positive infinity..
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
![The Most Useful Curve in Mathematics [Logarithms]](/img/1.gif)
*I hope you did enjoy today's video :) Even though the results of the video were rather trivial, it did still not take away from the surprise!* Other than that, here are all of today's relevant links
Interested in more surprising mathematics and various other STEM topics? Why not try out Brilliant then? =D brilliant.org/FlammableMaths
Life is Complex Merch! :D teespring.com/life-is-complex
Better Quotient Rule: ruclips.net/video/oCuQIumw8zc/видео.html
Love from india bro
Yeah no - completely trivial, I could do that blindfolded.
*gulps
@@integralboi2900 Yes, and this implies f'/f = 1/(1 - x) or, f = 0, the former of which implies that h = g, where g piecewise equals -ln(1 - x) + A when x < 1, -ln(x - 1) + B when x > 1, and h piecewise equals ln(-f) + C when f < 0, ln(f) + D when f > 0.
If f < 0 and x < 1, then ln(-f) + C = -ln(1 - x) + A implies -f·exp(C) = 1/(1 - x)·exp(A), which implies f = K/(1 - x) with K < 0 and x < 1.
If f < 0 and x > 1, then ln(-f) + C = -ln(x - 1) + B, which implies f = K/(x - 1) with K < 0 and x > 1.
If f > 0 and x < 1, then by analogous argument, f = K/(1 - x) with K > 0 and x < 1.
If f > 0 and x > 1, then by analogous argument, f = K/(x - 1), with K > 0 and x > 1.
Combining all the possible cases for the most general solution, if f' = (x·f)', then f = K/(1 - x) with arbitrary real K for all nonzero x, which actually just forces the solution in the video.
In general, if f'=(xf)', then f=(1-x)f'
(See the comment above for an explicit answer, I also accidentally deleted my old comment)
Noce
Papa flammy’s chalkboards are the closest humanity will ever come to a frictionless surface
I hate these chalkboards that won’t stay still. Honestly I’m going to have to stop watching these videos for no other reason.
then u realize that chalks work with friction :D
U cannot write without friction
Me, a physics guy:
What is a friction?
x/(1-x)=1/(1-x) - 1
So that why they have the same derivative
I mean you see it right away, 14 minutes for that...
thats what i was also wondering, is obvious
that's also what I thought XD
just add a 0
@Larbitos O_o
0=1-1
x/(1-x)
= (x+0)/(1-x)
= (x+1-1)/(1-x)
=-(1-1-x)/(1-x)
=-(-1)/(1-x) + (1-x)/(1-x)
=1/(1-x) +1
Papa Flammy: "Never forget the plus C"
Also Papa Flammy:
🤡
@@sarthak8802 shave the stache bro
Proof( done before watching the video)
Well, the difference between the functions (1/1-x) and (x/1-x) is 1, a constant. So, on differentiating both sides and remembering that differentiation is additive, we get that the difference between the derivatives is equal to the derivative of 1, i.e, 0. So, the derivatives of the functions are equal.
Quod erat demonstrandum.
Anyhow, cool video!:)
I was about to comment the same. But I guess Euler did everything first and took the credit.
@@sahilbaori9052 😂😂
btw it felt really obvious to me that they just differ by 1. Idk why was he so confused with that 😂
@Cookie Monster Well, I wanted to leave something to keep young Gauss occupied. 😉
I'm only not liking this because the number of likes is currently NICE
One very trivial math fact that nevertheless surprised me: roll two dice with possibly differing number of sides (but with normal consecutive numbering starting with 1 -- this excludes for example, tens ten dice or the doubling cubes used in backgammon) The odds both dice show the same number depends only the number of sides on the die with the greater number of sides. The number of sides on the other die simply doesn't matter.
I agree it sounds wrong, but it only takes a few seconds to realize why it's right.
I need a proof or an explanation, i mean imagine two dice: 4 side and a 6 side. If i throw the 4 side first i have 1/6 chance to match, but if I throw the 6 first then I have 1/4 chance to match, did I miss something?
@@EvidLekan If you roll the six-sided first and it comes up 5 or 6, what chance does the four-sided have to match?
Anyway, consider both dice rolled together. There are 4×6=24 possible rolls. Only four of those have both dice the same. That is 4/24 chance which reduces to 1/6.
If I roll the dice with the greater number of sides first (let’s call that m and the smaller number of sides is n), then there’s a n/m chance that I roll a number that exists on the dice with n sides. The probability that the n-sided dice matches the number on the m-sided dice, is 1/n. So there’s a 1/m chance that they match. If I roll the n-sided dice first, there’s a 1/m chance that the m-sided dice matched the number rolled on the n-sided dice. Adding those probabilities together, we get a 2/m chance of them matching.
Is this correct?
This has the same energy as "57 is not prime"
true, lol 😂
Lol
Grothendieck would like to know your location
i mean it only takes a second as it's 60-3
5+7=12 which is divisible by 3.
Where do you get off
He is 4 quintillion universes ahead of us.
Put this guy in Papas basement back
Get a load of THIS guy
Load get a of guy this
He gets off at t’
And that's why kids you should never forget ur +C
I’ve shown students the same issue when integrating certain functions using different methods. For example, try integrating x/(x + 1)² by substitution and by parts.
How I looked at it is if you do long division for 1/(1-x), you get 1+x/(1-x). This shows that they are equivalent after being shifted by a constant.
Something similar happens with certain trig integrals. If you have an integral that can be solved in multiple different ways, there's a chance the results will look completely different, but are actually only off by a constant value.
Geometric series? Just say 1 = (1-x)/(1-x). Thus x/(1-x) +1 = (x+1-x)/(1-x) = 1/(1-x). Thus, the two functions differ by a constant (1) and must have the same derivative.
I remember something similar happening once when using two different methods of integration I got the solutions ln|2x| + c and ln|x| +c and thinking something went wrong before realizing that I could use log rules to show that ln|2x| and ln|x| differ by a constant.
The chalkboard's counterweight is probably too heavy. Add some weight to the chalkboard, or if it is possible, remove some weight from the counterweight, then it'll stop moving.
I know, but it's not that easy, since it's fixed to the ground 3:
just so you know two equation can result in the same derivative. it demonstrate this idea pretty well through antiderivative of that +C part of antiderivative. I dont see it is so hard to get tho. +1 is basically moving that whole graph up 1 unit vertically. it does not affect the shape of the graph, therefore same derivative
3:46 experimental mathematicians be like: ...
x/(1-x) = 1/(1-x) - 1
so the two functions are
1/(1-x) and 1/(1-x) - 1
since the other one just differs by a constant their derivative will be the same, right?
Although the info used in the video is still pretty useful
Yeah this surprised me too. Really like the videos where you share random surprising maths stuff. Nice vid papa :)
Uhh what about outside the radius of convergence of the geometric bitch tho
Can't agree more with simple stuff being the most surprising. There are so many complicated ideas I'm perfectly fine with but it's always the simple stuff which slows me down
I have to admit I had a similar confusion when I realised d/dx(log(a x))=1/x for all a. I had to double check that before it sank in.
Ah yes I remember coming across this problem when I learnt logs too. Something like because the log(ax) = log(a)+log(x) and differentiating drops the log(a) because it’s a constant right?
I think it's a matter of actually subtracting them to find out if they differ by a constant, rather than looking at how the expressions seem to relate to each other.
"It took me a second"... Since you had to derive several times, you are very fast!
loved this vid, it always puts a smile on my face when things like that happen and it always keeps my fascination sparked with mathematics!
Reminds me of having to come to terms with the fact that d/dx ln(2x) is still 1/x...
I didn't like it but the derivative rule said it was so, but then I thought about it some more. ln(ax) = ln(a) + ln(x). d/dx ln(a) = 0, so it checks out.
I was also confused with ln(ax)-ln(bx)=ln(a)-ln(b) for any strictly positive a, b and x.
Oh god, what a mind fuck LOL!
Get a load of THIS guy
I was surprised when I learned that (-b±sqrt(b^2-4ac))/(2a) is equivalent to (2c)/(-b∓sqrt(b^2-4ac)) when solving quadratic equation!
The first one is pretty easy. Just subtract the functions. 1/(1-x)-x/(1-x)=(1-x)/(1-x)=1 for x≠1. Still suprising.
Feeling first with a hundred others .
Probably thousands now
Only 621 views
@@nihar1206 dang it
The most easy way to see this, is the following: Shift the two functions by 1 into negative x-direction, then you get -1/x and -(x+1)/x= -1/x - 1. Since the second one is just shifted by 1 in negative y-direction, the derivatives are equal.
An ex-apple, it's no more, it's shuffled off its mortal strudel, it's pining for the orchards
that means that f(x)=1/(1-x) holds the functional identity (x*f(x))'=f'(x).
You can generalize this identity with any function g(x) such that (g(x)*f(x))'=f'(x)and get f(x)=1/(1-g(x)).
That way you get some more suprising identities such ass (x^2/(1-x^2))'=(1/(1-x^2))' or even choose crazier fucntions g(x).
Anyways this can be explained by adding 1 and substracting 1 the same way you did in the video, but I still find it interesting!
very nice video
If you do polinom by polinom division: x / ( -x+1 ) you get that it is: -1 + 1/(1-x)
Looks like police has finally found flammable as he ass giving out too many hearts
1/(1-x) - C = (1 - C(1-x))/(1-x) = (1 - C + Cx)/(1-x)
Shifting 1/(1-x) down by 1 gives x/(1-x)
Shifting 1/(1-x) down by 2 gives (-1 + 2x)/(1-x)
Shifting 1/(1-x) up by 1 gives (2 - x)/(1-x)
These are all vertical translations and thus preserve the derivative.
When you had to use log tables at school because slide rules were not allowed, you multiplied two values by finding the logs of the two values, adding them and finding the antilog of the result. Therefore, it should not come as a surprise that [log(ax)]' = [log(x)]'
Turns out there is another way to multiply two numbers using cosine tables:
cos A cos B = (cos (A-B) + cos (A+B))/2
- you just need to scale down cos A and cos B first.
I love little things like this where it just looks wrong, but turns out to be quite clear.
Life is a matrix of a, -b, b and, a
That is so accurate....
I really liked your little rant at the beginning, it was really interesting :)
God damn it, years and years of maths, I never gave this a proper thought! Thanks for opening my eyes papa flammy.
DAMN i saw the thumbnail and checked for myself and it was true, so i had to click. that's a sick fact bro
Is this related to derivatives of arctan and arccot on different ranges?
I like this guy.
This really is flammable maths, my brain is on fire after this that is
I mean two derivatives are the same if the functions are off only by a constant, in this case the constant is -1. Adding -(1-x)/(1-x) to the first equation gives the second, verifying this.
Very interesting fact is that either 1/(1-x) or x/(1-x) generates the same finite group under nesting (no multiplicative neither additive group).
Your pronunciation of French is spot-on : "Legendre".
LIFE IS *MATRIX/COMPLEX*
x/(1 - x) = (x - 1)/(1 - x) + 1/(1 - x) = -1 + 1/(1 - x). Therefore, if x |-> f(x) = x/(1 - x), x |-> g(x) = 1/(1 - x), x |-> h(x) = -1, then f = h + g, and since D[h] = 0 almost everywhere (in fact, in this particular case, actually everywhere), D[f] = D[g].
The fact that (log')(a·x) = (log')(x) is relevant in measure theory, because this idea gives rise to the idea of the Haar measure, which has the characteristic of being scale invariant.
But if you subtract the two functions you get 1? I dunno seems really trivial! ... Any rational fucntion y = (ax + b)/(cx + d )can, by division be written as a y axis translation of a function of the form e/cx + d...and thus will have the same gradient.
And by extension, given any polynomials f(x), g(x) with degree n, there exists a polynomial h(x) of degree at most n-1 such that the derivative of f(x)/g(x) and h(x)/g(x) are the same.
0:04 what is what the meme said? is too small to read :(
So basically every poll had an option that got 69%
So x is just a constant in disguise. Pretty neat
You were so red to today, but still loved the video. Sometimes trivial things are so amazing because they have been sitting under our nose for so long
I was wondering about the value of x where all of this make sense and the domain of the fuction log with product property...for example the geometric series formula works for certain value of x and you can only use the product property of the log function if the numbers are positive, or am i wrong or missing something?
11:03 your chalkboard had erectile dysfunction? Brilliant
the x's don't cancel at 2:23, because x/(y/z) != (x/y)/z, and your situation is the latter case not the former
cotangent squared and cosecant squared are clearly different functions and *surprise* they have the same derivatives. :) Fun fact, the same is true for the hyperbolic versions of them. Now about the Riemann zeta function...
Great video
Lol i thought today the same thing that sec²x and tan²x have same derivatives(slopes), then i realized after few seconds, it's obvious 😂🤣🤣, what a coincidence and i saw this video now. Sry for being late, but honestly it was timed correct😂
Damn, you are right! :0
I thought about it, and then it became clear to me:
1/(1-x)=(1-x)/(1-x) + x/(1-x)=1 + x/(1-x)
So they are clearly having the same derrivative only using simple transformations and no Taylor Expansions
Maybe all our intuition is wrong after all and your chalkboard never goes up but we just think it does.
When he lifts his chalkboard, in fact the entire universe is moving down relative to it, and the chalkboard remains stationary.
@@seandimmock5813 true it just depends on the perspective
@@seandimmock5813 isn't that just gravity but earth not the universe
Pre-watch, from the thumbnail:
That's really neat!! Never would have dreamed this up, but once you write it down & stare at it for a bit, it becomes obvious. Here's how.
Claim: f'(x) = g'(x), where f(x) = 1/(1-x) and g(x) = x/(1-x)
Well that claim is equivalent to
f'(x) - g'(x) = 0; (f(x) - g(x))' = 0
But f(x) - g(x) = 1/(1-x) - x/(1-x) = (1-x)/(1-x) = 1
And differentiating both sides gives the result.
f'(x) - g'(x) = 0
PS. Like your shirt says, life is antisymmetric.
Fred
Hehe :D Glad you learned something new Fred! :)
Hey man, math never ceases to amaze! Great work!
i miss when randoms would walk in on your videos lol. Now you have both magic chalk and a magic chalkboard. I'm liking the progression
Lol
I actually encountered something like this myself! I was dealing with trigonometric functions and two 'different' functions seemed to have the same derivative. I later realized that the two functions just differed by half.
If y make the integral of 1/(1-x)^2, y obtain 1/(1-x) + c. But how to obtain x/(1-x) + c ?
All you have to remember is to add (or subtract) your constant of integration in front (or back) of your function and you can make any number of the derivatives of the functions equal to each other. Happy derivable derivations day.
Have a Flammy flam-a-boi-le math day.
If f(X)=Y=X/(X-1) this is the same as XY=X+Y.
Also f(x) is an involution. (meaning it is own inverse!)
(1/(1-x)) -1 = (x/(1-x)), and so if you differentiate, the constant goes away, and the derivatives would be the same
Thank you for this video. It is Not trivial but perhaps elementary. How do you integrate 1/x when you do not yet know about the number e? Historically that is called the quadrature (squaring) of the hyperbola.
Reminds me of the doubly infinite series:
...+x^3+x^2+x+1+x^-1+x^-2+x^-3+... =
(1+x+x^2+...) + (x^-1+x^-2+...) =
1/(1-x) + x^-1/(1-x^-1) =
1/(1-x) + 1/(x-1) =
1/(1-x) - 1/(1-x) =
0
So all integer powers of any number add to zero!? I believe attributed to Euler (who else).
Not too difficult to spot the error, but funny.
How do I get on your level?
Just curious, how did you happen to come across this problem? I had to do this exact problem a few days ago, and did a double take when I saw the thumbnail.
I think it is a pretty common problem, often used as an example. I had the exact same problem in university some years ago.
What do you say in the intro? “Welcome back to navieo” ?
“Welcome back to another video”
Awesome video, math boi!
For f not equal to g, ..., oh, never mind, Papi Flami covered that. :-|
I never would have come to the conclusion just by eyeballing the expression that x/(1-x) = 1/(1-x) - 1 but it is. Here's proof:
x/(1-x) = x/(-x + 1)
= x/-(x -1)
= ((x-1)+1)/-(x-1)
= -(1 + 1/(x-1))
= - 1 + 1/-(x-1)
= -1 + 1/(1 + (-x))
= 1/(1-x) -1
I FINALLY UNDERSTAND CALCULUS AND I UNDERSTAND YOUR VIDEOS!!!!!
1/(1-x) = x/(1-x)+1 now take the derivative of both sides and you get d/dx(1/(1-x)) = d/dx(x/1-x) + d/dx(1) = d/dx(x/1-x)) Proof complete :D
Amazing video. Simple fun!
Although it was trivial, and easy to see that the trick fir the proof was to add -1+1 in the numerator, thisbwas still a cool video. Trivial is truly not the opposite to interesting
It is kinda obvious when you think about it.
1 = (1-x)/(1-x) = 1/(1-x) - x/(1-x)
Therefore
1 + x/(1-x) = 1/(1-x)
This was nice, I miss highschool math.
Corporates need you to find the differences between these 2 functions.
*Looks at their derivatives*
They're the same
If i was confronted with whether two simple functions have the same derivative, my first thought might have been to, i don't know, compute both derivatives.
I dont get the remember part at the beginning. Why (f/g)' = f'/g' ? This is obviously not true since this would mean that the derivative of x = x/1 would be 1/0 and not 1 ?! Or am i missing something?
Often identities involving derivatives are only true when all functions involved are continuous and can be resolved. As you've produced a divide by 0 error the expression may no longer hold. This function y=1/0 doesn't have any conditions under which it is true - it can't be resolved (you can't graph it).
The reason for this is that when these identities are found, it is assumed that derivatives act like fractions, which if you want to be mathematically rigorous is not the case - specifically, it's not the case when not all of the functions involved are continuous or can be resolved. There may be other cases where it isn't true but I don't know what they are if there are any.
This is my best guess based on my current understanding, it may not be applicable to this specific identity, or may even be completely wrong. Alternatively, I have found several situations where you can end up cancelling terms in a problem and getting something like 1=1/0 or 2=1/0 etc, 0 just doesn't behave well in maths. This may simply be another case of 0 doing weird things that don't make sense. essentially it just means that whatever you did isn't valid - not because you did it wrong, but simply because doing it lead to a divide by 0 error.
It isn’t a statement that it hold for every f and g, the goal was to found f and g for which the equation hold.
This is just the same as f(x) and f(x)+C have the same derivative.
the derivative mapping
d/dx: {[f]|f: K → K} → {f: K → K} with the
equivalence relation
g ∈ [f] ⇔ ∃ c ∈ K: f(x) = c + g(x) for all x ∈ K
is injective
Good sh!t brodah 😳💥💯
Damn thats pretty sneaky
The part about the geometric series confused me greatly. If you're approaching zero maybe they get close but not if you're approaching positive infinity..
Life is
MATRIX
Just cancel the dominators so 1¹=x¹ therefore x=1 😎😎
Please explain the t-shirt 🥺
@Pranoy Kumar oh ya, it can be. THANKS
It looks like an orthogonal matrix or a rotation matrix
The only thing I can think of is 'life is a cycle'
@@holomurphy22 I initially thought it meant c^2, cuz a^2 + b^2..... ya know the guy called Pithogaros right😅
@@sbstuff9696 Lol its still somehow related to it though ;)
Life is complex, because C is isomorphic to X, with x being 2x2 matrices of that form.
Please review JEE mains math Qns
Just do this: 1/(1-x) - x/(1_x) = (1-x)/(1-x) =1 . So they indeed are apart by a conatant
Papa,how to get good at math?
I honestly pity folks that don't see the beauty. Of course, they feel sorry for me, because, IMHO, poetry is largely over-blown fluff.
A TRIVIAL SURPRISE!
Really surprising!🤩
No, not really. They differ by a constant (1) and so have the same derivative.
Life is... orthogonal?
Complex
X/(1-x) is the same as (x-1)/(1-x)+1/(1-x). So basically just translation
Where are you from ???
We can say : 1' = 0, so ((1-x)/(1-x))'=0 + and finaly (1/(1-x))' - (x/(1-x))'=0. Done