Proof that 0^0=0/0=0 (if we want to define it in any way) lets assume 0^0=a (a is any number) and 0/0=b then 0^0=0^(1-1)=0^1/0^1=0/0. now 0/0=b =>b=0/0=(0+0)/0=0/0+0/0=b+b =>b=0 =>0/0=0^0=0.
3 месяца назад
@@neutronenstern. Of course it would be more natural to make the quotient of to equal magnitudes equal to 1.
Substitute x = u² and it becomes a "double-sided" limit in u (although it's a bit of cheating; the 0s in the indeterminate form "0^0" never become double-sided).
@@Happy_Abe But is that continuous at x = 0? I think the u^2 form is, but I have doubts about the ¦x¦ form (haven't thought about it deeply - just 'skin reaction').
@@dlevi67 I just entered both variants into desmos. It appears that both can be "patched" to become continuous by defining f(0) = 0 , but neither is _differentiable_ at 0 .
@@ValidatingUsername I mean, other things raised to the power of zero are equal to one because no multiplications have taken place, and a product of no factors is just equal to the multiplicative identity (1) for the same reason that a sum of no terms is equal to the additive identity (0). Sure, you're doing repeated multiplication with zeroes, but by raising it to the power of zero, there aren't actually any zeroes being multiplied that can "get rid of" the implicit starting value of one.
@@angeld23 I’ve never heard of the additive identity but that’s hilarious. I know there are claims to have proofs for thinks like N^0 always equals 1 and I’ve memorized that and the reasons why but I don’t agree with it fundamentally, which also ties into my comment but also being meant as a kind of math humour.
Hey bprp! Remember the A level math exam you did a few months back? I wanted to ask you since you were not able to get to the last question of that exam, have you thought about doing that question on a separate video?
Sar how can any periodic function can be expressed as a superposition of sine and cosine function of different time periods with suitable coefficients this line is written in a book in oscillation chapter and they are clubing this statement with Simple Harmonic Motion eqn
I don't see the point. With the change of variables y = ln(ln(1/x)) → infinity or x = exp(- e^y), it is an example of the same very kind he started from: (exp(-e^y))^(1/y). One can simply take his original one (exp(-y^2))^(1/y), set x = (exp(-y^2), so y = √(ln(1/x)), and here you are: x^{1/√(ln(1/x))} →0 as x→0. I am glad if you see any fun here.
I think there's a problem when you apply De L'Hopital's rule, because the definition states that you can apply it if the limit of f(x) is equal to the limit of g(x) and is an indeterminate form, but in your case the limits do not approach the same value (different signs). There's also the problem that after applying De L'Hopital's rule the limit is not a Real value, contradicting the definition. Correct me if im wrong.
That's a very good observation. However, l'Hôpital's Rule doesn't require the top and bottom to be the same sign. It does also apply to the case where the top is going to infinity and the bottom is going to negative infinity (or vice-versa).
@@narfwhals7843 No, because that's not an indeterminate form. A limit of the form 0/infty always goes to 0, and a limit of the form infty/0+ always goes to infty. L'Hôpital's Rule applies to limits of the form 0/0 or ±infty/±infty; those are indeterminate forms that can go to any value, so they require L'Hôpital's Rule or some other tool to find out what value they go to.
I know you probably wouldn’t notice this but I got another problem I want you to solve! Prove that sqrt(1 + x^2) + sqrt(1 + x^2 + x * sqrt(3)) >= sqrt(3) geometrically. I got this problem in my geometry textbook in a paragraph about the law of cosine. Please help me solve it! 🙏 🙏
1:00 isn't A^-infinity equal to 1? -Infinity is 1/infinity, thus zero, this would make A^0 which is equal to one. I don't know if I'm the one who got it wrong please someone confirm If I'm right or wrong
@@yurenchu you have taken the limit of the derivative, which is the same as the derivative IF IT EXISTS. You can have a derivative without a limit of the derivative at that point. I calculated the actual derivative. Try to take the derivative of x^2 cos(1/x^2) at 0
So now we also have 1^(+infinity) --> 0 , because lim_{x--> 0⁺} f(x)^g(x) where f(x) = e^[-1/ln(-ln(x))] g(x) = -ln(x) apparently equals 0 . EDIT: Corrected mistake, by adding a minus-sign.
i plotted y=x^(x/0) on desmos and got a constant graph having y=0 but coudnt justify it... please explain what it is.. i also did it with e and got the same real graph
Strictly speaking, your function is undefined because it involves division by 0. However, Desmos is probably interpreting numbers divide by 0 as some kind of infinity. What I see when I plot this in Desmos is that the output is 0 everywhere on certain intervals. In particular, for x < -1 and for 0 < x < 1. I don't see any values for x > 1 or -1 < x < 0. If Desmos interprets a positive number divided by 0 to be positive infinity, then we can make sense of the right half of the graph. If x > 1, then we have a number greater than 1 raised to an "infinite" power, which blows up to infinity. So for all x > 1, we should get "infinity". Now, for 0 < x < 1, if you raise any of these numbers to the "infinity" power, you get 0, since numbers between 0 and 1 get smaller and smaller as you raise them to larger and larger powers. If Desmos interprets a negative number divided by 0 to be negative infinity, then we can make sense of the left half of the graph. Recall that a^(-b) = 1/(a^b). So essentially, we have x^(-"infinity") = 1/(x^"infinity"). If x < -1, then as you take higher and higher powers, the x^"infinity" will oscillate wildly but blast off to infinity and negative infinity. However, in either case, 1/("positive or negative infinity") is 0. On the other hand, for -1 < x < 0, we would get x^"infinity" should be 0 since larger and larger powers will make the overall number shrink. But 1/0 would blast off to infinity or negative infinity. So, again, technically your functions is undefined everywhere. But if we play fast and loose, and treat infinity and limits like numbers, we can make sense of the graph.
So basically 0^0 can approach 0 if the exponent grows much much more slowly than the denominator. You could actually make the limit anything you want no? Just need to choose functions that diminish appropriately.
If you're using real-valued functions with a positive base function, yes, you could make the limit any nonnegative number you want (though negative numbers as limits are impossible unless you allow for complex numbers). If your exponent function is positive and approaches 0 significantly slower than the base function, then you could end up with 0 or any number between 0 and 1. If your exponent function is negative and approaches 0 significantly slower than the base function, then you could end up with any number greater than 1, or positive infinity. And if the exponent function approaches 0 at the same rate or faster than the base function, then you get 1 as the limit.
I think you simply proved that 0^0 is indeterminate, and the limit will vary depending on how it's written. I mean, the limit as x goes to 0+ of x^0 is 1, and the limit as x goes to 0+ of 0^x = 0. You've shown us it could be e. In fact, you can make an expression and have the limit be any value you wish. I see wanting a defined value for this as a bit foolish, but math isn't supposed to be stodgy so if it's fun why not have fun?
0^0 is not indeterminate. "Indeterminate" is a term that we apply to specific forms of _limits_ . So if a _limit_ takes the form of 0^0 , then we say this limit form is indeterminate; we must determine the correct value of the limit by a more elaborate means. Other indeterminate limit forms are 0/0 and 1^infinity . However, in most (if not all) contexts outside limits, the expression 0^0 has the value of 1, just as in most contexts (outside limits) the expression 1^infinity has the value of 1 . (In contexts outside limits, the expression 0/0 is _undefined_ .)
Proof that 0^0=0/0=0 (if we want to define it in any way) lets assume 0^0=a (a is any number) and 0/0=b then 0^0=0^(1-1)=0^1/0^1=0/0. now 0/0=b =>b=0/0=(0+0)/0=0/0+0/0=b+b =>b=0 =>0/0=0^0=0.
Ok, let's accept that 0/0 = 0. If we insist that a/b + c/d = (ad + bc)/(bd) should still hold, then 1 = 1+0 = 1/1 + 0/0 = (1*0 +1*0)/(1*0) = 0/0 = 0. Boom, we get 1 = 0. How do you explain this?
Sorry, but that proof is wrong. Because x^(c-d) = (x^c)/(x^d) is not valid when x = 0 . Consequences of this invalid approach: x^(-3) = 1/(x^3) x^(-3) = x^(2-5) = (x^2)/(x^5) ==> 1/(x^3) = (x^2)/(x^5) Suppose x = 0 . Then lefthandside becomes 1/(x^3) = 1/(0^3) = 1/0 = ±infinity and righthandside becomes (x^2)/(x^5) = (0^2)/(0^5) = 0/0 = 0 (last step according to you reasoning) so that means that ±infinity = 0 ?
When we talk about 0 to 0, we want both zeros to be equal. Otherwise we can define a number of constants for this. lim x -> inf (a^-x)^(1/x) ≈ 1/a Take any a and you get a series of numbers. So technically this approach of taking different zeros is wrong.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Finally, 0^0 approaches 0 (after 6 years): ruclips.net/video/X65LEl7GFOw/видео.html
Proof that 0^0=0/0=0 (if we want to define it in any way)
lets assume 0^0=a (a is any number) and 0/0=b
then 0^0=0^(1-1)=0^1/0^1=0/0.
now
0/0=b
=>b=0/0=(0+0)/0=0/0+0/0=b+b
=>b=0
=>0/0=0^0=0.
@@neutronenstern. Of course it would be more natural to make the quotient of to equal magnitudes equal to 1.
@@neutronenstern. b+b=2b whatever b is.
this should be a yearly series, 1 episode every year!
Hmmm, I can try!
So... Big 0 means it reaches 0 the slowest
1/ln(-ln(x)) isn’t always defined. It only makes sense for 0
Substitute x = u² and it becomes a "double-sided" limit in u (although it's a bit of cheating; the 0s in the indeterminate form "0^0" never become double-sided).
@@yurenchu can use |x| at that point too.
@@Happy_Abe But is that continuous at x = 0? I think the u^2 form is, but I have doubts about the ¦x¦ form (haven't thought about it deeply - just 'skin reaction').
@@dlevi67 I just entered both variants into desmos. It appears that both can be "patched" to become continuous by defining f(0) = 0 , but neither is _differentiable_ at 0 .
@@yurenchu Thank you!
I think maths people and muggles have different thoughts on the word 'cute'
So cute proving nothing multiplied no times is nada.
@@ValidatingUsername I mean, other things raised to the power of zero are equal to one because no multiplications have taken place, and a product of no factors is just equal to the multiplicative identity (1) for the same reason that a sum of no terms is equal to the additive identity (0). Sure, you're doing repeated multiplication with zeroes, but by raising it to the power of zero, there aren't actually any zeroes being multiplied that can "get rid of" the implicit starting value of one.
@@angeld23 I’ve never heard of the additive identity but that’s hilarious.
I know there are claims to have proofs for thinks like N^0 always equals 1 and I’ve memorized that and the reasons why but I don’t agree with it fundamentally, which also ties into my comment but also being meant as a kind of math humour.
@@ValidatingUsernameSo funny to see a person thinking that rising to power is multiple multiplication. Tell me, please, how you compute, say, 2^√2.
@@angeld23Will I surprise you by saying that some things raised to the power of zero can be equal to any positive number?
I can get on board with freestyle mathematics, I hope to see the Olympics adopt this pioneering new sport. 9:17
Maximum cuteness. 😍
Hey bprp! Remember the A level math exam you did a few months back? I wanted to ask you since you were not able to get to the last question of that exam, have you thought about doing that question on a separate video?
All this work to turn the limit back into e^lnx :3
Ikr 😂
Congratulations! This is a beauty!
The GOAT 🐐
This was definitely "cute" in a mathemagical way :D
That was a great and thought-out approach
Great video. 0⁰ = 1. Thanks 😁
Sar how can any periodic function can be expressed as a superposition of sine and cosine function of different time periods with suitable coefficients this line is written in a book in oscillation chapter and they are clubing this statement with Simple Harmonic Motion eqn
Actually this is just another e^(-inf)=0.
Substitute y=ln(1/x), you get e^( -y/ln(y) ) -> 0 as y -> inf
True that. And like I said, my previous example was a bit more complicated. I like this one more! : )
Excellent. I was going to comment that, so I'm glad someone already did. It's an easier way to do the limit!
Sensational video.
you bring me hope
Damn I've always been wondering about this, but now I finally have an answer!
3:45 Is the inverse of the Ackermann function not technically the slowest?
Could you please compute the integral of [ a/(x^4 - a^2) ]dx
Can you create an equation that involves all of the trig functions?
Let me see what I can do!
that is indeed the cutest limit I have ever seen
This makes me so happy!
We can use the cube root of ln rather than nested logs to get a smaller 0 that still works.
It seems to be two sided in four complex directions. However, there could be directions where the limit is different from either side.
I'd like to see you and SyberMath debate what 0^0 is equal to. For the record, he still believes it's equal to 1 (if you're not taking the limit).
I don't think they would disagree. You can define 0^0 =1 in many contexts - just not all... and that's why in a general sense it's undefined.
I don't see the point. With the change of variables
y = ln(ln(1/x)) → infinity
or x = exp(- e^y), it is an example of the same very kind he started from: (exp(-e^y))^(1/y).
One can simply take his original one
(exp(-y^2))^(1/y),
set x = (exp(-y^2), so
y = √(ln(1/x)), and here you are:
x^{1/√(ln(1/x))} →0 as x→0.
I am glad if you see any fun here.
I think there's a problem when you apply De L'Hopital's rule, because the definition states that you can apply it if the limit of f(x) is equal to the limit of g(x) and is an indeterminate form, but in your case the limits do not approach the same value (different signs). There's also the problem that after applying De L'Hopital's rule the limit is not a Real value, contradicting the definition. Correct me if im wrong.
That's a very good observation. However, l'Hôpital's Rule doesn't require the top and bottom to be the same sign. It does also apply to the case where the top is going to infinity and the bottom is going to negative infinity (or vice-versa).
@justintroyka8855 does it also apply if one goes to 0 and the other to +- infinity?
@@narfwhals7843 No, because that's not an indeterminate form. A limit of the form 0/infty always goes to 0, and a limit of the form infty/0+ always goes to infty. L'Hôpital's Rule applies to limits of the form 0/0 or ±infty/±infty; those are indeterminate forms that can go to any value, so they require L'Hôpital's Rule or some other tool to find out what value they go to.
I thought it would be harder to solve, heh. Great one!
The key to having comparatively large zeroes is to draw the zero bigger!
I took a level further maths and maths so I can understand your videos
Now how about 0⁰→-1? Even if it's complex, there should be a simple one somewhere.
lim_{t --> ∞} [f(t)]^[g(t)]
where
f(t) = e^(-t)
g(t) = iπ/(t+1)
I know you probably wouldn’t notice this but I got another problem I want you to solve!
Prove that sqrt(1 + x^2) + sqrt(1 + x^2 + x * sqrt(3)) >= sqrt(3) geometrically.
I got this problem in my geometry textbook in a paragraph about the law of cosine.
Please help me solve it! 🙏 🙏
The graph of x^(1/ln(-ln(x))) is so weird, i don't understand it lol
Can you solve the indefinite integral of sqrt(1+sqrtx) pls?
If f,g are analytic near a with f(a)=g(a)=0 then the limit as z goes to a of f(z)^g(z) will always be 1
This is the cutest math ever 🥰😇
1:00 isn't A^-infinity equal to 1? -Infinity is 1/infinity, thus zero, this would make A^0 which is equal to one. I don't know if I'm the one who got it wrong please someone confirm If I'm right or wrong
so, nothing not raised to the power of anything still equals nothing. Thanks for clearing that up!
Can you please
Prove that the piecewise function
f(x)=e^(-1/x^2 ) for x≠0
0 for x=0
Is differentiable at x=0
Thanks in advance
By definition, f'(0) = e^(-1/x^2)/x, x->0, the limit clearly exists and is equal to 0, so the function is differentiable at 0 with derivative 0.
@@minecrafting_il Actually, the derivative of f(x) is
f'(x) = 2 * (e^(-1/x²)) / x³
But indeed,
lim_{x-->0} e^(-1/x²) = 0
and
lim_{x--> 0} f'(x) =
= lim_{x--> 0} 2*(e^(-1/x²))/(x³)
... substitute t = 1/x² ...
= lim_{t--> +infinity} 2*(e^(-t))*(t√t)
= lim_{t--> +infinity} (2t√t)/(e^t)
... using L'Hopital: [2t√t]' = 3√t , [e^t]' = e^t ...
= lim_{t--> +infinity} (3√t)/(e^t)
... using again L'Hopital: [3√t]' = (3/2)/√t , [e^t]' = e^t ...
= lim_{t--> +infinity} (3/2)/(√t * e^t)
= (3/2)/(+infinity)
= 0
so f(x) is continuous and differentiable at x=0 , with f'(x) = 0 .
@@yurenchu you have taken the limit of the derivative, which is the same as the derivative IF IT EXISTS. You can have a derivative without a limit of the derivative at that point. I calculated the actual derivative.
Try to take the derivative of x^2 cos(1/x^2) at 0
@@minecrafting_il I know but how did you show that this limit approaches zero
What about 0^x when x aproches 0 ?
What about x^(1/x)? That seems to approach zero pretty easily.
That doesn't have the form of 0^0 , but rather the form of 0^(infinity) .
@@yurenchu ahhh, makes sense. Thank you!
@@dotcomgamingd5564 You're welcome!
Perfect, I enjoyed😍😍😍😍😍👌🔥
Very nice
So now we also have 1^(+infinity) --> 0 , because
lim_{x--> 0⁺} f(x)^g(x)
where
f(x) = e^[-1/ln(-ln(x))]
g(x) = -ln(x)
apparently equals 0 .
EDIT: Corrected mistake, by adding a minus-sign.
Wow I just watched that other one
Why doesn't Wolfram Alpha show this result for exp(1/ln(-ln(x))) ?
Now let's see Paul Allen's limit!
Yay Finally!!!!!!!
this was so cute
0^0 is 1
Plug it in the exponential power series for evidence
That's not a proof. But defining it as 1 makes sense.
Is it good?
What if i take ln(-ln(ln(-ln(x))))
i plotted y=x^(x/0) on desmos and got a constant graph having y=0 but coudnt justify it... please explain what it is.. i also did it with e and got the same real graph
Strictly speaking, your function is undefined because it involves division by 0.
However, Desmos is probably interpreting numbers divide by 0 as some kind of infinity.
What I see when I plot this in Desmos is that the output is 0 everywhere on certain intervals. In particular, for x < -1 and for 0 < x < 1. I don't see any values for x > 1 or -1 < x < 0.
If Desmos interprets a positive number divided by 0 to be positive infinity, then we can make sense of the right half of the graph. If x > 1, then we have a number greater than 1 raised to an "infinite" power, which blows up to infinity. So for all x > 1, we should get "infinity". Now, for 0 < x < 1, if you raise any of these numbers to the "infinity" power, you get 0, since numbers between 0 and 1 get smaller and smaller as you raise them to larger and larger powers.
If Desmos interprets a negative number divided by 0 to be negative infinity, then we can make sense of the left half of the graph. Recall that a^(-b) = 1/(a^b). So essentially, we have x^(-"infinity") = 1/(x^"infinity"). If x < -1, then as you take higher and higher powers, the x^"infinity" will oscillate wildly but blast off to infinity and negative infinity. However, in either case, 1/("positive or negative infinity") is 0. On the other hand, for -1 < x < 0, we would get x^"infinity" should be 0 since larger and larger powers will make the overall number shrink. But 1/0 would blast off to infinity or negative infinity.
So, again, technically your functions is undefined everywhere. But if we play fast and loose, and treat infinity and limits like numbers, we can make sense of the graph.
@@MuffinsAPlenty thank u
But 1 to the infinity is a 1 but the graph shows you 0 . Why???
Why dont u also try plotting the exact same graph on mathway plzz.. Its Different
So basically 0^0 can approach 0 if the exponent grows much much more slowly than the denominator. You could actually make the limit anything you want no? Just need to choose functions that diminish appropriately.
If you're using real-valued functions with a positive base function, yes, you could make the limit any nonnegative number you want (though negative numbers as limits are impossible unless you allow for complex numbers).
If your exponent function is positive and approaches 0 significantly slower than the base function, then you could end up with 0 or any number between 0 and 1. If your exponent function is negative and approaches 0 significantly slower than the base function, then you could end up with any number greater than 1, or positive infinity. And if the exponent function approaches 0 at the same rate or faster than the base function, then you get 1 as the limit.
Interesting, though is much much more complicated than the first example.
No solo 0 elevado a la 0 es 1
Very nice
as x->0 x^x ->1 simple.
Cute!
So cute
that's cute
Thanks!
Freestyle video lmao
=0÷0
=0^1 × 0^-1
:a^m×a^n=a^m+n
=O^1-1
=0^0
:0^0=0
=0
😅
oooh
uh, how hard is it to draw a vertical line at "0"?
find out in this masterclass of mathematical engineering.
I mean prove not proof. Sorry for my lousy English and spell check.
I think you simply proved that 0^0 is indeterminate, and the limit will vary depending on how it's written. I mean, the limit as x goes to 0+ of x^0 is 1, and the limit as x goes to 0+ of 0^x = 0. You've shown us it could be e. In fact, you can make an expression and have the limit be any value you wish. I see wanting a defined value for this as a bit foolish, but math isn't supposed to be stodgy so if it's fun why not have fun?
He's not looking to define a value for 0^0. He's exactly showing different examples of getting the limit to be something different.
0^0 is not indeterminate. "Indeterminate" is a term that we apply to specific forms of _limits_ . So if a _limit_ takes the form of 0^0 , then we say this limit form is indeterminate; we must determine the correct value of the limit by a more elaborate means. Other indeterminate limit forms are 0/0 and 1^infinity .
However, in most (if not all) contexts outside limits, the expression 0^0 has the value of 1, just as in most contexts (outside limits) the expression 1^infinity has the value of 1 . (In contexts outside limits, the expression 0/0 is _undefined_ .)
hello po
completely wrong, you cant treat reverse infitines as zeros, since those infinities are not the same, x^x is the only way
damn
lol
Proof that 0^0=0/0=0 (if we want to define it in any way)
lets assume 0^0=a (a is any number) and 0/0=b
then 0^0=0^(1-1)=0^1/0^1=0/0.
now
0/0=b
=>b=0/0=(0+0)/0=0/0+0/0=b+b
=>b=0
=>0/0=0^0=0.
:/
@@CptFedora why the long face
Ok, let's accept that 0/0 = 0.
If we insist that a/b + c/d = (ad + bc)/(bd) should still hold, then 1 = 1+0 = 1/1 + 0/0 = (1*0 +1*0)/(1*0) = 0/0 = 0. Boom, we get 1 = 0. How do you explain this?
@@陳彥廷-v2u well if we accept that( a+a)/a=2,then 0=1=2.
yea saying 0/0=0 or 0/0=1 both gives problems.
Sorry, but that proof is wrong. Because x^(c-d) = (x^c)/(x^d) is not valid when x = 0 .
Consequences of this invalid approach:
x^(-3) = 1/(x^3)
x^(-3) = x^(2-5) = (x^2)/(x^5)
==>
1/(x^3) = (x^2)/(x^5)
Suppose x = 0 . Then lefthandside becomes
1/(x^3) = 1/(0^3) = 1/0 = ±infinity
and righthandside becomes
(x^2)/(x^5) = (0^2)/(0^5) = 0/0 = 0 (last step according to you reasoning)
so that means that ±infinity = 0 ?
You have too much time apparently
When we talk about 0 to 0, we want both zeros to be equal. Otherwise we can define a number of constants for this.
lim x -> inf (a^-x)^(1/x) ≈ 1/a
Take any a and you get a series of numbers. So technically this approach of taking different zeros is wrong.
Actually, (a^-x)^(1/x) = 1/a for any positive real values of a and x , so it's not an "approaching" limit.