Does this CALCULUS Proof BREAK Math?
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- Опубліковано 5 лют 2025
- We use inverse trigonometric functions (arctan) and differentiation to prove that π = 0! We review the main concepts in detail and work carefully through the steps of the proof. Does this proof break math?
The concepts in this video are fundamental in high school math (including AP Calculus in the USA) and college math, in homework, assignments, tests and exams, and combine beautifully in this fun video!
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The trouble is arctan(1/x) is not continuous at x = 0, lim x → 0+ arctan(1/x) = π/2 and lim x → 0- arctan(1/x) = -π/2 so arctan(1/x) is not differentiable at x = 0. There is a right side derivative and a left side derivative which are not equal. Consequently, arctan(x) + arctan(1/x) is not truly constant on R. It must be divided into 2 sections: on (0, ∞) arctan(x) + arctan(1/x) = π/2 and on (-∞, 0) arctan(x) + arctan(1/x) = -π/2. Something has gone amiss!
Hi @Jack_Callcott_AU thank you so much for sharing this very detailed and insightful explanation! 🥳 I hope you have an amazing day/evening/night! 😊
@@MathMasterywithAmitesh It was an interesting problem for me. That is why I spent some time on it. Thanks you for the interesting maths.
Let's look at a right triangle ABC, with side lengths a, b, c and angles α, β, π/2, such that a, b and c are opposite to the angles α, β, π/2 respectively.
In this case, we know that
α+β+π/2 = π
==> α+β = π/2
We also know that
tan(α) = a/b
tan(β) = b/a
==>
arctan(a/b) = α
arctan(b/a) = β
==>
arctan(a/b) + arctan(b/a) = α+β = π/2
Notice that if x=a/b, then 1/x=b/a
Meaning
arctan(x) + arctan(1/x) = π/2
So far we have shown that the function constracted in the video geometricly represebt the sum of the two non- right angle angles in a right triangle, and thus constant
---------------
To see where the problem lies, we should remember that tan(x) is defined on the trigonometric unit circle. In particular, a possitive angle x corresponds to the 1st Quadrant, while a negative angle corresponds to the 4th quadrant.
That means that for -π/20+
It's correct that arctan(x) is a continuous function on all R, but arctan(1/x) is not. Moreover,
lim arctan(x) ≠ lim arctan(x)
x-->-inf x-->+inf
And hence
lim arctan(x) ≠ lim arctan(x)
x-->0- x-->0+
The function f has a jump type discontinuity at 0. It shouldn't be too surprisung, since it is undefined there
Hi @joelklein3501 Wow!!! Thank you so much for sharing such an extremely detailed and insightful comment! 😊 I absolutely loved reading your comment, and it beautifully and clearly explains the trigonometric identity from scratch in a way that a precalculus student can understand, as well as explains the identity thoroughly from the perspective of calculus! 🥳 The formatting is very clear and easy to read for anyone as well. 😊 It's the most amazing mathematical comment I have ever received on this channel, thank you SO much for taking the time to share this! 🙏 I hope you have an amazing day/evening/night and Happy New Year!!!!! 🥳🎉🎊
@@MathMasterywithAmitesh Haha, I'm glad to hear that you appreciated it so much. I enjoyed the video, and never actually noticed this arctan identity, so I had fun sketching an explanation. Hope to see more insightful videos from you
Good day/evening/night and happy new year!
@@joelklein3501 Thank you so much! 😊 Yes, it's an unusual arctan identity and I loved your explanation right from the basics using right triangles! (Also, although this is not as good as your explanation since it relies on another identity, we can also derive it from the tan angle sum formula tan(a + b) = [tan(a) + tan(b)]/[1 - tan(a)tan(b)].)
Thank you so much for your interest - I am looking forward to producing lots more videos this year that I hope you will enjoy watching! 🥳
I think the mistake is where the result of the limit was set equal to the result of the derivative. Limits calculate a functions value at a point, while derivatives calculate the slope at a point. This difference leads to the result of
pi = 0.
Hi @encounteringjack5699 thank you so much for your comment and for sharing your thoughts! 😊 In the video, I don't think we set the limit equal to the result of the derivative. The crux of the argument is (and of course it's a wrong argument somewhere 😝)
(1) "If f(x) = arctan(x) + arctan(1/x), then f'(x) = 0, which implies that f(x) is a constant function" - We don't equate the derivative to the limit, we just use f'(x) = 0 for all x to conclude that f(x) is a constant function
(2) "Since f(x) is a constant function, lim_{x -> ∞} f(x) = lim_{x -> -∞} f(x), which implies that π/2 = -π/2, or π = 0" - A constant function has the same limits at ∞ and -∞
Of course, there are other parts of the argument than (1) and (2) but that's the crux. Do you see where the error may be? 😊
I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@@MathMasterywithAmitesh I see. The problem is that there was an assumption that, because the derivative is zero, then the limit as it goes to positive and negative infinity must be the same.
However, that is not true. You could have a piece-wise function happening with the function. As long as the derivative along those areas are zero, the derivative as derived holds true.
This can be concluded by the fact the derivatives are opposite in the sign (+,-). Showing that the rates of change are equal, which means any variation in x, apart from x equals 0, the result will be a constant.
A graph of the function also shows a piece-wise function, with x = 0 not being defined.
@@encounteringjack5699 Hi! Thanks so much for sharing your detailed thoughts! 😊 Yes, you got it with your second paragraph and your last sentence! 🥳 (I will just add regarding the third paragraph, maybe you meant to say this, but in this case the derivative is 0 everywhere, there isn't any case of the derivatives having opposite sign, but the function does have opposite sign on different pieces 😊)
Still thinking about the problem
The way you think maths is amazing
Thank you so much!!! I am so happy to read your comment! 😊
Saying that the derivative is 0 then the function is constant, is integrating. You cannot integrate over an interval where the function is not continue.
Hi @meurdesoifphilippe5405 thank you so much for your comment and for sharing your insight! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
the function you have taken is not continuous at all points, therefore f'(x)=0 does not imply x=c
for example, a function floor(x) has it's derivative to be 0(for the points which it is continuous and differentiable), but floor(x) is not a constant function
Hi @KrishnaChandra108 thanks so much for your comment and for sharing your thoughts! 😊 The function floor(x) is very interesting! I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
If you consider the function f[x]=arctan[x]+arctan[1/x] ,then you have the following properties : for x < 0 f[x] = - π/2 , for x= 0
f[x] is not defined ,for x > 0 , f[x] = π/2 . This makes it clear that you may get the contradictory statement that π = 0.
It is in any case a good example , showing that one can get very strange results by using an equality outside its range of
application.
Hi @renesperb thank you so much for sharing and for your comments, as always! 😊 I also like this example because it can be studied using both calculus and/or trigonometric identities (and the latter is accessible to precalculus students). I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Wow! This is tempting...
HINT for those who want to find the mistake:
Just look at a few values of this funktion for absolute small values.
Hi @WolfgangFeist thank you so much for your comment and for sharing the hint! 😊 I always love reading your comments! I like this example because the function arctan(x) + arctan(1/x) can be studied by precalculus students as well and is related to a lesser known/used trigonometric identity. 😊 I hope you have an amazing day! 🥳
Before watching, my guess is divides by zero (possibly by taking tan(pi/2) or something)
Hi @derelbenkoenig thanks so much for sharing your thoughts! 😊 Yes, that's a very common "false proof" method, but not in this case (the reason is different here since we don't actually evaluate tan(π/2) but rather evaluate a limit lim_{x -> ∞} arctan(x) = π/2 (for example) which is valid). Did you have further thoughts after watching? I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@MathMasterywithAmitesh yep indeed I watched and the problem was more complicated than I expected. I'll have to read more of others comments to understand better
Hi @derelbenkoenig ! Thanks for letting me know! If you have any further questions about the video, please feel free to reply here and I'd be happy to try to answer. I think there are comments here which explain it in some detail (although some comments are a bit more precise than others). 😊
Very cool 😁
Hi @mukundchandra69 thank you so much for your comment! I am so happy you enjoyed the video! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
Lim(x→0) does not exist so we can't say that the function is a constant function
Hi @mahima_abhishek_ thank you so much for your comment and for sharing! 😊 I hope you have an amazing day/evening/night and Happy New Year!!! 🥳🎉🎊
@MathMasterywithAmitesh wish you the same 😊
X,2x+5=8