Cases 1 and 2 are specific instances of case 3, no? A constant function is symmetric about a/2 (and about any other domain point you choose), and a function only defined at x = a/2 is symmetric about a/2.
In other words it doesn't matter which direction along the x axis you integrate. Either forward or backwards over the same interval will give the same result. Area under the curve does not depend on which way you add up the little pieces.
Two quick notes: - Case 1 is actually just a specific example of case 3. (i.e. a constant function is just a specific example of a function that is symmetric about a point.) - Case 4 doesn’t require symmetry at all. It’s true for ALL integrable functions that ₀ ∫ ᵃ f(x) = ₀ ∫ ᵃ f(a-x) . You can see this because the proof in the video doesn’t use anything about symmetry, it works for any integrable function. For example, take f(x) = x. It’s not a symmetric function. But ₀ ∫ ᵃ x = a ²/2 , and ₀ ∫ ᵃ (a-x) = a ² - (a ²/2) = a ²/2. So case 4 is a property that is always true for all integrable functions, not just symmetric ones.
@ That's right, f(x) is an odd function (i.e. f(-x) = -f(x) ). x² on the other hand is an even function (f(-x) = f(x) ) Note that the above property of integrals holds for all functions, even ones that are neither odd nor even. For example, take f(x) = eˣ . ₀ ∫ ᵃ eˣ = eᵃ - 1 , and ₀ ∫ ᵃ eᵃ⁻ˣ = -eᵃ⁻ᵃ + eᵃ⁻⁰ = eᵃ - 1 .
Nice little thought experiment. As some of the other comments already mentioned, the "Case 3" already covers all the other possible cases (i.e, the statement is true if and only f is symmetric about a line x = a/2) Proof that functions symmetric about a line x=a/2 fulfill this statement: 1. Assume that f is symmetric about a line x = a/2 2. Therefore f( (a/2) - r ) = f( (a/2) + r ) for all r 3. For all real x there exists real r such that x = (a/2) + r 4. Plugging this into step 2. gives: f( a - x ) = f( x ) Proof that functions not symmetric about a line x=a/2 don't fulfill this statement: 1. Assume that f is not symmetric about a line x = a/2 2. Therefore there exists real r such that f( (a/2) - r ) != f( (a/2) + r ) 3. We can rewrite this as f( a - [ (a/2) + r ] ) != f( (a/2) + r ) 4. But now selecting x = (a/2) + r, we get: f( a - x ) != f(x) Could probably prove this a bit more elegantly, but didn't spend too much time to make it look nicer.
When f is symmetrical with respect to the vertical line x = a/2. For example: f(x) = sin(x) with a = pi. Edit after watching the video: Case 1 and 2 are just special cases of Case 3. The 4th case is completely unrelated to the question originally posed.
Your case 3 covers all possibilities. f(x) = f(a-x) (if and only if) g(t) = f(a/2 + t) is even (with appropriate domains, i.e. domains shifted by a/2) =>) If f(x) = f(a-x), then g(t) = f(a/2 + t), so g(-t) = f(a/2 - t) = f( a - (a/2 + t) ) =(f's property) f(a/2 + t) = g(t), so g is even.
A couple of questions: 1) Are you saying ANY two functions that have the same integral over a set range from 0 to a are the same? Then are sin and cos over the range 0 to 2(pi) the same function (since the integral of both is zero)? (obviously not, right?) 2) You use zero as a lower bound. Couldn't any value work as a lower bound?
Question: When is f(x) = f(a - x) always true? Answer: in the following cases: - when f(x) is a constant function - when a = 0 and f(x) is an _even_ function - when a ≠ 0 and f(x) is an even function that has been horizontally shifted over a distance a .
If you substitute t=x - a/2, then the Class 3 functions will be symmetric about the y-axis (x=0). What functions does that include? I can think of functions based on |y|; functions based on even-numbered powers of y (y^2, y^4, y^6...), including inverse powers (1/y^2, 1/y^4, 1/y^6...); and functions based on cosine (y) and secant (y). Are there others?
case 4 is a special case of what's widely referred to as the "king property" of integrals, meaning: the integral of f(x)dx from a to b equals the integral of f(a+b-x)dx from a to b, where a and b are real numbers. Have you got any clue where the name "king property" actually comes from?
Hi. Quick question. What if f is a periodic function (and even maybe for it to work) with the period being a factor of a? Wouldn't f(x) = f(x+(-a)) = f(a-x)? Or am I making an error?
The periodicity of the function is not important as to whether f(x) = f(a-x) . For example, consider f(x) = tan(x) . This function has a periodicity of a = π , but generally, tan(x) does not equal tan(π-x) . (The only values of x for which it would be true, are multiples of π .) The reason why the equation "f(x) = f(a-x)" works for f(x) = cos(x) but not for f(x) = tan(x), is because cos(x) has a vertical line of symmetry, and tan(x) doesn't.
time=4:56 in this case, the symmetry is about a line not a point. What is about a point is the position of that axis of symmetry. regards! (factorial) [edited] well...I'm always to much hurry 🤣 I saw the error & immediately complain. Then I continued watching and realized that you correct the issue by telepathy without the need to read my post ! 😆🤣🤣
I know it’s tough to fit all details on the thumbnail, but it needs to be conveyed f cannot be identity function if you want question to be a challenge.
Make a video showing all integers solutions for n which | (3n+16) / (4-n) | = k² | k € Q I made this question by myself and the solutions for n are n = -10, -5, -3, 0, 3, 32 I even put the equation on a C compiler and these are the solutions fr
I don't agree with the 2nd case. In fact, we have to define the meaning of "a". If we meant "for any a", then the 2nd case doesn't work, as there are many values for "a", where f(x)=single_defined_value != f(a-x)=undefined. If we meant "some a value exists, so that f(x)=f(a-x)", then the 2nd case is valid, but at the same time almost any non-monotonic function works, as there exist an "a" value where f(x1)=f(x2) => a=x2+x1. So in a good way we have to define the initial question more accurately and don't juggle with different definitions in different cases.
I really like you videos. But a quick question in this one: why aren't there a 5th case? It's not clear for me that a function of other "type" could or could not be a solution. Never stop learning!
Actually, Prime Newtons' written x is the _correct_ (and usual) way in mathematics to display the _variable_ x , which is distinguishably different from the multiplication symbol × . The people who write variable x as indistinguishable from the multiplication symbol × are the ones who write their variable x incorrectly.
I really appreciate how you teach us with friendly and dramatic gestures.
So basically King's rule in definite integrals where x can be replaced by a+b-x is reflecting the graph about (a+b)/2 !
Yes
Yes, if the integral looks difficult, 90% of the time it's just this rule.
Cases 1 and 2 are specific instances of case 3, no? A constant function is symmetric about a/2 (and about any other domain point you choose), and a function only defined at x = a/2 is symmetric about a/2.
In other words it doesn't matter which direction along the x axis you integrate. Either forward or backwards over the same interval will give the same result. Area under the curve does not depend on which way you add up the little pieces.
Two quick notes:
- Case 1 is actually just a specific example of case 3. (i.e. a constant function is just a specific example of a function that is symmetric about a point.)
- Case 4 doesn’t require symmetry at all. It’s true for ALL integrable functions that ₀ ∫ ᵃ f(x) = ₀ ∫ ᵃ f(a-x) . You can see this because the proof in the video doesn’t use anything about symmetry, it works for any integrable function.
For example, take f(x) = x. It’s not a symmetric function. But ₀ ∫ ᵃ x = a ²/2 , and ₀ ∫ ᵃ (a-x) = a ² - (a ²/2) = a ²/2. So case 4 is a property that is always true for all integrable functions, not just symmetric ones.
I've seen some people call case 4 the King's Property of integrals.
f(x) = x is odd, like sin(x), right?. f(x) = x, -f(-x) = -(-x) = x = f(x)
@@geostorm8192 yes
@ That's right, f(x) is an odd function (i.e. f(-x) = -f(x) ). x² on the other hand is an even function (f(-x) = f(x) )
Note that the above property of integrals holds for all functions, even ones that are neither odd nor even. For example, take f(x) = eˣ .
₀ ∫ ᵃ eˣ = eᵃ - 1 , and ₀ ∫ ᵃ eᵃ⁻ˣ = -eᵃ⁻ᵃ + eᵃ⁻⁰ = eᵃ - 1 .
Nice little thought experiment.
As some of the other comments already mentioned, the "Case 3" already covers all the other possible cases (i.e, the statement is true if and only f is symmetric about a line x = a/2)
Proof that functions symmetric about a line x=a/2 fulfill this statement:
1. Assume that f is symmetric about a line x = a/2
2. Therefore f( (a/2) - r ) = f( (a/2) + r ) for all r
3. For all real x there exists real r such that x = (a/2) + r
4. Plugging this into step 2. gives: f( a - x ) = f( x )
Proof that functions not symmetric about a line x=a/2 don't fulfill this statement:
1. Assume that f is not symmetric about a line x = a/2
2. Therefore there exists real r such that f( (a/2) - r ) != f( (a/2) + r )
3. We can rewrite this as f( a - [ (a/2) + r ] ) != f( (a/2) + r )
4. But now selecting x = (a/2) + r, we get: f( a - x ) != f(x)
Could probably prove this a bit more elegantly, but didn't spend too much time to make it look nicer.
Case 4 is a very interesting property, thanks for sharing!
Case 4 is true for any f(x) , not just for functions that satisfy f(x) = f(a-x) .
Now you need to prove that there is no case 5. Which I don’t know how to prove. Proving that no more cases exist is very tricky
Case 4 is the same of case 1, cause definite integral results a constant
Thanks for giving our minds a good workout!
Im not too sure about case 4. The integrals are the same, but that doesnt mean the functions are the same.
When f is symmetrical with respect to the vertical line x = a/2. For example: f(x) = sin(x) with a = pi.
Edit after watching the video: Case 1 and 2 are just special cases of Case 3. The 4th case is completely unrelated to the question originally posed.
Your case 3 covers all possibilities.
f(x) = f(a-x) (if and only if) g(t) = f(a/2 + t) is even (with appropriate domains, i.e. domains shifted by a/2)
=>) If f(x) = f(a-x), then g(t) = f(a/2 + t), so g(-t) = f(a/2 - t) = f( a - (a/2 + t) ) =(f's property) f(a/2 + t) = g(t), so g is even.
My initial guess is all even, a-periodic functions. What has the integral have to do with the question?
A couple of questions:
1) Are you saying ANY two functions that have the same integral over a set range from 0 to a are the same? Then are sin and cos over the range 0 to 2(pi) the same function (since the integral of both is zero)? (obviously not, right?)
2) You use zero as a lower bound. Couldn't any value work as a lower bound?
This was a great video, opened my eyes!
Question: When is
f(x) = f(a - x)
always true?
Answer: in the following cases:
- when f(x) is a constant function
- when a = 0 and f(x) is an _even_ function
- when a ≠ 0 and f(x) is an even function that has been horizontally shifted over a distance a .
Isn't case 3 the same as case 4? Case 4 simply calculates the area bounded by case 3.
All these cases come down to the third one.
@boguslawszostak1784 I suppose even case 1 could be considered the same as case 3, but I wouldn't have classified it as such
If you substitute t=x - a/2, then the Class 3 functions will be symmetric about the y-axis (x=0). What functions does that include? I can think of functions based on |y|; functions based on even-numbered powers of y (y^2, y^4, y^6...), including inverse powers (1/y^2, 1/y^4, 1/y^6...); and functions based on cosine (y) and secant (y). Are there others?
case 4 is a special case of what's widely referred to as the "king property" of integrals, meaning:
the integral of f(x)dx from a to b equals the integral of f(a+b-x)dx from a to b, where a and b are real numbers.
Have you got any clue where the name "king property" actually comes from?
Thanks. Never heard of the name before now. Just learned something .
@@PrimeNewtons Never stop learning! Cause those who stop learning, ... 😀
I make video about math topics but I can’t attract many students until now congratulate you because of high quality teaching ❤
Hi. Quick question. What if f is a periodic function (and even maybe for it to work) with the period being a factor of a? Wouldn't f(x) = f(x+(-a)) = f(a-x)? Or am I making an error?
The periodicity of the function is not important as to whether f(x) = f(a-x) . For example, consider f(x) = tan(x) . This function has a periodicity of a = π , but generally, tan(x) does not equal tan(π-x) . (The only values of x for which it would be true, are multiples of π .)
The reason why the equation "f(x) = f(a-x)" works for f(x) = cos(x) but not for f(x) = tan(x), is because cos(x) has a vertical line of symmetry, and tan(x) doesn't.
Wow, that was unexpected. 🙂
time=4:56
in this case, the symmetry is about a line not a point. What is about a point is the position of that axis of symmetry.
regards! (factorial)
[edited] well...I'm always to much hurry 🤣
I saw the error & immediately complain.
Then I continued watching and realized that you correct the issue by telepathy without the need to read my post ! 😆🤣🤣
g(t)=f(t-a/2) is an odd func of t
Simple answer: a = 2x
I know it’s tough to fit all details on the thumbnail, but it needs to be conveyed f cannot be identity function if you want question to be a challenge.
ln(x)/x dx
Make a video showing all integers solutions for n which | (3n+16) / (4-n) | = k² | k € Q
I made this question by myself and the solutions for n are n = -10, -5, -3, 0, 3, 32
I even put the equation on a C compiler and these are the solutions fr
I don't agree with the 2nd case. In fact, we have to define the meaning of "a". If we meant "for any a", then the 2nd case doesn't work, as there are many values for "a", where f(x)=single_defined_value != f(a-x)=undefined. If we meant "some a value exists, so that f(x)=f(a-x)", then the 2nd case is valid, but at the same time almost any non-monotonic function works, as there exist an "a" value where f(x1)=f(x2) => a=x2+x1. So in a good way we have to define the initial question more accurately and don't juggle with different definitions in different cases.
I really like you videos. But a quick question in this one: why aren't there a 5th case? It's not clear for me that a function of other "type" could or could not be a solution. Never stop learning!
Your x is not an x. Never liked teachers who developed their own writing style and expected their students to decipher
Your parents gasped “that’s not human” when you popped out, but did not make a big fuss about it like their whining kid does.
That is a standard and very common way of writing x when doing maths.
Looks like n or pi or x, but my screen is small
Actually, Prime Newtons' written x is the _correct_ (and usual) way in mathematics to display the _variable_ x , which is distinguishably different from the multiplication symbol × . The people who write variable x as indistinguishable from the multiplication symbol × are the ones who write their variable x incorrectly.
I showed this video to my friend and now her boyfriend wants to try it too. Waiting for the report💞
А ну, иди отсюда. Это место только для легенд, а не для ботов