Prima facie, f(x) must be quadratic, ie of the form ax² + bx + c (ax² + bx + c) + (ax² - 2ax + a + bx - b + c) = x² 2ax² + (2b - 2a)x + (2c + a - b) = x² Immediately we get a = 1/2, b = 1/2 and c = 0, so: f(x) = x(x + 1) / 2 (ie the sum of digits from 1 to x) However, we are told that f(11) = 50, not 11 * 12 / 6 = 66 as initially expected. This extra -16 term must be negated in the f(10) term to get back to f(x) + f(x - 1) = x². This extra term must therefore be 16(-1)ˣ f(x) = x(x + 1) / 2 + 16(-1)ˣ f(41) = 41 * 42 / 2 - 16 = 845
An interesting method is to try and solve the generic functional equation. By trying a quadratic, you find that f(x) = x(x+1)/2 is a solution. Generic solutions are of the form f(x) = x(x+1)/2 + g(x) where g(x) + g(x-1) = 0. This means that g(x) could be a square wave function oscillating between A and -A where A is real
You can express the arithmetic series that you want to evaluate as: 12+13+14+...+41 It's probably easier. S=41*(41+1)/2-11*(11+1)/2 = 795 795 + 50 = 845
f(x) + f(x-1) = x2 f(x-1) + f(x-2) = x2 - 2x + 1 (I)-(II) f(x)-f(x-2) = 2x-1 If we take only odd numbers as we only know f(11) : lets def y such as x=2y+1 or y=(x-1)/2 x=2y+1 => x-2=2(y-1)+1 f(2y+1)-f(2(y-1)+1)=2x-1=4y+1 We can define another function g(y)=f(2y+1) and of course g(y-1)=f(2(y-1)+1) so we can write g(y) - g(y-1) = 4y+1 Then, we can break the recurrence by writing g(y) = g(0) + sum(i=1 to y) [4i+1] = g(0) + 4y(y+1)/2 + y = g(0)+2y^2+3y We have f(11)=g(5)=g(0)+2*25+3*5 =g(0)+65 = 50 Then g(0)=-15 And of course f(41)=g(20)=g(0)+2*400+3*20 = -15+800+60=845
Can do quicker with triangle numbers. The sum of consecutive triangle numbers are the squares, thus f(n)=T(n)+a*(-1)^n for some real a. T(11)=66, so a=16, so f(41)=845.
Just dealing with integers. Using f(x)+f(x-1)=x^2 and f(x-1)+f(x-2)=(x-1)^2, subtract to get the difference equation f(x)-f(x-2)=(2x-1). This results in two series for odds and evens. If f(0)=A and f(1)=B then for evens f(x)=A+x(x+1)/2 and for odds f(x)=B+(x-1)(x+2)/2. A+B=1. For this problem f(11)=B+65=50, so B=-15. Then f(41)= -15+860 = 845.
I manipulated the original equation and I noticed that f(x+1) - f(x-1) = 1 + 2x and f(x+2) - f(x-2) = 2 + 4x You can then quite easily prove by induction that f(x+n) - f(x-n) = n+2nx. Considering that 41 =26+15 and 11 = 26-15, you can write f(41) = f(11) + 15 +2*26*15 = 845.
You can actually simplify by starting with f(41) and going down to f(11) so that all negatives are removed. Secondly, there is no need to add 12+13, etc. it is actually a sum of consecutive numbers from 12 to 41.
very interesting equation. nice to see some different anserw than x=1. good video, keep doing what you do, it’s nice to watch you solving math problems :D
Hello sir , I am watching you since 2021 . I was in 10 grade now I am in 11 grade . In 11 grade I have to learn calculus and other higher chapters . I am asking you that if you would like to make a (A to Z) series on calculus as a beginner. Can you? I asked you cause I understand your method.
I do this: f(x) + f(x-1) = x^2 --> f(x) = x^2 - f(x-1) And from here, I said to my self: come on, a simple recursive function, just use python xD def f(x): return 50 if x == 1 else x ** 2 - f(x - 1) print(f(41)) output > 845. Just a joke... nice video, keep going!
Very nice! But i think it would have been nicer if f(11)=66,because then we will have nice function f(x)=0.5(x^2+x). I wonder what is f(x) that satysfies both f(x)+f(x-1)=x^2 and f(11)=50. 💯👍
Prima facie, f(x) must be quadratic, ie of the form ax² + bx + c
(ax² + bx + c) + (ax² - 2ax + a + bx - b + c) = x²
2ax² + (2b - 2a)x + (2c + a - b) = x²
Immediately we get a = 1/2, b = 1/2 and c = 0, so:
f(x) = x(x + 1) / 2 (ie the sum of digits from 1 to x)
However, we are told that f(11) = 50, not 11 * 12 / 6 = 66 as initially expected. This extra -16 term must be negated in the f(10) term to get back to f(x) + f(x - 1) = x². This extra term must therefore be 16(-1)ˣ
f(x) = x(x + 1) / 2 + 16(-1)ˣ
f(41) = 41 * 42 / 2 - 16 = 845
Wow!
That was awesome
No! It must Not! Also 16cos(px) will Do
th-cam.com/video/JwUlmBNXFv4/w-d-xo.html
Prima facie ??
@@michaelempeigne3519 ??? = ?
An interesting method is to try and solve the generic functional equation. By trying a quadratic, you find that f(x) = x(x+1)/2 is a solution. Generic solutions are of the form f(x) = x(x+1)/2 + g(x) where g(x) + g(x-1) = 0. This means that g(x) could be a square wave function oscillating between A and -A where A is real
I basically did this, with my g(x) being Acos(pi*x), with A=16 for this example
You can express the arithmetic series that you want to evaluate as:
12+13+14+...+41
It's probably easier.
S=41*(41+1)/2-11*(11+1)/2 = 795
795 + 50 = 845
wait why did you add the 50?
@@oldrichmarek419 50 is f(11).
f(x) + f(x-1) = x2
f(x-1) + f(x-2) = x2 - 2x + 1
(I)-(II) f(x)-f(x-2) = 2x-1
If we take only odd numbers as we only know f(11) :
lets def y such as x=2y+1 or y=(x-1)/2
x=2y+1 => x-2=2(y-1)+1
f(2y+1)-f(2(y-1)+1)=2x-1=4y+1
We can define another function g(y)=f(2y+1) and of course g(y-1)=f(2(y-1)+1)
so we can write g(y) - g(y-1) = 4y+1
Then, we can break the recurrence by writing
g(y) = g(0) + sum(i=1 to y) [4i+1] = g(0) + 4y(y+1)/2 + y = g(0)+2y^2+3y
We have f(11)=g(5)=g(0)+2*25+3*5 =g(0)+65 = 50 Then g(0)=-15
And of course f(41)=g(20)=g(0)+2*400+3*20 = -15+800+60=845
f(x) slightly different for odds vs evens. Odds: (x^2 + x)/2 - 16 evens: (x^2 + x)/2 + 16
Can do quicker with triangle numbers. The sum of consecutive triangle numbers are the squares, thus f(n)=T(n)+a*(-1)^n for some real a. T(11)=66, so a=16, so f(41)=845.
is that geometric series.
Just dealing with integers. Using f(x)+f(x-1)=x^2 and f(x-1)+f(x-2)=(x-1)^2, subtract to get the difference equation f(x)-f(x-2)=(2x-1). This results in two series for odds and evens. If f(0)=A and f(1)=B then for evens f(x)=A+x(x+1)/2 and for odds f(x)=B+(x-1)(x+2)/2. A+B=1. For this problem f(11)=B+65=50, so B=-15. Then f(41)= -15+860 = 845.
Method: Telescoping sum equating to the sum over consecutive 1 (mod 4) terms
( f(41) + f(40) ) - ( f(40) + f(39) ) + ... + ( f(13) + f(12) ) - ( f(12) + f(11) )
= Sum{ 4m-3 }
= 4 Sum{ m } - Sum{ 3 }
= 4 Sum{ m } - 4 Sum{ m } - 45
= 4*21*11 - 4*3*7 - 45
= 795
= f(41) - 50
f(41) = 845
I manipulated the original equation and I noticed that f(x+1) - f(x-1) = 1 + 2x and f(x+2) - f(x-2) = 2 + 4x You can then quite easily prove by induction that f(x+n) - f(x-n) = n+2nx. Considering that 41 =26+15 and 11 = 26-15, you can write f(41) = f(11) + 15 +2*26*15 = 845.
That's the way 👍🏿
I was under a lot of stress when you kept the evaluation in negatives until last step 😂
I am now amazed. Thanks Math. Thanks Syber.
Np. Thank you
You can actually simplify by starting with f(41) and going down to f(11) so that all negatives are removed. Secondly, there is no need to add 12+13, etc. it is actually a sum of consecutive numbers from 12 to 41.
a[n] = a[1] + (n-1)d
so n = (a[n] - a[1])/d + 1
and S[n] = (a[1] + a[n])*n/2
a[1] = 25, a[n] = 81, d = 4
very interesting equation. nice to see some different anserw than x=1. good video, keep doing what you do, it’s nice to watch you solving math problems :D
Thank you!
The answer is 1,397 😀
f(x) + f(x-1) = x^2
f(11) = 50
f(11) + f(11-1) = (11)^2
50 + f(10) = 121
f(10) = 121 - 50 = 71
f(40) = 4 * f(10)
= 4 * 71
= 284
f(41) + f(41-1) = (41)^2
f(41) + f(40) = 1,681
f(41) + 284 = 1,681
f(41) = 1,681 - 284 = 1,397 QED
I respect your mathematical principles. You are among a very few to use proper mathematical notifications.
Thank you! 🥰
@@SyberMath But it’s true. Mathematical notations must be unambiguous.
Thank you so much
No problem
Genial, Merci 🐾
what was the initial function you used?
Hello sir ,
I am watching you since 2021 . I was in 10 grade now I am in 11 grade . In 11 grade I have to learn calculus and other higher chapters . I am asking you that if you would like to make a (A to Z) series on calculus as a beginner. Can you? I asked you cause I understand your method.
Look up 3 blue 1 brown "The essence of calculus" series
@@cadenorris4009 thanks 😄
I do this:
f(x) + f(x-1) = x^2 --> f(x) = x^2 - f(x-1)
And from here, I said to my self: come on, a simple recursive function, just use python xD
def f(x):
return 50 if x == 1 else x ** 2 - f(x - 1)
print(f(41))
output > 845.
Just a joke... nice video, keep going!
you got me there! 😁
thanks
if you use 42 you would have answered the meaning of life
919? Its a prime number, obviously important.
1+10+20=50
Very nice! But i think it would have been nicer if f(11)=66,because then we will have nice function f(x)=0.5(x^2+x). I wonder what is f(x) that satysfies both f(x)+f(x-1)=x^2 and f(11)=50. 💯👍
f(x) here has to be 0.5(x+1)(x)+(-1)^x*16. If you decrease f(11) by 1 then f(12) must go up by 1 then f(13) must go down by 1, etc.
@UCR1JCG-fok9hCnDH3CJk0Hg if you want a function that works for all real x you can replace (-1)^x with a sin curve.
@@sapphosfriend9558 nice!
Recursion
nice
Sum of ((-1)^k)×k²
hi
👍