Solving a Functional Equation | f(x)+f(x-1)=x^2
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- Опубликовано: 29 сен 2024
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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
ruclips.net/video/JwUlmBNXFv4/видео.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
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.
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).
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 👍🏿
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 😄
f(x) slightly different for odds vs evens. Odds: (x^2 + x)/2 - 16 evens: (x^2 + x)/2 + 16
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.
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
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
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!
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
Sum of ((-1)^k)×k²
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.
I am now amazed. Thanks Math. Thanks Syber.
Np. Thank you
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
I was under a lot of stress when you kept the evaluation in negatives until last step 😂
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
what was the initial function you used?
1+10+20=50
if you use 42 you would have answered the meaning of life
919? Its a prime number, obviously important.
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!
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.
Recursion
Genial, Merci 🐾
hi
nice
👍
Thank you so much
No problem