An Interesting Functional Equation

[@spelunkerd]

The cards referring to other videos and your channel are obstructing our view of the last page, at 8:48. It's easy to correct that in retrospect by going to KZbin studio and dragging the cards down to the bottom. You've left ample room below for those cards so it should be easy. Thanks for the video!

[@trojanleo123]

Agreed. Hoping the content creator heeds your advice.

[@richardhole8429]

​@@trojanleo123I simply touched the screen and they went away.

[@conanthecribber]

@@richardhole8429 not on my PC they didn't.

[@sudiptoits]

Easy question We could do, f(x) = x^2-f(x-1) So f(41) = 41^2 - f(40) = 41^2-40^2 + f(39) and so on, so f(41) = 41^2-40^2+39^2-38^2… + 15^2-14^2+13^2-12^2+f(11) = (41-40)(41+40) + (39-38)(39+38) +… + (13-12)(13+12)+ f(11) = (81+77+73+…+25) + 50 Using AP formulae you can calculate this to be 845

[@sonicbreaker00]

If you write the first few terms like f(1), f(2), f(3), ... in terms of f(0), then you can immediately see that the general solution is: f(N) = N(N+1)/2 + [(-1)^N]*f(0). Now use f(11) = 50 to obtain f(0) = 16. Therefore, f(N) = N(N+1)/2 + [(-1)^N]*16. Thus, f(41) = 41*42/2 - 16 = 845.

[@paulortega5317]

If the domain is restricted to integers.

[@florianbasier]

@@paulortega5317 true, but we only care about integers here

[@Patrik6920]

Wolfram alpha has a nasty solution for this _gasping for air_ if the unknow function f(x)+f(x-1)=x^2 the function f(x) is f(x) = c (-1)^(x - 1) + 1/2 (-1)^(x - 1) ((-1)^(x + 1) x^2 + (-1)^(x + 1) x), havent plotted this so i dont know what the constant C are (pretty sure it can be simplified alot) .. it is the solution to f(x)=x^2-f(x-1), where the function is unknown edit: seems C is (x^2)/2 + x/2 -> C=(x^2+x)/2 = x(x+1)/2

[@alexandreocadiz9967]

@@florianbasier yes but it seems that f(x) = x(x+1)/2 + 16*cos(x*pi) is a valid answer to the equation for real numbers.

[@pavelsolaris1901]

Excellent job

[@rickdesper]

It's easy to do this in < 15 seconds using Excel. 😀 And spreadsheet manipulation is a useful skill to develop. I like solutions where it's explained why you take the steps you do, instead of just saying "look what happens if we try to make an alternating sum using pairs of consecutive values of f()". It would be useful to set up a difference equation, i.e. f(n+1) = (n+1)^2 - f(n). This would quickly and naturally lead to the kind of alternating sum you want.

[@Blaqjaqshellaq]

Another approach you can take is to consider that f(13) - f(12)=12 + 13, and so on for these sums. So f(41) - 50=12 + 13 + 14 +... + 40 + 41. The triangular sum 1 + 2 + 3 +... +n=n*(n+1)/6, so f(41) - 50=(41*42)/2 - (11*12)/2=861 - 66=795, so f(41)=795 + 50=845.

[@次野先生]

f(x+1)-f(x-1)=2x+1를 구하고 x=12,14,"""40를 차례로 대입하여 변변더하면 f(41)-f(11)=25+29+""81 이하 우변 등차수열의 합795 답845

[@wannabeactuary01]

Loved this. Admittedly when you see things where you want to shout at the screen then that makes it even better. My moment is at 6:03. You dd all the hard work and showed: 12^2 - 13^2 + 14^2 -15^2 +... +40^2 - 41^2 = -(12 +13 +14 +15 +... + 40 + 41)

[@ZannaZabriskie]

Correct, but: DO NOT SUM the couples! If you don’t, you have: 12+13+14+15+…+41+50 That’s is S(41)-S(11) + 50 easy peasy!

[@yuvalmagen100]

beautiful

[@dan-florinchereches4892]

Well this looks hairy to start with. But if you pair up the function and say (1) f(x)=x^2-f(x-1) then substituting the expression for f(x-1) back f(x)=x^2-(x-1)^2+f(x-2)= 2n-1 + f(x-2) this means f(13)=2*13-1 +f(11) f(15)=2*15-1 +f(13) .... f(41)=2*41-1+f(39) number of terms =(41-11)/2 = 15 adding everything up f(13)+f(15)+...+f(41)=f(11)+f(13)+...+f(39)+ 2*(13+15+...+41) -1*15 reducing like terms f(41)=2*(13+14+15+...+40+41-14-16-18-...-40)-15=2*(1+2+3+...+41-1-2-...-12-2*(7+8+9+...+20))-15=2*(42*41/2-13*12/2-2*(1+2+...+20-1-2-...-6))-15=2*(41*21-13*6-2*(21*20/2-6*7/2))-15=2*(861-78-2*(210-21))-15=2*(783-2*189)-15=2*405-15=810-15=795 edit : I missed the f(11)=50 from the total which should be 845. Totally forgot about that way of calculating the arithmetic series using first and last term. Makes sense to keep if on top of your mind... most exercises back in the day were mindless use of formula for sum from a0 to an

[@saeedjaffer532]

It is clear that the function is a polynomial f(x) =ax^2+bx+c solving for a, b, c a=b=1/2, c=0 then f(x) =1/2x^2+1/2x then f(11)=66 not 50, as given. Then f(41)=861. If f(x) +f(x-1)was=x^2-32 then f(11)=50 is correct and f(41)=845

[@boguslawszostak1784]

But is NOT a polynomial

[@cutcrew2743]

For this sequence we could set f(1) to -15. It is interesting to see that f(41) is nearly the same for small starting values. Here are some tuples of f(1), f(41) and f(1000): (-15, 845, 4934), (0, 860, 4949), (1, 861, 4950), (10, 870, 4959).

[@mathpro926]

Nice video

[@adandap]

This is equivalent to a difference equation a_(n+1) + a_n = n^2, which can be solved by multiplying by (-1)^(n+1) and summing from 11 to 41.

[@alipourzand6499]

Lots of nice methods in this one! Hope it will make cents! ☺

[@xinpingdonohoe3978]

Cents as in a small amount of money?

[@MrRyanroberson1]

I originally came up with a solution that's only a little off the true answer, but I think my method is still a useful demonstration: f(x) + f(x-1) = x^2 f(x+1) + f(x) = x^2 + 2x + 1 => f(x+1) - f(x-1) = 2x+1 now we offset by 1 to find f(x) = f(x-2) + 2x - 1 f(x-2) = f(x-4) + 2(x-2) - 1 => f(x) = f(x-4) + 4x - 6 and repeating this strategy of expanding the function on the right side to keep doubling the gap... f(x) = f(x-8) + 8x - 28 f(x) = f(x-16) + 16x - 120 f(x) = f(x-32) + 32x - 496 therefore, in particular, f(43) = f(11) + 32x43 - 496 = 930 and finally f(43) = f(41) + 85; 930-85 = 845

[@konstantinvorontsov9811]

f(x) = x(x+1)/2 + 16(-1)^x is a short form for f(2z) = 2z^2 + z + 16 f(2z+1) = 2z^2 + 3z - 15 Interesting that no common formula for f(x) exists for real x, only for integer x

[@downrightcyw]

How do you find out the general form of f(x) = x(x+1)/2 + 16(-1)^x ?

[@eli37co]

From the general equation you can say f(x) is ax^2+bx + c. a=0.5, b=0.5 anc c= 0. There for f(x) is 0.5x^2 + 0.5x. f(11) is 66 not 50.

[@hasanpezuk8757]

Yes question is wrong f(11) = 66 and (50 + 16) f(41) = 861 (845 + 16 )

[@GolumTR]

How do you know this is the unique function such that f(x) + f(x-1) =x^2 ?

[@rickdesper]

@@hasanpezuk8757 Question is wrong?

[@Mediterranean81]

f(x) + f(x-1) = x^2 f(x) + f(x+1) = x^2 + 2x + 1 Subtract f(x+1)-f(x+1) = 2x + 1 Divide by 2 f’(x) = x+0.5 f(x) = (x^2 + x)/2 + c f(11) = 66 + c 50 - 66 = c -16 = c So f(x) = (x^2 + x - 32)/2 f(41) = (41^2 + 41 - 32)/2 = 845

[@saeedjaffer532]

If f(x) =(x^2+x-32)/2 then f(x-1)=(x^2-x-32)/2 and f(x) +f(x-1)=x^2-32 not x^2 only. This is due to the mistake of f(11)=50 as given . It should be 66. There's no need to give it. We can assume f(x) is a polynomial of x and solving to have f(x) =(x^2+x)/2. Then f(11)=66

[@RashmiRay-c1y]

You are assuming that1/2[ f(x+1) -f(x-1)] = [f(x+1)-f(x-1)]/{(x+1)-(x-1)] = df/dx which is not necessarily true. It works out in this case.

[@abhirupkundu2778]

@@RashmiRay-c1y Yea, it works only if the denominator approached zero. In this case, it was approaching two if we checked the limit, and hence, this guy's approach was mathematically wrong.

[@boguslawszostak1784]

Subtract f(x+1)-f(x+1) = 2x + 1 ...really? f(x+1)-f(x-1) = 2x + 1 You don’t even know if such a differentiable function exists (and actually it doesn’t), yet you’re already calculating its derivative! But your formula f(x+1)-f(x-1) = 2x + 1 is promising; don't give up!

[@pavelsolaris1901]

This is incorrect reasoning

[@timeonly1401]

Given f(x) + f(x-1) = x² Solving, we get f(x) = x² - f(x-1) Given f(11) = 50. Then: f(12) = 12² - 50 f(13) = 13² - 12² + 50 f(14) = 14² - 13² + 12² - 50 .... f(41) = (41² - 40²) + (39² - 38²) +-... + (13² - 12²) + 50 = 50 + Sum [from k=6 to 20] { (2k+1)² - (2k)² } = 50 + Sum [from k=6 to 20] { (4k² + 4k + 1) - (4k²) } = 50 + Sum [from k=6 to 20] { 4k + 1 } = 50 + 4* Sum [from k=6 to 20] { k } + Sum [from k=6 to 20] { 1 } = 50 + 4* Sum [from k=6 to 20] { k } + 15 = 65 + 4* Sum [from k=6 to 20] { k } = 65 + 4 * ( Sum [from k=1 to 20] { k } - Sum [from k=1 to 5] { k } ) = 65 + 4 * [ (20)(21)/2 - (5)(6)/2 ] = 65 + 4 * ( 210 - 15 ) = 65 + 780 = 845 Done!!

[@alexandreocadiz9967]

This problem smells more like an arithmetic progression problem than a functional equation one. But it was fun anyway ^^

[@zerosir1852]

1. If f(x)=(-x²+13x+7) then find f(0), f(1), f(2), ......,f(5). 2. If f(x)=(x²+x+11) then find f(0), f(1), f(2), ......,f(9). 3. If f(x)=(x²+x+41) then find f(0), f(1), f(2), ......,f(39). You will get some clue.

[@ConradoPeter-hl5ij]

f(x).+ f(x-1) = x² and, f(11)=50 f(41)= ? /////////////////////////////////////////////////// Also, there is another function who makes the same: g(x) + g(x-1) = x² => g(x) = x² - [g(x-1)] => g(x) = x² - [(x-1)² - g(x-2)] => g(x) = x² - (x-1)² + g(x-2) => g(x) = x² - (x-1)² + (x-2)² - g(x-3) => g(x) = x² - (x-1)² + (x-2)² - [(x-3)² -g(x-4)] => g(x) = x² - (x-1)² + (x-2)² - (x-3) +g(x-4) ... => g(x) = x² - (x-1)² + (x-2)² - f(x-3) + ... + (x-2k)² - (x-2k-1)²+...+ 3² - 2² + 1² -0² => g(x) = x² + (-1)¹(x-1)² + (-1)²(x-2)² + (-1)³(x-3)²+ (-1)⁴(x-4)²+...+(-1)ⁿ(x-n)²+...+ 3² -2² + 1²-0² _______________________ | => g(x)= Σ (-1)ⁿ(x-n)² | -------------- *obs.: x²-(x-1)² = 2x-1= x+(x-1) => g(x) = Σ(-1)ⁿ(x-n)² = x+(x-1)+(x-2)+(x-3)+(x-4)++...+5+4+3+2+1 => g(x) = Σ(-1)ⁿ(x-n)² = = (x+1)+(x+1)+(x+1)+(x+1)+(x+1) \____________________________/ (x/2) times => g(x) = Σ(-1)ⁿ(x-n)² = (x+1)(x/2) =Σ(n) this function do the same: ___________________ => | g(x) = [(x+1)(x/2)] | ------------ Demonstration: g(x) + g(x-1) = x² [(x+1)(x/2)]+{(x)[(x-1)/2]}=x² (x/2)[(x+1)+(x-1)]=x² (x/2)[2x]=x² x²=x² ////////////////////////////////////////////////// Coming back to the topic question: f(x) + f(x-1) = x² and, f(11)=50 f(11)+f(10)=11² f(10)=121-50 f(10)=71 and f(12)+f(11)=12² f(12)=144-50 f(12)=94 So, making a table base for f(x): x. f(x) 0. 16 1. -15 2. 19 3. -10 4. 26 5. -1 6. 37 7. 12 8. 52 9. 29 10. 71 11. 50 12. 94 13. 75 14. 121 15. 104

[@ConradoPeter-hl5ij]

Oops. I lose the anwser while trying to send the message. Wait, I will make that again

[@ConradoPeter-hl5ij]

let, l(x)= f(x) - g(x) f(x) + f(x-1)=x² g(x) + g(x-1)=x² =>f(x) + f(x-1) = g(x) + g(x-1) =>f(x) - g(x) = - [f(x-1) - g(x-1)] => l(x)= - l(x-1) this is intersting

[@ConradoPeter-hl5ij]

making a table base for g(x): x. g(x) 0. 0 1. 1 2. 3 3. 6 4. 10 5. 15 6. 21 7. 28 8. 36 9. 45 10. 55 11. 66 12. 78 13. 91 14. 105 15. 120

[@ConradoPeter-hl5ij]

making a table base for l(x): x. l(x) 0. 16 1. -16 2. 16 3. -16 4. 16 5. -16 ... ... 2k. 16 2k+1. -16 ... ... n (-1)ⁿ(16)

[@ConradoPeter-hl5ij]

Therefore, f(41) - g(41) = l(41) f(41) - {(41+1)(41/2)} = (-1)⁴¹(16) f(41) = (-16) + {(42)(41/2)} f(41) = -16 + {(21)(41)} f(41) = -16 + (20+1)(40+1) f(41) = -16 + [800+(20+40)+1] f(41) = -16 + [800+60+1] f(41) = -16 + 861 f(41) = 845 the anwser is f(41) = 845. 😊

[@rushillakdawala4402]

Elementary JEE level question.

[@vafamoshtagh5130]

Solving for f(x) + f(x-1) = x^2, I get f(x) = (x^2+x)/2, resulting in f (11) = 66 not 50. f (41) = 861 not 845. Any ideas?

[@eoyount]

You're missing a constant term. f(x) = (x^2 + x)/2 + C. Then when you plug in x=11, you get C = -16 and everything works out.

[@mscardholder]

​@@eoyountit is not constant, but alternating for even and odd. So it should be 16*(-1)^x or 16cos(pi*x)

[@eoyount]

@@mscardholder You're right. I forgot it was a sum so the constants wouldn't cancel out. Need that alternating sign to get them to cancel.

[@ChinoBlack24]

could someone tell whats the derivate of e^x?

[@giuseppemalaguti435]

41^2-40^2+39^2-38^2.....+15^2-14^2+13^2-f(12)=81+77+73...+29+169-144+f(11)=81+77+73+....+25+50=845

[@kateknowles8055]

f(11) +f(10) = 121 f(12) + f(11) = 144 f(12) - f(10) = 144-121 = 23 f(13) + f(12) = 169 f(13) - f(11) = 169 - 144 = 25 f (x) - f(x-2) = 2x-1 f (13) = 25 + f(11) = 25 + 50 = 75 f(x) = 2x -1 + f (x-2) f (15) = 29 + f(13) = 29 + 75 = 104 f(x) = 4x -6 +f(x-4) f (15) = 54 + f(11) = 54+ 50 = 104 f(x) = nx +f(x-n) - n(n-1)/2 ls not viable f(19) = 70 + 104 =174 by checking with n = 3 and x = 15 we have f(15) = 45 +f (12) -3 so f(12) = 104 - 42 = 62 f(11) = 50 , this is given f(10) is computed from 121 and 50 . It is 71. Checking validity of a formula for even values of x, because they are haywire. f(12) is computed from 144 and 50. It is 94. f(12) - f(10) = 23 = 94 -71 To be rigorous , it is safe to prove by induction what is intuitive here. These snail steps are safe also if double-checked. 🐌🐌🐌🐌🐌🐌 f(17) = 4x17-6 + f(13) = 62 + 75 = 137 (confessing I slipped a bit here) f( 14) is computed from 196- f(13) . f(14) = 196 - 75 = 121 It is also f(14) = 56-6 + f(10) = 50 +71 = 121 But it is f(41) which is the goal. f(41) = 41x4-6 + f(37) It is 158 + 4x37-6 + f(33). It is 8 x 39-12 +f(33)= It is 8x 41 - 28 +f(33) Now jumping down in eights: 🐇🐇🐇🐇🐇🐇🐇🐇 f(33) = 33 x 8 -28 + f(25) f(25) = 25x8 -28 + f(17) = 200-28 + 137 = 309 f(33) = 264 -28 + 314 = 236+309 = 545 f(41) = 8x41-28 + 550 = 328-28 +545 . It is 845 Could develop more algebra to this series, but I am going to follow this podcast now that the kangaroo is about to bound away 🦘 bye-bye

[@zzjdkjx]

f(x)=ax^2+bx+c ax^2+bx+c+a(x-1)^2+b(x-1)+c=x^2 2ax^2+(2b-2a)x+2c-b+a=x^2 2a=1,2b-2a=0,2c-b+a=0 a=1/2,b=1/2,c=0 f(x)=x^2/2+x/2 f(41)=41X41/2+41/2=861 f(41)=861 OK

[@cyruschang1904]

f(x) + f(x - 1) = x^2 f(x) = x^2 - f(x - 1) f(12) = 12^2 - 50 f(13) = 13^2 - (12^2 - 50) f(14) = 14^2 - 13^2 + (12^2 - 50) f(15) = 15^2 - 14^2 + 13^2 - (12^2 - 50) f(41) = 41^2 - 40^2 + 39^2 ... + 13^2 - 12^2 + 50 = 81 + 77 + 73 + 69 + ... + 25 + 50 = (81 + 25)(15)/2 + 50 = 53(15) + 50 = 845

[@oscaramorim7234]

👍💯

[@btb2954]

f(x)=x^2-f(x-1) f(12)=12^2-50 f(13)=13^2-12^2+50 >> (13-12)(13+12)+50 >> 13+12+50 f(15)= 15^2-14^2+13^2-12^2+50 >> (15-14)(15+14)+(13-12)(13+12)+50 >> 15+14+13+12+50 this continues for every odd n so f(41) = 41+40+39...+12+50 = 41(42)/2 - 11(12)/2 + 50 = 845

[@boguslawszostak1784]

"by brutal force", but works

[@sergiykanilo9848]

f(x)=x^2-f(x-1)=x^2-(x-1)^2+f(x-2) =(x+(x-1))*(x-(x-1))+f(x-2) =x + (x-1) + f(x-2) f(41) =41+40+39+38+...+13+12+f(11) =(41+12)*(41-12+1)/2+50 =485

[@FabienLegendre-v7u]

f(x)=x²-f(x-1) f(41)=41²-f40) f(41)=41²-40²+f(39) f(41)=41²-40²+39²-38²+f(37) f(41)=41²-40²+.......+13²-12²+f(11) =(41-40)(41+40)+..+(13-12)(13+12) +f(11) =81+77+73+...+29+25+f(11) Un=U0+4n U0=25 81=25+4n........n=14 f(41)=(14+1)(81+25)/2+f(11) f(41)=795+f(11)=795+50=845 Or : .....f(n)=(n²+n+32×(-1)^n)/2

[@dominiquelarchey-wendling5829]

f(x)+f(x-1)-f(x-1)-f(x-2)=x²-(x-1)² f(x)-f(x-2)=2x-1 g(x) := f(2x+1) g(x+1)-g(x)=4(x+1)+1 g(x+n)-g(x)=4(n(n+1)/2+nx)+n f(41)-f(11)=g(20)-g(5)=g(15+5)-g(5)=4(120+75)+15=795 f(41)=845

[@phill3986]

👍👏❤️😊👏👍

[@ArminVollmer]

In Mathematica, it's a oneliner: f[x_]:=f[x]=x^2-f[x-1];f[11]=50;f[41] yields 845.

[@mircoceccarelli6689]

f( 41 ) = 50 + S( 14 ) S( n ) = g( 0 ) + g( 1 ) + ... + g( n ) g( n ) = 25 + 4n , 0

[@lifeofheaven-co4fr]

Omg crazy😊 🤪

[@ConradoPeter-hl5ij]

f(x) + f(x-1) = x² g(x) = (x+1)(x/2) also, g(x)+g(x-1)=x² then, f(x) + f(x-1) = g(x) - g(x-1) f(x) - g(x) = (-1)*[ f(x-1) - g(x-1)] making l(x)= f(x) - g(x) l(x) = (-1)*[ l(x-1) ] l((x) = -l(x-1) making f(0): f(0) =1² - ...(8² - {9² - [10² - (11² - 50)]}) f(0)= 1-(4-(9-(16-(25-(36-(49-(64-(81 -(100-(121-50)))))))))) => f(0)=16 searching l(0): l(0)=f(0) - g(0) l(0)=16 - 0 => l(0)=16 then, l(1) = -16 and l(2) = 16 ... therefore, l(n)= (-1)ⁿ(16) also, l(n)=f(n) - g(n) f(n)=l(n) + g(n) therefore, ■ f(n)=[(-1)ⁿ(16)] + [(n+1)(n/2)] making n=41 f(41)= -16 +[(42)(41/2)] f(41)= -16 +(21)(41) f(41)= -16 + 861 f(41)= 845