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.

[@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

[@alipourzand6499]

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

[@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.

[@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.

[@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!

[@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.

[@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

[@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?

[@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!

[@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

[@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

[@mircoceccarelli6689]

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