p-x-1-p-x-1-4x-2-2x-10-p-x-

Question Number 198850 by mustafazaheen last updated on 25/Oct/23

p (x + 1) + p (x - 1) = {4 x}^{2} - 2 x + 10

p (x) = ?

Answered by Rasheed.Sindhi last updated on 25/Oct/23

L e t P (x) = {a x}^{2} + b x + c

P (x + 1) + P (x - 1)

= (a {(x + 1)}^{2} + b (x + 1) + c) - (a {(x - 1)}^{2} + b (x - 1) + c)

= a ({(x + 1)}^{2} + {(x - 1)}^{2}) + b ((x + 1) + (x - 1)) + 2 c

= 2 a (x^{2} + 1) + 2 b (x) + 2 c

= 2 {a x}^{2} + 2 a + 2 b x + 2 c = {4 x}^{2} - 2 x + 10

2 a = 4, 2 b = - 2, 2 a + 2 c = 10

a = 2, b = - 1, c = 5 - a = 5 - 2 = 3

P (x) = 2 x^{2} - x + 3

Commented by mr W last updated on 25/Oct/23

a r e t h e r e o t h e r p o s s i b i l i t i e s ?

f o r e x a m p l e

p (x) = 2 x^{2} - x + 3 + f (x)

w i t h f (x) = - f (x - 2)

b u t i^{'} m n o t s u r e i f s u c h a f (x) e x i s t s .

Commented by mr W last updated on 25/Oct/23

g r e a t!

t h a t m e a n s

p (x) = 2 x^{2} - x + 3 + k \sin \frac{(2 n + 1) π x}{2}

i s a l s o a s o l u t i o n .

Commented by Rasheed.Sindhi last updated on 25/Oct/23

I^{'} m a l s o n o t s u r e \boldsymbol s i r .

Commented by Frix last updated on 25/Oct/23

f (x) = \sin \frac{(2 n + 1) π x}{2}

Commented by witcher3 last updated on 25/Oct/23

f (x) = - f (x - 2)

f (x - 2) = - f (x - 4) = - f (x)

\Rightarrow f (y) = f (y + 4), y periodic 4

\sin (\frac{π}{2} x)

\sin (\frac{π}{2} (x - 2)) = \sin (\frac{π x}{2} - π) = - \sin (\frac{π x}{2}) . . worck

Commented by mr W last updated on 26/Oct/23

g r e a t!

g r e a t!

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