Solve this: 53 Suppose px = a0+a1x+a2x2+ +anxn If px≤ex-1-1 - Maths -

Question

Solve this:

53   S u p p o s e   p x   =   a 0 + a 1 x + a
2 x 2 + + a n x n   I f   p x ≤ e x - 1 - 1   f o
r   a l l   x ≥ 0 p   P r o v e   t h a t
  a 1 + 2 a 2 + + n a n ≤ 1

Aarushi Mishra · Accepted Answer

px≤ex-1-1Put x=1p1≤e1-1-1p1≤1-1p1≤0But a≥0, a∈ℝ⇒p1=0Let fx=ex-1-1f1=0px≤ex-1-1px≤fxPut x=1+hp1+h≤f1+hp1+h-0≤f1+h-0p1+h-p1≤f1+h-f1Divide by h both sidesp1+h-p1h≤f1+h-f1hNow take limitlimh→0p1+h-p1h≤limh→0f1+h-f1hlimh→0p1+h-p1h≤limh→0f1+h-f1hBy definition, limh→0p1+h-p1h=p'1 and limh→0f1+h-f1h=f'1p'1≤f'1Now f'x=ex-1f'1=e1-1=1p'x=a1+2a2x+
+nanxn-1p'1=a1+2a2+ +nan⇒a1+2a2+ +nan≤1⇒a1+2a2+
+nan≤1

Solve this: 53 . S u p p o s e p x = a 0 + a 1 x + a 2 x 2 + . . . . . . . + a n x n . I f p x ≤ e x - 1 - 1 f o r a l l x ≥ 0 . p P r o v e t h a t a 1 + 2 a 2 + . . . . + n a n ≤ 1 .

Solve this:

$53 . S u p p o s e p (x) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . + a_{n} x^{n} . I f |p (x)| |\leq| |e^{x - 1} - 1| f o r a l l x \geq 0 . p P r o v e t h a t |a_{1} + 2 a_{2} + . . . . + n a_{n}| \leq 1 .$