Functional Equation

Calculus Level 5

Given a function f : R → R f :\mathbb{R} \rightarrow \mathbb{R} such that

f ( x ) = 2014 x + ∫ 0 x 201 4 t f ( x − t ) d t , f(x)=2014x+\int_{0}^{x} 2014^t f(x-t) \ dt,

find the value of f ( 1 2 ) . f \left(\dfrac{1}{2}\right).


The answer is 2873.98.

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Pratik Shastri by Pratik Shastri , 35, India

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1 solution

f ( x ) = 2014 x + ∫ 0 x 2014 t f ( x − t ) . d t f ( x ) − 2014 x = ∫ 0 x 2014 t f ( x − t ) . d t f ( x ) − 2014 x = 2014 x ∫ 0 x f ( t ) 2014 t . d t . . . . . . ( 1 ) f ′ ( x ) − 2014 = l n ( 2014 ) [ f ( x ) − 2014 x ] + f ( x ) L e t f ( x ) = y : d y d x − ( l n ( 2014 ) + 1 ) y = 2014 − ( 2014 l n ( 2014 ) ) x S o l v i n g t h i s d i f f e r e n t i a l e q u a t i o n a n d p l u g i n g x = 1 2 f\left( x \right) \quad =\quad 2014x\quad +\quad \int _{ 0 }^{ x }{ { 2014 }^{ t } } f\left( x-t \right) .dt\\ f\left( x \right) \quad -\quad 2014x\quad =\quad \int _{ 0 }^{ x }{ { 2014 }^{ t } } f\left( x-t \right) .dt\\ f\left( x \right) \quad -\quad 2014x\quad =\quad { 2014 }^{ x }\int _{ 0 }^{ x }{ \frac { f\left( t \right) }{ { 2014 }^{ t } } } .dt\quad ......(1)\\ f'\left( x \right) \quad -\quad 2014\quad =\quad ln(2014)[f\left( x \right) -2014x]\quad +\quad f\left( x \right) \\ \\ Let\quad f\left( x \right) =y\quad :\\ \\ \frac { dy }{ dx } \quad -(ln(2014)+1)y\quad =\quad 2014\quad -\quad (2014ln(2014))x\\ \\ Solving\quad this\quad differential\quad equation\quad and\quad pluging\quad x\quad =\quad \frac { 1 }{ 2 }

f ( 1 2 ) = 2873.9 \boxed{\quad f\left( \frac { 1 }{ 2 } \right) =2873.9}

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f ( x ) = 2014 x + ∫ 0 x 2014 t f ( x − t ) d t f ( x ) − 2014 x = ∫ 0 x 2014 t f ( x − t ) d t f ( x ) − 2014 x = 2014 x ∫ 0 x f ( t ) 2014 t d t f ( 0 ) = 0 f ′ ( x ) − 2014 = l n ( 2014 ) [ f ( x ) − 2014 x ] + f ( x ) u s i n g a = 2014 a n d b = l n ( a ) + 1 = l n ( a e ) a n d b − l n ( a ) = 1 f ′ ( x ) = l n ( a e ) f ( x ) + a − a l n ( a ) x f ′ ( x ) = b f ( x ) + a − a l n ( a ) f\left( x \right) \quad =\quad 2014x\quad +\quad \int _{ 0 }^{ x }{ { 2014 }^{ t } } f\left( x-t \right) dt\\ f\left( x \right) \quad -\quad 2014x\quad =\quad \int _{ 0 }^{ x }{ { 2014 }^{ t } } f\left( x-t \right) dt\\ f\left( x \right) \quad -\quad 2014x\quad =\quad { 2014 }^{ x }\int _{ 0 }^{ x }{ \frac { f\left( t \right) }{ { 2014 }^{ t } } } dt\\ f\left( 0 \right) =0\quad \\ f'\left( x \right) \quad -\quad 2014\quad =\quad ln(2014)[f\left( x \right) -2014x]\quad +\quad f\left( x \right) \\ using\quad \\ a=2014\quad and\quad b=ln\left( a \right) +1=ln\left( ae \right) \quad and\quad b-ln\left( a \right) =1\\ f'\left( x \right) =\quad ln\left( ae \right) f\left( x \right) +a-aln\left( a \right) x\\ f'\left( x \right) =\quad bf\left( x \right) +a-aln\left( a \right)

Taking Laplace transformation

s F ( s ) = b F ( s ) + a s − a l n ( a ) s 2 F ( s ) = a { s − l n ( a ) s 2 ( s − b ) } F ( s ) = a { ( b − l n a ) b 2 [ 1 s − b − 1 s ] + l n ( a ) b s 2 } sF\left( s \right) =bF\left( s \right) +\frac { a }{ s } -\frac { aln\left( a \right) }{ { s }^{ 2 } } \\ F\left( s \right) =a\left\{ \frac { s-ln\left( a \right) }{ { s }^{ 2 }\left( s-b \right) } \right\} \\ F\left( s \right) =a\left\{ \frac { \left( b-lna \right) }{ { b }^{ 2 } } \left[ \frac { 1 }{ s-b } -\frac { 1 }{ s } \right] +\frac { ln\left( a \right) }{ b{ s }^{ 2 } } \right\}

Taking Inverse Laplace transformation

f ( x ) = a { ( b − l n a ) b 2 [ e b x − 1 ] + l n ( a ) b x } f ( 1 2 ) = a { ( b − l n a ) b 2 [ e 0.5 b − 1 ] + l n ( a ) 2 b } f ( 0.5 ) = 2014 { 1 [ l n ( 2014 e ) ] 2 [ e 0.5 l n ( 2014 e ) − 1 ] + l n ( 2014 ) 2 l n ( 2014 e ) } f ( 0.5 ) = 2014 { [ 2014 e − 1 ] [ l n ( 2014 e ) ] 2 + l n ( 2014 ) 2 l n ( 2014 e ) } f ( 0.5 ) = 2873.98 f\left( x \right) =a\left\{ \frac { \left( b-lna \right) }{ { b }^{ 2 } } \left[ { e }^{ bx }-1 \right] +\frac { ln\left( a \right) }{ b } x \right\} \\ f\left( \frac { 1 }{ 2 } \right) =a\left\{ \frac { \left( b-lna \right) }{ { b }^{ 2 } } \left[ { e }^{ 0.5b }-1 \right] +\frac { ln\left( a \right) }{ 2b } \right\} \\ f\left( 0.5 \right) =2014\left\{ \frac { 1 }{ { \left[ ln\left( 2014e \right) \right] }^{ 2 } } \left[ { e }^{ 0.5ln\left( 2014e \right) }-1 \right] +\frac { ln\left( 2014 \right) }{ 2ln\left( 2014e \right) } \right\} \\ \\ f\left( 0.5 \right) =2014\left\{ \frac { \left[ \sqrt { 2014e } -1 \right] }{ { \left[ ln\left( 2014e \right) \right] }^{ 2 } } +\frac { ln\left( 2014 \right) }{ 2ln\left( 2014e \right) } \right\} \\ \\ f\left( 0.5 \right) =2873.98

Rishabh Deep Singh - 4 years, 1 month ago

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