Function

is it any new thing?

does not f = sin satisfy the condition?

can u please check if the application of taylor expansion was valid here

i applied it without being much aware of it

http://www.targetiit.com/iit-jee-forum/posts/equations-pls-help-18306.html

sorry, i was not making myself clear. I meant fⁿ(0) = 0.

the given function is infinitely differentiable everywhere. But that does not guarantee that it has a taylor expansion that converges to f everywhere. There are only few examples of such functions (that are known as entire functions in complex analysis)

sir can u please show how fⁿ(x) = 0

what i can see is fⁿ(0) = 0

so by maclaurine expansion (or putting a =0 in taylor expansion)

f(x) = f(0) + f'(0)x/1! + f''(0) x²/2! +.....
f(x) = 0

You can google taylor expansion. this is just indicative, this is not a rigorous proof by any means! The taylor expansion may not be valid everywhere.

Also, you can see that the function is infinitely differentiable and for any fⁿ(x) = 0.

That indicates by Taylor expansion f(x) = f(0) = 0.

f^''+f=0

multiply both sides by f^' and integrate wrt x

so (f'(x))² + f(x)² = 0

so both are zero

gr8 work by qwerty

sorry for wasting ur time friends...i got it now!
thanx a lot

f^" + f = 0

integrating both sides from 0 to x

₀∫^x f dx = - f ^'(x) [ f'(0) = 0 ]

now consider only the first quadrant ,

and consider an infintesimally small x coordinate = h ,

if we assume f ^' (x) to be zero in this small region , then the function is increasing from 0 to h , ad since f(0) = 0 , the graph from 0 to h lies above x axis , and hence area under the graph should be +ve

but we wrote ₀∫^x f dx = - f ^'(x)
and we assumed f^' > 0 , so ₀∫^x f dx will come out -ve ,which is contradictory ,

similarly if we assume f'(x) < 0 we again get contradiction

hence f(x) will neither be increasing nor decreasing in first quadrant

similarly we can prove for second quadrant . hence f is neither increasing noe decreasing

hence proved

F''(x) + F(x) = 0

On putting x = 0 we get, F''(x) = 0
So, F(x) = 0 - F''(x) = 0
and so.... maybe

see i m a student of Xth so i m sorry if there's any mistake and i m not in touch with these problems but only tried.....!!

i never said tht f"(x) + f(x) = 0 alone says f''(x) = f'(x)

i said f"(x) + f(x) = 0 and f'(x)=[f(x+h)-f(x)]/h together say f"(x)=f'(x)

dude ur counter example itself is wrong f(x) can't be sin x....since d(sin x)/dx≠0 when x=0[1]

see wat i did was substitute for f(x) by -f"(x) in definition of f'(x)...it gives f"(x)=f'(x)....isn't it?

In fact it can be proved that

using it we can give an alternative proof of the identity

sin(x+ y) = sin x cos y + cos x siny

consider G(x) = [F'(x)]² + [F(x)]²

G'(x) = 0 so G(x) is constant

G(0) = 0

so G(x) = 0 for all real x

as G(x) = [F'(x)]² + [F(x)]²

F is zero for all real x

@ kunl how u got F''(x) = F'(x) take F = sin and check

using F''(x)+F(x)=0 and definition of F(x) we get F"(x)=F'(x)which is ultimately equal to -F(x).
also we are given F'(0)=F"(0)=-F(0)=0
combining this with the result F''(x)=F'(x)=-F(x) the only way this is possible is when F(x)=0 for all real x.

F''(x)+F(x)=0
Putting x=0
F''(0)+F(0)=0
But given that F(0)=0
So F''(0)=0
Rahulmishra F''(0)=0 butF''(x) need not to be equal to zero.And please check your working also.