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Differential Calculus

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7 Mar 2009 12:08:26 IST

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Let $f: \mathbb{R}\to\mathbb{R}$ be a function that is differentiable infinitely many times and all its derivatives are also continuous. If $f$ satisfies the differential equation
$f^{\prime\prime}(x)+f^\prime(x)-f(x)=0$
for all $x\in [0,\,2009]$ and $f(0)=f(2009)=0$ , find $f(1004)$ .

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Ankit Rana

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7 Mar 2009 21:15:46 IST

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Wait a minute sir, posting the solution

Ankit Rana

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7 Mar 2009 21:48:22 IST

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Let $f: mathbb{R} omathbb{R}$ be a function that is differentiable infinitely many times and all its derivatives are also continuous. If $f$ satisfies the differential equation
$f^{primeprime}(x)+f^prime(x)-f(x)=0$
for all $xin [0,,2009]$ and $f(0)=f(2009)=0$ , find $f(1004)$ .

Apply Rolle's theorem for the interval [0,2009]

so for some p we have f'(p) = 0 since f(2009)=f(0)=0

Now apply mean value theorem for the interval [p,2009]

hence for some q we have

f''(q) = f'(2009) - f'(p) / 2009-p

= f'(2009)/ [2009-p] since f'(p) = 0

Now

f(x) = f ' (x) + f '' (x)

f ' (x) = f '' (x) + f ''' (x)

f '' (x) = f '''' (x) + f '''' (x)

.........

For x = 2009

f(2009) = f ' (2009) + f '' (2009)

but since f(2009) = 0

f ' (2009) = - f '' (2009)

f '' (2009) = -f ''' (2009)

f ''' (2009) = - f '''' (2009)

.......

Hence we have f ' (2009) = f ''' (2009) = f ''''' (2009) = ......... = - f '' (2009) = - f '''' (2009) = ......

But f '' (2009) = f ''' (2009) + f '''' (2009)

and also f ''' (2009) = - f '''' (2009) we have f '' (2009) = 0

Therefore f ' (2009) = f ''' (2009) = f ''''' (2009) = ......... = - f '' (2009) = - f '''' (2009) = ...... = 0

Hence f '' (q) = 0

But f(q) = f ' (q) + f '' (q) = f ' (q)

differentiate above eq. to get f ' (q) = f '' (q) = 0

hence f(q) = 0

Now we can prove that f(q) = f ' (q) = f '' (q) = ...=0 just as we did for x = 2009

Now apply mean value theorem between [p,q] to get a point r such that f(r) = f' (r) = f '' (r) = ..... = 0 just as we did for point q

Same thing can be done with the interval [0,p]

Thus by induction we can prove that all the point between 0 and 2009 are zero

Thus f(x) = 0 for x = [0,2009]

Hence f(1004) = 0

Ankit Rana

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7 Mar 2009 21:53:19 IST

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enjoyed solving this sum....hope my solution is right

Anant Kumar

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7 Mar 2009 23:17:23 IST

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Ankit... your solution is correct but a bit lengthy. Indeed, we prove that $f(x)\equiv0$ for all $x\in[0,\,2009]$ . To see this we first note that if $f$ is constant, it obviously has to be zero identically.
So, let $f$ be non-constant. Then it must attain a maximum and minimum in $[0,2009]$ .
Claim 1: $f$ cannot have a positive maximum. If it had so, then at the point of maximum, $f^\prime =0$ , and $f^{\prime\prime}<0$ so that $f^{\prime\prime}+f^\prime - f<0$ , a contradiction. Hence $f\leq 0$ for all $x\in[0,2009]$ .

Claim 2: $f$ cannot have a negative minimum. If it had so, then at the point of minimum, $f^\prime=0$ , and $f^{\prime\prime}>0$ so that $f^{\prime\prime}+f^\prime - f>0$ , again a contradiction. Hence, $f\geq 0$ for all $x\in[0,2009]$

Together, Claim 1 and Claim 2 imply that $f\equiv 0$ for all $x\in[0,2009]$ .

Ankit Rana

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8 Mar 2009 00:13:40 IST

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absolutely right sir, this method must have striked b4 the lengthy one

sometimes previous problems do affect the thinking (in the last prob. intermediate value theorem was used so my thinking was diverted)

such thinking (the one in ur solution) does help in many cases specially when we deal with problems based on parity and odd/even

just prove that one term cannot be odd and even at the same time just as in the above prob. required to prove that f(x) cannot be =ve and -ve aat the same time and hence it ought to be zero.

nice solution short and sweet (lol) :D :D :D

Hari Shankar

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8 Mar 2009 11:12:14 IST

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@Ankit: We have f(2009) = f'(2009)+f"(2009). Since it is given f(2009) = 0, We have f'(2009) = -f"(2009). This much is clear.

Now, further, how do we arrive at f"(2009) = -f"'(2009) and the rest of the equations? In fact fⁿ(x) = -2 f^n-1(x).

Hari Shankar

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8 Mar 2009 12:35:51 IST

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Sorry, it should be fⁿ(x) = -n f^n-1(x)

Ankit Rana

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8 Mar 2009 15:12:09 IST

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hmm I thought for x=2009 f ' (x) = - f '' (x)

so diff. both sides to get f '' (x) = -f ''' (x)

but now I get it if two fuctions are equal at some point (or if we have some relation between them) then the equality may not be true for its derivative

plz. help sir, how would u prove it using Rolle's and lagrange's mean value theorem

Hari Shankar

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8 Mar 2009 15:53:05 IST

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I dont know if it can be solved on these lines. I did not know how to do it. I have seen this problem in the book Berkeley Problems in Mathematics (or something like that) and the solution is the same as the one kaymant sir has given.

Anant Kumar

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8 Mar 2009 18:56:00 IST

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Now that Bhatt sir has pointed it out, the line of action followed by Ankit is incorrect ( probably I got carried away with the flow).
However, we can also apply brute force. We first solve the differential equation (which is easy to do):
The auxiliary equation is $lambda^2 + lambda -1=0$ . So the eigenvalues are
$lambda_1=dfrac{-1+ sqrt{5}}{2}$ and $lambda_2=dfrac{-1- sqrt{5}}{2}$
So the general solution is $f(x)=Ae^{lambda_1x} + Be^{lambda_2 x}$
To find the solution satisfying the given boundary conditions, we see that
$f(0)=A+B =0 quad Rightarrow B=-A$
and $f(2009)=A(e^{lambda_12009}-e^{lambda_22009})=0quad Rightarrow A=0$ , since the quantity in the bracket is non-zero. And so $A=B=0$ . As such $f(x)equiv 0$ for all $xin [0,2009]$

Hari Shankar

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8 Mar 2009 21:14:17 IST

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thats what i tried to do first. i assumed a solution of the form ae^bx. But I have forgotten differential equations, so i was not sure whether this is the only solution.

Of course, the problem did not require it. Very nice problem, really.

'',' Off to a nu start!!

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8 Mar 2009 21:16:35 IST

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but the soln given by anant sir was quite gud!! really enjoyed dis 1....

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