Find The Range of Function.

  • Thread starter klen
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  • #1
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A function f :[0,1]→ℝ is twice differentiable with f(0)=f(1)=0 such that

f''(x) - 2f'(x) + f(x) ≥ e^x, x ε [0,1].

Then for 0<x<1 which of the following is true
0<f(x)<∞, -1/2<f(x)<1/2, -1/4<f(x)<1, -∞<f(x)<0.

I tried to solve it in the following way:
I tried to take the e^x to the left side and found that the equation becomes,
[(e^-x)f'(x) - (e^-x)f(x)]' ≥ 1, thus on integration,

(e^-x)f'(x) - (e^-x)f(x) -f'(0) ≥ x which is same as

[(e^-x)f(x)]' ≥ x on integration

(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.
I don't know how to proceed from here.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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hi klen! :smile:

(try using the X2 button just above the Reply box :wink:)
(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.

or you could say let g(x) = e-xf(x) …

then g''(x) ≥ 1 :wink:

(btw, how did you get on with your series question?)
 
  • #3
41
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Hi Tiny Tim,

Thanks for the help.

in the series question, we can break the sum as groups of four, like this:
[tex] S_n = \sum_{k=1}^{4} {(-1)^{k(k+1)/2}}{k^2} + \sum_{k=5}^{8} {(-1)^{k(k+1)/2}}{k^2} + ...[/tex]
and so on,
in each of these sums the first two are negative terms and last two are positive,
so the sum evaluates to,
[tex] S_n = \sum_{k=0}^{n} {(32k-12)}[/tex]
this gives the sum to be
[tex] S_n= {16n^2+4n}[/tex]
which is possible only for first and fourth values.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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excellent! :smile:

(good LaTeX too :wink:)
 
  • #5
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in each of these sums the first two are negative terms and last two are positive,
so the sum evaluates to,
[tex] S_n = \sum_{k=0}^{n} {(32k-12)}[/tex]

How do you get to this step? :confused:
 
  • #6
41
1
In these sums for the first terms we have k=1,5,9...
so k is of form k=4i-3, i=1,2,3....
similarly for the next successive terms k is of form 4i-2, 4i-1, 4i.
Since the first two terms are positive and last two are negative we have,
[tex]S_n= \sum_{i=1}^n {-(4i-3)^2} + \sum_{i=1}^n {-(4i-2)^2} + \sum_{i=1}^n {(4i-1)^2} + \sum_{i=1}^n {(4i)^2}[/tex]
which after solving comes,
[tex]S_n= \sum_{i=1}^n {(32i-12)}[/tex]
 
  • #7
3,812
92
In these sums for the first terms we have k=1,5,9...
so k is of form k=4i-3, i=1,2,3....
similarly for the next successive terms k is of form 4i-2, 4i-1, 4i.
Since the first two terms are positive and last two are negative we have,
[tex]S_n= \sum_{i=1}^n {-(4i-3)^2} + \sum_{i=1}^n {-(4i-2)^2} + \sum_{i=1}^n {(4i-1)^2} + \sum_{i=1}^n {(4i)^2}[/tex]
which after solving comes,
[tex]S_n= \sum_{i=1}^n {(32i-12)}[/tex]

That was cool! :cool:

Thanks! :smile:
 

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