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I've tried somehow using Taylor to try and figure this one out.

Unfortunately, I couldn't seem to get a solid answer. Thank you very much for your help!

Let f be a continuos function, $$f:R\rightarrow R$$ $$f(x), \ f'(x) \ , \ f''(x) \ \text{are continuous} $$

and let the integral $$\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx \space \text{exist and be finite}$$

Find the value of $$f(0)\ \text{and}\ |f'(0)|$$

egreg
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Simba
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  • Do not use displaystyle in titles. Please add your thoughts. – Did May 13 '13 at 15:42
  • sorry, the only reason ive done that is because my last post was edited by someone and he changed it to displaystyle. – Simba May 13 '13 at 15:44
  • Integral $\int g$ to exist it is necessary that $g$ is finite everywhere. Only possible if$ f(x) = 1 + 4x^2 + kx^n$ which means f(0) = 1 and f'(0) = 0 – alekhine May 13 '13 at 15:46
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    You're wrong on two counts, alekhine. You don't need the function to be bounded at the origin. And you didn't expand correctly either. See my answer. – not all wrong May 13 '13 at 16:06
  • @eyal: Which post was that? – Did May 13 '13 at 16:47
  • @Did it was this one http://math.stackexchange.com/questions/389527/int-01-fracfxxp-exists-and-finite-implies-f0-0 and sorry for the delayed answer. thank you. – Simba May 17 '13 at 12:04
  • @eyal Nobody introduced displaystyle in your other post. – Did May 17 '13 at 12:47
  • @Did i think i dont understand what do you mean by displaystyle ? i thought u ment that i inserted lyx code in the headline.. what did u mean? – Simba May 17 '13 at 12:53
  • LaTeX has two styles for maths, textstyle (encoded by $...$) and displaystyle (often encoded by $$...$$). The latter is made for displayed maths, the former for intext maths. One should restrict to textstyle in titles of posts on MSE. – Did May 17 '13 at 13:09

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Let $a=f(0)-1, b=f'(0)$. Then continuity of the second derivative gives $f(x)= 1+a+ bx+O(x^2)$ and hence the integrand is $$\frac{(a+bx+O(x^2))^2-4x^2}{x^{3.5}} = \frac{a^2+2abx+(b^2+a\times\cdots-4)x^2+O(x^3)}{x^{3.5}}$$ A function like $x^c$ is integrable at the origin only if $c>-1$ (check) and hence the first term in the numerator which may be nonzero is the cubic term.

You should be able to figure out what $a,|b|$ are now.

not all wrong
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  • can u please explaine the how why the equaition or the LHS and RHS are equal? i cant seem to understand how u got there – Simba May 13 '13 at 16:10
  • $(a+b+c)^2= a^2+ 2ab + (b^2+2ac) +(2bc+ c^2)$ right? That's just expanding a square out and grouping the terms together. That's all I did, except that I noticed that we don't care about terns that have $x^3$ or higher in them. – not all wrong May 13 '13 at 16:15
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    @eyal Knowing the answer is helpful I guess :P, you will find $ f(0) = 1 $ and $ |f'(0)| = 2 $, you may go along the given solution. – smiley06 May 13 '13 at 17:36
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    @smiley06 I don't leave bits out from spite, y'know! The idea is to get people over the stumbling blocks, rather than doing everything for them; it's more helpful in the long run P: – not all wrong May 13 '13 at 17:47
  • @Sharkos Ok i can show that $f(0) = 1 $ but im still not sure how to show that $|f'(0)=2| $ or maybe im doing something wrong? any hints? guidence? thank you! – Simba May 20 '13 at 12:27
  • What can the coefficient of $x^2$ in my numerator be? Now since $a=0$ you have an equation involving $b$. But $b=f'(0)$... – not all wrong May 20 '13 at 14:36