1thanx bhaiya......was missing u and ur posts.......
1) If f : R→R is a continuous and differentiable function such that
\int_{-1}^{x}{f(t)dt} + f'''(3) \int_{x}^{0}{dt} = \int_{1}^{x}{t^{3}dt} - f'(1) \int_{x}^{2}{t^{2}dt} + f''(2) \int_{3}^{x}{tdt},
then find the value of f'(4).
2) If mod(x) < 1, then prove that
1 - 2x1 + x +x2 + 2x - 4x31 - x2 + x4 + 4x3 - 8x71 - x4 + 8 + ......∞ = 1 + 2x1 + x + x2
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3 Answers
62The first question
\int_{-1}^{x}{f(t)dt}+f'''(3)\int_{x}^{0}{dt}=\int_{1}^{x}{t^{3}dt}-f'(1)\int_{x}^{2}{t^{2}dt}+f''(2)\int_{3}^{x}{tdt} \\\int_{-1}^{x}{f(t)dt}-f'''(3)\int_{0}^{x}{dt}=\int_{1}^{x}{t^{3}dt}+f'(1)\int_{2}^{x}{t^{2}dt}+f''(2)\int_{3}^{x}{tdt}
Now differentiate wrt to x to get...
f(x)=f'''(3)+x^3+f'(1)x^2+f''(2)x
This is a cubic.. so differentiate 3 times to get
f'''(x)=6 so we get f'''(3)=6
similary find f'(1) and f''(2)
Hence we have found f..
Remaining work can be carrried out nicely by you :)
62The second post has a small mistake in the first term.. it should be 1-x+x^2 in the denominator..
the the LHS can be thought of as the derivative of
-\ln(1-x+x^2)+\ln(1-x^2+x^4)+\ln(1-x^4+x^8)+......=-\ln\left( (1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)+......\right)
=-\ln\left( \frac{1+x^3}{1+x}\times\frac{1+x^3}{1+x}\times\frac{1+x^3}{1+x}\times....\right)
=-\ln\left( \frac{1+x^3}{1+x}\times\frac{1+x^6}{1+x^2}\times\frac{1+x^{12}}{1+x^4}\times....\right)
Now we multiply and divide inside the logarithm by
=\frac{1-x^3}{1-x} to trigger a chain of collapses which reaches....
=-\ln\frac{1-x}{1-x^3} = ln(1+x+x^2)
Now if we differentiate both sides... we get the required answer
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