Unable to find answer
- Thread starter mimie
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Suppose:
[MATH]f(x)=\sin(2x)\ln(x+1)[/MATH]
[MATH]f(0)=0[/MATH]
Then:
[MATH]f'(x)=\frac{\sin(2x)}{x+1}+2\cos(2x)\ln(x+1)[/MATH]
[MATH]f'(0)=0[/MATH]
[MATH]f''(x)=-\frac{\sin(2x)}{(x+1)^2}+\frac{4\cos(2x)}{x+1}-4\sin(2x)\ln(x+1)[/MATH]
[MATH]f''(0)=4[/MATH]
[MATH]f'''(x)=\frac{2\sin(2x)}{(x+1)^3}-\frac{6\cos(2x)}{(x+1)^2}-\frac{12\sin(2x)}{x+1}-8\cos(2x)\ln(x+1)[/MATH]
[MATH]f'''(0)=-6[/MATH]
And so near \(x=0\) we have:
[MATH]f(x)\approx\frac{4x^2}{2!}-\frac{6x^3}{3!}=2x^2-x^3[/MATH]
Here is a plot of the given function (in black) the approximating polynomial the 3 of us have derived (in green) and the approximating polynomial given in the problem statement (in red):
[MATH]f(x)=\sin(2x)\ln(x+1)[/MATH]
[MATH]f(0)=0[/MATH]
Then:
[MATH]f'(x)=\frac{\sin(2x)}{x+1}+2\cos(2x)\ln(x+1)[/MATH]
[MATH]f'(0)=0[/MATH]
[MATH]f''(x)=-\frac{\sin(2x)}{(x+1)^2}+\frac{4\cos(2x)}{x+1}-4\sin(2x)\ln(x+1)[/MATH]
[MATH]f''(0)=4[/MATH]
[MATH]f'''(x)=\frac{2\sin(2x)}{(x+1)^3}-\frac{6\cos(2x)}{(x+1)^2}-\frac{12\sin(2x)}{x+1}-8\cos(2x)\ln(x+1)[/MATH]
[MATH]f'''(0)=-6[/MATH]
And so near \(x=0\) we have:
[MATH]f(x)\approx\frac{4x^2}{2!}-\frac{6x^3}{3!}=2x^2-x^3[/MATH]
Here is a plot of the given function (in black) the approximating polynomial the 3 of us have derived (in green) and the approximating polynomial given in the problem statement (in red):