Using a Power Series to Approximate a Definite Integral

[DivGradCurl]

I think I got pretty close to the answer to this problem. However, I just can't obtain the right approximation at the end. Please, help me find where I made a mistake.

Thanks.

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Use a power series to approximate the definite integral to six decimal places.

\int _0 ^{1/3} x^2 \arctan \left( x^4 \right) dx

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\frac{d}{dx} \left[ \arctan \left( x^4 \right) \right] = \frac{4x^3}{1+x^8}

\frac{1}{1+x^8}=\frac{1}{1-\left( -x^8 \right) } = \sum _{n=0} ^{\infty} \left( -x^8 \right) = \sum _{n=0} ^{\infty} \left( -1 \right) ^n x^{8n}

\frac{d}{dx} \left[ \arctan \left( x^4 \right) \right] = 4 \sum _{n=0} ^{\infty} \left( -1 \right) ^n x^{8n+3}

\arctan \left( x^4 \right) = 4 \sum _{n=0} ^{\infty} \int \left( -1 \right) ^n x^{8n+3} dx = \mathrm{C} + \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{8n+4}}{8n+4}

x^2 \arctan \left( x^4 \right) dx = \mathrm{C} + \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{8n+6}}{8n+4}

\int x^2 \arctan \left( x^4 \right) dx = \mathrm{C} + \sum _{n=0} ^{\infty} \int \frac{\left( -1 \right) ^n x^{8n+6}}{8n+4} = \mathrm{C} + \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{8n+7}}{\left( 8n+4 \right) \left( 8n+7 \right)} = \mathrm{C} + \frac{x^7}{28} - \frac{x^{15}}{180} + \frac{x^{23}}{460} - \frac{x^{31}}{868} + \cdots

\int _0 ^{1/3} x^2 \arctan \left( x^4 \right) dx = \left[ \frac{x^7}{28} - \frac{x^{15}}{180} + \frac{x^{23}}{460} - \frac{x^{31}}{868} + \cdots \right] _0 ^{1/3} = \frac{1}{2^2\cdot 3^7\cdot 7} - \frac{1}{2^2\cdot 3^{17} \cdot 5} + \frac{1}{2^2\cdot 3^{23} \cdot 5 \cdot 23} - \frac{1}{2^2 \cdot 3^{31} \cdot 7 \cdot 31} + \cdots

b_1 = \frac{1}{2^2\cdot 3^{17} \cdot 5} \approx 3.9 \times 10^{-10} < 10^{-6} \Longrightarrow \int _0 ^{1/3} x^2 \arctan \left( x^4 \right) dx \approx \frac{1}{2^2\cdot 3^7\cdot 7} - \frac{1}{2^2\cdot 3^{17} \cdot 5} \approx 0.000016

The correct answer would be 0.000065.

 

[NateTG]

You appear to be ignoring the x^2 when you're determining the coefficients for your power series.

 

[DivGradCurl]

Oh... I did not keep the 4 factor all the way down. Thanks.

 

[nrqed]

I think I got pretty close to the answer to this problem. However, I just can't obtain the right approximation at the end. Please, help me find where I made a mistake.

Thanks.

--------------------------------------------------------------

Use a power series to approximate the definite integral to six decimal places.

\int _0 ^{1/3} x^2 \arctan \left( x^4 \right) dx

--------------------------------------------------------------

\frac{d}{dx} \left[ \arctan \left( x^4 \right) \right] = \frac{4x^3}{1+x^8}

\frac{1}{1+x^8}=\frac{1}{1-\left( -x^8 \right) } = \sum _{n=0} ^{\infty} \left( -x^8 \right) = \sum _{n=0} ^{\infty} \left( -1 \right) ^n x^{8n}

\frac{d}{dx} \left[ \arctan \left( x^4 \right) \right] = 4 \sum _{n=0} ^{\infty} \left( -1 \right) ^n x^{8n+3}

\arctan \left( x^4 \right) = 4 \sum _{n=0} ^{\infty} \int \left( -1 \right) ^n x^{8n+3} dx = \mathrm{C} + \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{8n+4}}{8n+4}



x^2 \arctan \left( x^4 \right) dx = \mathrm{C} + \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{8n+6}}{8n+4}

Why not simply use the Taylor series of Arctan? It's simply

ArcTan (x^4) = x^4 - {x^{12} \over 3} + {x^{20} \over 5} - {x^{28} \over 7} + \ldots



Notice that you willl run into a probllem with your constant of integration "C" since it will get multiplied by x^2 later and you would have to integrate it as well! So you must get rid of it. The way to do so is of course to impose that Arctan(0) =0.

But again, using the Taylor series of ArcTan seems simpler to me.


Pat