SCIP in C++11 ― 1.3.3節 その4

連分数：問題1.37～1.39

再帰使うと、以前シルヴァーマンで反復で書いた時よりはだいぶすっきりするな。

シルヴァーマンで使ったコードを修正したい誘惑に駆られるが、

末尾再帰（5章）を装備するまでは、メモリ消費が読めないので却下。

1.37bの反復版はあまり興味ないのでパス。

                                 

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const double continuedFraction

(const function<double(int)>& numeratorSequence,

 const function<double(int)>& denominatorSequence,

 const int orderMax)

{

    function<double(int)> partialContinuedFraction;

    partialContinuedFraction=

        [&partialContinuedFraction,

         &numeratorSequence,&denominatorSequence,orderMax]

        (const int order){

        if(order==orderMax){return(numeratorSequence(order)

                                   /denominatorSequence(order));}

        else{return(numeratorSequence(order)

                    /(denominatorSequence(order)

                      +partialContinuedFraction(order+1)));}

    };

    return(partialContinuedFraction(0));

}

const double denominatorE_2(const int order){

    if(1==order%3){return(static_cast<double>((order/3+1)*2));}

    else{return(1.0);}

}

const double tanCF(const double x, const double orderMax)

{

    const function<double(int)>

        numerator=[x](const int order){

        if(0==order){return(x);}

        else{return(-x*x);}

    };

    const function<double(int)>

        denominator=[](const int order){

        return(static_cast<double>(order*2+1));

    };

    return(continuedFraction(numerator,denominator,orderMax));

}

int main(int argc, char** argv)

{

    int orderMax(15);

    const double goldenRatio((sqrt(5.0)+1)/2);

    for(int order=0;order<orderMax;++order){

        const double goldenRatioCF

            (1.0/continuedFraction([](const int i){return(1.0);},

                                   [](const int i){return(1.0);},

                                   order));

        const double relativeError

            (abs((goldenRatio-goldenRatioCF)/goldenRatio));

        if(relativeError<0.0001){

            cout<<setprecision(16)<<"Excersize 1.37:"<<endl

                <<"k="<<order+1<<endl

                <<"golden ratio with CF="<<goldenRatioCF<<endl

                <<"exact value="<<goldenRatio<<endl

                <<"relative error="<<relativeError

                <<endl<<endl;

            break;

        }

    }

    cout<<"Excersize 1.38:"<<endl<<"e with CF="

        <<continuedFraction([](const int i){return(1.0);},

                           denominatorE_2,orderMax)+2.0<<endl

        <<"e="<<exp(1.0)<<endl<<endl;

    const double pi(3.1415926535897932384626);

    cout<<"Excersize 1.39:"<<endl

        <<"tan(pi/4) with CF="<<tanCF(pi/4.0,orderMax)<<endl

        <<"tan(pi/3) with CF="<<tanCF(pi/3.0,orderMax)<<endl;

    return(0);

}

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出力

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Problem 1.37:

k=10

golden ratio with CF=1.618181818181818

exact value=1.618033988749895

relative error=9.136361346630259e-05

Problem 1.38:

e with CF=2.718281828458563

e=2.718281828459045

Problem 1.39:

tan(pi/4) with CF=0.9999999999999999

tan(pi/3) with CF=1.732050807568877