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Senin, 17 Januari 2011

bukti ketidaksamaan Cauchy-Schwarz untuk Integral

Ketidaksamaan Cauchy-Schwarz untuk integral :

$(\int_{a}^{b}f(x)g(x)dx)^2\leq \int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx$

Bukti:

perhatikan integral lipat dua dari

F(x,y)=(f(x)g(y)-f(y)g(x))^2

atas persegi panjang

$R={(x,y):a\leq x\leq b,a\leq y\leq b}$

maka

$\int_{a}^{b}\int_{a}^{b}[f(x)g(y)-f(y)g(x)]^2dydx\geq 0$

$\int_{a}^{b}\int_{a}^{b}[f(x)g(y)-f(y)g(x)]^2dydx=\int_{a}^{b}\int_{a}^{b}(f^2(x)g^2(y)+f^2(y)g^2(x)-2f(x)g(x)f(y)g(y))dydx\Rightarrow$

$\int_{a}^{b}\int_{a}^{b}[f(x)g(y)-f(y)g(x)]^2dydx=\int_{a}^{b}\int_{a}^{b}f^2(x)g^2(y)dydx+\int_{a}^{b}\int_{a}^{b}f^2(y)g^2(x)dydx-2\int_{a}^{b}\int_{a}^{b}f(x)g(x)f(y)g(y)dydx\Rightarrow$

$\int_{a}^{b}\int_{a}^{b}[f(x)g(y)-f(y)g(x)]^2dydx=\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(y)dy+\int_{a}^{b}f^2(y)dy\int_{a}^{b}g^2(x)dx-2\int_{a}^{b}\int_{a}^{b}f(x)g(x)dx\int_{a}^{b}f(y)g(y)dy\geq 0$

$2\int_{a}^{b}f(x)g(x)dx\int_{a}^{b}f(y)g(y)dy\leq \int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(y)dy+\int_{a}^{b}f^2(y)dy\int_{a}^{b}g^2(x)dx$

$2\int_{a}^{b}f(x)g(x)dx\int_{a}^{b}f(x)g(x)dx\leq \int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx+\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx$

$2\int_{a}^{b}f(x)g(x)dx\int_{a}^{b}f(x)g(x)dx\leq 2\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx$

$(\int_{a}^{b}f(x)g(x)dx)^2\leq \int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx$

Jumat, 07 Januari 2011

Bukti Hubungan Barisan Finobacci dan Rasio Emas

Sebelumnya saya sudah posting hubungan Barisan Finobacci dan Rasio Emas,kali ini saya akan mencoba membuktikannya dengan menggunakan induksi matematika.
misalkan:

$s_n...........f_n=\frac{1}{\sqrt{5}}(\phi ^{n}-(-1)^{n}\phi ^{-n})$
untuk n=1
$f_1=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2})=1(benar)$
untuk n=2

$f_2=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^2-(\frac{1-\sqrt{5}}{2})^2)=1(benar)$
asumsikan
$f_k=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k-(-1)^k(\frac{1-\sqrt{5}}{2})^k)$
akan dibuktikan kebenaran f(k) mengimplikasikan kebenaran f(k+1)

$f_k_+_1=f_k+f_k_-_1=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k-(-1)^k(\frac{1-\sqrt{5}}{2})^k)+\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^k^-^1-(-1)^k^-^1)(\frac{1-\sqrt{5}}{2})^k^-^1)=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k+(\frac{1+\sqrt{5}}{2})^k^-^1-((\frac{\sqrt{5}-1}{2})^k+(\frac{\sqrt{5}-1}{2})^k^-^1)$

$f_k_+_1=f_k+f_k_-_1=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k^-^1)(\frac{3+\sqrt{5}}{2})-(\frac{\sqrt{5}-1}{2})^k^-^1\frac{\sqrt{5}-3}{2})=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k^-^1(\frac{1+\sqrt{5}}{2})^2-(\frac{\sqrt{5}-1}{2})^k^-^1(\frac{\sqrt{5}-1}{2})^2)$

$f_k_+_1=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^k^+^1-(\frac{\sqrt{5}-1}{2})^k^+^1)=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^k^+^1-(-1)^k^+^1(\frac{1-\sqrt{5}}{2})^k^+^1)=\frac{1}{\sqrt{5}}(\phi ^k^+^1-(-1)^k^+^1)(\phi )^{k+1})$

jadi, terbukti bahwa f(k) mengimplikasikan kebenaran f(k+1).Berdasar prinsip induksi matematika,f(n) benar untuk setiap bilangan asli n

Barisan Finobacci dan rasio emas

suatu barisan terkenal {f} yang disebut barisan finobacci,didefinisikan

f_1=f_2=1, f_n_+_2=f_n_+_1+f_n

jadi, barisan 1 sampai dengan 10 adalah

1,1,2,3,5,8,13,21,34,55,89,144

misalkan

$\phi =\frac{1}{2}(1+\sqrt{5})\approx 1,618034$

orang-orang yunani menyebut bilangan ini dengan rasio emas (golden Ratio)

dapat ditunjukkan dengan induksi matematika,bahwa hubungan keduanya:

$f_n=\frac{1}{\sqrt{5}}(\phi ^{n}-(-1)^{n}\phi ^{-n})$
dapat pula dibuktikan,bahwa

$\lim_{n \to \infty }\frac{f_n_+_1}{f_n}=\phi$

hal yang menarik lainnya,bahwa akar dari

$x^{2}-x-1=0\rightarrow x=\Phi dan x=\frac{-1}{\phi }$

Senin, 03 Januari 2011

Integral Trigonometri

-Integral cos x
misal f(x)=sin x

$f(x)=sinx\rightarrow \frac{df(x)}{dx}=cosx\rightarrow df(x)=cosxdx\rightarrow$

$\int df(x)=\int cosxdx\rightarrow \int cosxdx=\int df(x)=f(x)+C=sinx+C$

-Integral sin x
misal f(x)=cos x

$f(x)=cosx\rightarrow \frac{df(x)}{dx}=-sinx\rightarrow df(x)=-sinxdx\rightarrow$

$\int df(x)=-\int sinxdx\rightarrow -\int sinxdx=\int df(x)=f(x)+C=cosx+C$
$\int sinxdx=-cosx+C$

-Integral Tan x
misal:
u=cos x, maka du=-sin x dx

$\int tanxdx=\int \frac{sinx}{cosx}dx=\int \frac{sinxdx}{cosx}=\int \frac{-du}{u}$

$\int tanxdx=-ln|u|+C=-ln|cosx|+C$

-Integral sec x
misal:
u=sin x,maka du=cos dx

$\int secxdx=\int \frac{1}{cosx}dx=\int \frac{1}{cosx}.\frac{cosx}{cosx}dx=\int \frac{cosx}{cosx^2x}dx=\int \frac{cosx}{1-sin^2x}dx\rightarrow$

$\int secxdx=\int \frac{cosxdx}{1-sin^2x}=\int \frac{du}{1-u^2}=\frac{1}{2}\int (\frac{1}{1+u}+\frac{1}{1-u})du\rightarrow$

$\int secxdx=\frac{1}{2}(ln|1+u|-ln|1-u|)+C=\frac{1}{2}ln|\frac{1+u}{1-u}|+C\rightarrow$

$\int secxdx=\frac{1}{2}ln|\frac{1+sinx}{1-sinx}|+C$

-Integral cosec x
Misal:
u=cos ,maka du=-sin xdx

$\int cosecxdx=\int \frac{1}{sinx}dx=\int \frac{1}{sinx}.\frac{sinx}{sinx}dx=\int \frac{sinx}{sin^2x}dx=\int \frac{sinx}{1-cos^2x}dx\rightarrow$

$\int cosecxdx=\int \frac{sinxdx}{1-cos^2x}=\int\frac{-du}{1-u^2}=-\frac{1}{2}\int(\frac{1}{1+u}+\frac{1}{1-u})du\rightarrow$

$\int cosecxdx=-\frac{1}{2}(ln|1+u|-ln|1-u|)+C=-\frac{1}{2}ln|\frac{1+u}{1-u}|+C\rightarrow$

$\int cosecxdx=-\frac{1}{2}ln|\frac{1+cosx}{1-cosx}|+C$

-Integral cotan x
Misal:
u=sin x, maka du=cosx dx

$\int cotan xdx=\int \frac{cosx}{sinx}dx=\int \frac{du}{u}=ln|u|+C\rightarrow$

$\int cotan xdx=ln|sinx|+C$

integral subsitusi trigonometri (1)

$\int \frac{1}{\sqrt{a^2-x^2}}dx$
misalkan :
$x=asint\Rightarrow t=\sin^{-1}\frac{x}{a},-\frac{\Pi }{2}\leq t\leq \frac{\Pi }{2}$
dx=acostdt

maka,

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\int \frac{acostdt}{\sqrt{a^2-(asint)^2}}=\int\frac{acostdt}{\sqrt{a^2-a^2sin^2t}}=\int \frac{acostdt}{\sqrt{a^2(1-sin^2t)}}\Rightarrow$

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\int \frac{acostdt}{acost}=\int dt=t+C=\sin^{-1}\frac{x}{a}+C$

$\int \sqrt{a^2-x^2}dx$

misalkan
$x=asint\Rightarrow t=\sin^{-1}\frac{x}{a},-\frac{\Pi }{2}\leq t\leq \frac{\Pi }{2}$
dx=acostdt

maka,

$\int \sqrt{a^2-x^2}dx=\int \sqrt{a^2-(asint)^2}(acostdt)=\int \sqrt{a^2-a^2sin^2t}(acostdt)$

$\int \sqrt{a^2-x^2}dx=\int (acost)(acostdt)=a^2\int cos^2tdt\Rightarrow$

$\int \sqrt{a^2-x^2}dx=a^2\int (\frac{1}{2}cos2t+\frac{1}{2})dt=a^2(\frac{1}{4}sin2t+\frac{1}{2}t)+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{4}sin2t+\frac{1}{2}a^2t+C=\frac{a^2}{2}sintcost+\frac{1}{2}a^2t+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{2}sin(\sin^{-1}\frac{x}{a})cos(\sin^{-1}\frac{x}{a})+\frac{1}{2}a^2\sin^{-1}\frac{x}{a}+C$

$\int \sqrt{a^2-x^2}dx=\frac{a^2}{2}.\frac{x}{a}\sqrt{1-(\frac{x}{a})^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C=\frac{ax}{2}\sqrt{\frac{a^2-x^2}{a^2}}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

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