Konsultimet Mat avancuar Funksioni periodik Ushtrime te zgjidhura dhe udhezime.

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– Posted on 21/06/2013Posted in: USHTRIME MATEMATIKE

1.Gjeni perioden e funksioneve te meposhtme: a)y=sinx\cdot cosx   b)y=2cos^{2}x-1 , c)y=sin^{2}x   d)y=\frac{cosx}{sinx}   

zgjidhje

Funksionet sinkx, coskx, kane perioden T=\frac{2\pi }{k}, ndersa funksionet tgkx, cotgkx  kane perioden T=\frac{\pi }{k} 

a) y=sinx\cdot cosx=\frac{1}{2}\cdot 2sinx\cdot cosx=\frac{1}{2}sin2x\Rightarrow T=\frac{2\pi }{2}=\pi

b)y=2cos^{2}x-1=cos^{2}x-sin^{2}x=cos2x\Rightarrow T=\pi

c)cos2x=cos^{2}x-sin^{2}x=1-2sin^{2}x\Rightarrow sin^{2}x=\frac{1}{2}\left ( 1-cos2x \right )   Te vertetojme se perioda e ketij funksioni eshte pi.  Shenojme me T perioden e funksionit.f(x+T)=f(x)\Rightarrow \frac{1}{2}\left [ 1-cos2(x+T) \right ]=\frac{1}{2}\left ( 1-cos2x \right )\Leftrightarrow 1-cos2(x+T)=1-cos2x\Leftrightarrow cos2(x+T)=cos2x\Rightarrow 2(x+T)-2x=2\pi \Rightarrow T=\pi

d)Njelloj funksioni eshte cotgx.

2.Vertetoni se nese funksioni f eshte periodik me periode T, atehere funksioni g:y=A\cdot f(kx+b)  ku A, k, b jane konstante (k>0) eshte periodik me periode T/k.

zgjidhje

Funksioni f periodik kjo do te thote se f(x+T)=f(x)\Rightarrow f\left [ (kx+b)+T \right ]=f(kx+b) . E zeme se a eshte perioda e funksionit g d.m.th g(x+a)=g(x)\Rightarrow g(x+a)=A\cdot f\left [ k(x+a)+b \right ]=A\cdot f\left [ (kx+b)+ka \right ]=A\cdot f(kx+b)=g(x)\Rightarrow ka=T\Rightarrow a=\frac{T}{k}

Postimi tjeter per funksionet jo periodike

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