Question

Sa se calculeze S= g(0)+g(1)+...+g(2012) unde g:R-> R, g(x)= f`(x)-f``(x) iar f:R-> R, f(x) = e^x - e^(-x)

Answer 1

Explicație pas cu pas:

Cautam forma functiei g.

Pentru aceasta, derivam f o data si, respectiv de doua ori.

$f'(x)=(e^x-e^{-x})'=(e^x)'-(e^{-x})'=e^x-e^{-x}*(-x)'=e^x-e^{-x}*(-1)=e^x+e^{-x}$

$f''(x)=[f'(x)]'=(e^x+e^{-x})'=(e^x)'+(e^{-x})'=e^x+e^{-x}*(-x)'=e^x+e^{-x}*(-1)=e^x-e^{-x}$

Atunci g devine:

$g(x)=f'(x)-f''(x)=e^x+e^{-x}-(e^x-e^{-x})=e^x+e^{-x}-e^x+e^{-x}=e^{-x}+e^{-x}=2e^{-x}=\frac{2}{e^x}$

Si atunci suma noastra se transforma in:

$S=g(0)+g(1)+...+g(2012)\\S=\frac{2}{e^0}+\frac{2}{e^1}+...+\frac{2}{e^{2012}}$

Observam ca am dat peste o progresie geometrica cu primul termen $b_0=\frac{2}{e^0}$ si de ratie:

$q=\frac{b_1}{b_0}=\frac{\frac{2}{e^1}}{\frac{2}{e^0}}=\frac{2}{e^1}*\frac{e^0}{2}=\frac{1}{e}$

Aplicam formula sumei primilor n termeni ai unei progresii geometrice:

$S_n=b_1*\frac{q^n-1}{q-1}$, unde b₁ este primul termen al progresiei, q este ratia si n este numarul de termeni.

In cazul nostru, avem b₀ ca fiind primul termen, avem determinata ratia q si stim ca sunt 2013 termeni in progresie deoarece intre 1 si 2012 sunt 2012 termeni, la care adaugat si termenul b₀ (adica pornim numaratoarea de la 0, nu de la 1).

$S_{2012}=\frac{2}{e^0}*\frac{(\frac{1}{e})^{2013}-1}{\frac{1}{e}-1}=2*\frac{\frac{1}{e^{2013}}-1}{\frac{1-e}{e}}=2*\frac{\frac{1-e^{2013}}{e^{2013}}}{\frac{1-e}{e}}=2*\frac{1-e^{2013}}{e^{2013}}*\frac{e}{1-e}=2*\frac{1-e^{2013}}{e^{2012}(1-e)}$

Answer 2

f:R->R , f(x)=eˣ-e⁻ˣ

g:R->R , g(x)=f'(x)-f"(x)

f continua si derivabila pe Dmax=R (operatii cu functii elementare)

f'(x)=(eˣ-e⁻ˣ)'=(eˣ)'-(e⁻ˣ)'=x'eˣ-(-x)'e⁻ˣ=eˣ+e⁻ˣ ,∀x∈R

f"(x)=(eˣ+e⁻ˣ)'=(eˣ)'+(e⁻ˣ)'=x'eˣ+(-x)'e⁻ˣ=eˣ-e⁻ˣ ,∀x∈R

g(x)=f'(x)-f"(x)=eˣ+e⁻ˣ-(eˣ-e⁻ˣ)=eˣ+e⁻ˣ-eˣ+e⁻ˣ=2e⁻ˣ

g(0)=2e⁰

g(1)=2e¹=2e⁻¹

g(2)=2e⁻²

........................

g(2012)=2e⁻²⁰¹²

_____________"+"

g(0)+g(1)+g(2)+...+g(2012)=2e⁰+2e⁻¹+2e⁻²+...+2e⁻²⁰¹²=2(e⁻⁰+e⁻¹+e⁻²+...+e⁻²⁰¹²)

Obs. ca termenii sumei sunt termenii unei progresii geometrice cu primul termen b₁=e⁰

b₂=b₁*q=>q=b₂/b₁=e⁻¹/e⁰=e ,ratia q=e⁻¹

S₂₀₁₂=qⁿ-1/(q-1)=(e⁻²⁰¹³-1)/(e⁻¹-1)=(-1+1/e²⁰¹³)/(-1+1/e)=(-e²⁰¹³+1)/e²⁰¹³/(-e+1)/e=e(-e²⁰¹³+1)/e²⁰¹³(-e+1)=(-e²⁰¹³+1)/e²⁰¹²(-e+1)

g(0)+g(1)+g(2)+...+g(2012)=2(-e²⁰¹³+1)/e²⁰¹²(-e+1)