Functie trigonometrica
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DrAGos Calinescu
- Thales
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Functie trigonometrica
Gasiti formula pt: \( \sum_{k=1}^nk\cdot \sin k. \)
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Mateescu Constantin
- Newton
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Se demonstreaza prin inductie ca : \( \overline{\underline{\left\|\ \sum_{k=1}^n k\cdot\sin kx=\frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}\ ,\ x\ne 2t\pi\ ,\ t\in\mathbb{Z} \ \right\|}} \) .
Last edited by Mateescu Constantin on Sun Jan 17, 2010 4:40 pm, edited 1 time in total.
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Virgil Nicula
- Euler
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Si cum ai emis ipoteza ? (debutul celui de-al doilea pas al unui proces de inductie completa). 
Demonstratie. Gasim formula pentru \( f(x)\equiv\sum_{k=1}^nk\cdot \sin kx \) din care vom obtine \( \sum_{k=1}^nk\cdot \sin k=f(1) \).
Metoda I.
Pas 1 (direct sau prin inductie). \( \overline{\underline{\left\|\ \sum_{k=1}^n\sin kx=\frac {\sin\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}\ \right\|}} \).
Pas 2. Se aplica identitatea \( \sum _{k=1}k\cdot a_k=\sum _{s=1}^n\ \sum_{k=s}^na_k \) pentru \( a_k=\sin kx\ ,\ k\in\overline{1,n} \).
Metoda II.
Pas 1 (direct sau prin inductie). \( \overline{\underline{\left\|\ \sum_{k=1}^n\cos kx=\frac {\cos\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}\ \right\|}} \).
Pas 2 (prin derivare). \( \sum_{k=1}^nk\cdot \sin kx=-\left(\sum_{k=1}^n\cos kx\right)^{\prim}=\ \ldots\ \ldots \)
Metoda III (numere complexe). Notam \( z=cos x+i\cdot\sin x \). Atunci \( z^k=\cos kx+i\cdot\sin kx \), pentru orice \( k\in \overline {1,n} \).
Notam \( S_n\equiv\sum_{k=1}^nk\sin kx\in \mathbb R \), \( C_n\equiv\sum_{k=1}^nk\cos kx\in\mathbb R \). Asadar \( C_n+i\cdot S_n=\sum_{k=1}^nkz^k \) care se poate calcula
usor prin unul din cele doua procedee mentionate mai sus dupa care se identifica partile reale/imaginare etc.
Demonstratie. Gasim formula pentru \( f(x)\equiv\sum_{k=1}^nk\cdot \sin kx \) din care vom obtine \( \sum_{k=1}^nk\cdot \sin k=f(1) \).
Metoda I.
Pas 1 (direct sau prin inductie). \( \overline{\underline{\left\|\ \sum_{k=1}^n\sin kx=\frac {\sin\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}\ \right\|}} \).
Pas 2. Se aplica identitatea \( \sum _{k=1}k\cdot a_k=\sum _{s=1}^n\ \sum_{k=s}^na_k \) pentru \( a_k=\sin kx\ ,\ k\in\overline{1,n} \).
Metoda II.
Pas 1 (direct sau prin inductie). \( \overline{\underline{\left\|\ \sum_{k=1}^n\cos kx=\frac {\cos\frac {(n+1)x}{2}\cos\frac {nx}{2}}{\sin \frac x2}\ \right\|}} \).
Pas 2 (prin derivare). \( \sum_{k=1}^nk\cdot \sin kx=-\left(\sum_{k=1}^n\cos kx\right)^{\prim}=\ \ldots\ \ldots \)
Metoda III (numere complexe). Notam \( z=cos x+i\cdot\sin x \). Atunci \( z^k=\cos kx+i\cdot\sin kx \), pentru orice \( k\in \overline {1,n} \).
Notam \( S_n\equiv\sum_{k=1}^nk\sin kx\in \mathbb R \), \( C_n\equiv\sum_{k=1}^nk\cos kx\in\mathbb R \). Asadar \( C_n+i\cdot S_n=\sum_{k=1}^nkz^k \) care se poate calcula
usor prin unul din cele doua procedee mentionate mai sus dupa care se identifica partile reale/imaginare etc.
Last edited by Virgil Nicula on Sun Jan 17, 2010 9:57 pm, edited 1 time in total.
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Mateescu Constantin
- Newton
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Eu am demonstrat relatia direct prin inductie.
Etapa 1. Pentru \( n=1 \) egalitatea devine : \( \sin x=\frac{2\sin x-\sin 2x}{4\sin^2\frac x2}=\frac{2\sin x(1-\cos x)}{4\sin^2\frac x2}\ \Longleftrightarrow\ \sin^2\frac x2=\frac{1-\cos x}{2} \), adevarat.
Etapa 2. Presupunem egalitatea adevarata pentru \( n \) numere si o demonstram pentru \( n+1 \) numere.
Astfel ramane sa aratam numai ca \( \frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}+(n+1)\sin(n+1)x=\frac{(n+2)\sin(n+1)x-(n+1)\sin(n+2)x}{4\sin^2\frac x2} \)
\( \Longleftrightarrow^{x\ne 2t\pi}\ \underline{\underline{(n+1)\sin nx}}-\underline{\underline{\underline{n\sin(n+1)x}}}+(n+1)[\sin(n+1)x]\cdot 4\sin^2\frac x2=\underline{\underline{\underline{(n+2)\sin(n+1)x}}}-\underline{\underline{(n+1)\sin(n+2)x}} \)
\( \Longleftrightarrow\ \underline{\underline{(n+1)}}[\sin nx+sin(nx+2x)]+\underline{\underline{(n+1)}}[\sin(n+1)x]\cdot4sin^2\frac x2=2\underline{\underline{(n+1)}}\sin(n+1)x \)
\( \Longleftrightarrow\ 2\underline{\underline{\sin(n+1)x}}\cdot\cos x+\underline{\underline{sin(n+1)x}}\cdot 4\sin^2\frac x2=2\underline{\underline{\sin(n+1)x}} \). Daca \( \sin(n+1)x=0 \) atunci e evident. Altfel putem imparti
prin \( \sin(n+1)x\ :\ \cos+2\sin^2\frac x2=1\ \Longleftrightarrow\ sin^2\frac x2=\frac{1-\cos x}{2}\ ,\ \mbox{O.K} \)
Etapa 1. Pentru \( n=1 \) egalitatea devine : \( \sin x=\frac{2\sin x-\sin 2x}{4\sin^2\frac x2}=\frac{2\sin x(1-\cos x)}{4\sin^2\frac x2}\ \Longleftrightarrow\ \sin^2\frac x2=\frac{1-\cos x}{2} \), adevarat.
Etapa 2. Presupunem egalitatea adevarata pentru \( n \) numere si o demonstram pentru \( n+1 \) numere.
Astfel ramane sa aratam numai ca \( \frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}+(n+1)\sin(n+1)x=\frac{(n+2)\sin(n+1)x-(n+1)\sin(n+2)x}{4\sin^2\frac x2} \)
\( \Longleftrightarrow^{x\ne 2t\pi}\ \underline{\underline{(n+1)\sin nx}}-\underline{\underline{\underline{n\sin(n+1)x}}}+(n+1)[\sin(n+1)x]\cdot 4\sin^2\frac x2=\underline{\underline{\underline{(n+2)\sin(n+1)x}}}-\underline{\underline{(n+1)\sin(n+2)x}} \)
\( \Longleftrightarrow\ \underline{\underline{(n+1)}}[\sin nx+sin(nx+2x)]+\underline{\underline{(n+1)}}[\sin(n+1)x]\cdot4sin^2\frac x2=2\underline{\underline{(n+1)}}\sin(n+1)x \)
\( \Longleftrightarrow\ 2\underline{\underline{\sin(n+1)x}}\cdot\cos x+\underline{\underline{sin(n+1)x}}\cdot 4\sin^2\frac x2=2\underline{\underline{\sin(n+1)x}} \). Daca \( \sin(n+1)x=0 \) atunci e evident. Altfel putem imparti
prin \( \sin(n+1)x\ :\ \cos+2\sin^2\frac x2=1\ \Longleftrightarrow\ sin^2\frac x2=\frac{1-\cos x}{2}\ ,\ \mbox{O.K} \)
Last edited by Mateescu Constantin on Mon Jan 18, 2010 11:23 am, edited 2 times in total.
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Virgil Nicula
- Euler
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Mateescu Constantin
- Newton
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Virgil Nicula
- Euler
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Atunci exercitiul trebuia formulat astfel
Sorry. Formulata astfel, problema nu este interesanta deoarece ni se da rezultatul si nu apare "ipoteza inductiei" ca un act creator, devenind doar un exercitiu de rutina, plicticos. Ceea ce ma surprinde in mod neplacut este faptul ca numeroase exercitii care apar la capitolul "Inductia matematica" sunt de aceasta forma sau eventual se cere "cat face suma ... " ca apoi sa se solicite "verificarea prin inductie", ceea ce este la fel de "nociv" pentru a intelege profund aceasta metoda fecunda a inductiei matematice. Dupa parerea mea ar fi trebuit sa se solicite calcularea sumei respective, ramanand la alegerea elevului procedeul, eventual si metoda inductiei complete. Prin acest "discurs" nu ma adresez "problemistilor", ci indeosebi profesorilor de matematica (in actul predarii).Mateescu Constantin wrote: Sa se demonstreze prin inductie ca \( \overline{\underline{\left\|\ \sum_{k=1}^n k\cdot\sin kx=\frac{(n+1)\sin nx-n\sin(n+1)x}{4\sin^2\frac x2}\ ,\ x\ne 2m\pi\ ,\ m\in\mathbb{Z} \ \right\|}} \)