Funksiyaning davriyligi

Bu nima va nega kerak

Tabiat va texnikadagi ko'plab jarayonlar vaqt o'tishi bilan aynan takrorlanadi: Yerning o'z o'qi atrofida aylanishi (24 soat), fasllar almashishi (12 oy), yurak urishi kardiogrammasi yoki maishiy tarmoqdagi o'zgaruvchan elektr toki (50 Hz).

Matematikada bunday takrorlanuvchi jarayonlar davriy funksiyalar yordamida ifodalanadi:

• Funksiyaning bitta to'liq takrorlanish oralig'idagi (bir davrdagi) grafigini chizish orqali uning butun sonlar o'qidagi grafigini hosil qilish mumkin.
• DTM testlarida argumenti juda katta yoki manfiy bo'lgan ifodalarning qiymatini davrni ayirib tashlash orqali bir necha soniyada topish mumkin.

Ushbu darsda funksiyaning davriyligi, asosiy trigonometrik funksiyalarning davrlari va murakkab funksiyalarning davrini hisoblash usullarini o'rganamiz.

Asosiy g'oya

1. Davriy funksiya ta'rifi

Ta'rif: Agar $y = f(x)$ funksiya uchun shunday noldan farqli $T \neq 0$ son mavjud bo'lsaki:

1. Har qanday $x \in D(f)$ uchun $x + T \in D(f)$ va $x - T \in D(f)$ bo'lsa;
2. $f(x + T) = f(x)$ tenglik bajarilsa, $f(x)$ davriy funksiya, $T$ soni esa uning davri deyiladi.

Agar $T$ soni $f(x)$ ning davri bo'lsa, u holda ixtiyoriy butun $k \neq 0$ uchun $kT$ ham funksiyaning davri bo'ladi: $f(x + kT) = f(x), \quad k \in \mathbb{Z}.$

2. Asosiy davr (eng kichik musbat davr)

Davriy funksiyaning cheksiz ko'p davrlari mavjud. Ular orasidagi eng kichik musbat son $T_0 > 0$ funksiyaning asosiy davri deyiladi.

Masalan, $\sin x$ uchun $2\pi, 4\pi, 6\pi, -2\pi$ sonlari davr bo'ladi. Uning asosiy davri $T_0 = 2\pi$.

3. Asosiy trigonometrik funksiyalarning davrlari

Funksiya	Asosiy davri $T_0$
$y = \sin x$	$2\pi$
$y = \cos x$	$2\pi$
$y = \text{tg}\,x$	$\pi$
$y = \text{ctg}\,x$	$\pi$

4. $y = A f(kx + b)$ ko'rinishidagi funksiyalar davri

Agar $f(x)$ ning asosiy davri $T_0$ bo'lsa, u holda argument $k$ ga ko'paytirilganda grafik $Ox$ o'qi bo'ylab $|k|$ marta qisqaradi (yoki cho'ziladi). Shuning uchun yangi davr quyidagicha topiladi: $T = \frac{T_0}{|k|}.$

Xususan:

• $y = A \sin(kx + b)$ va $y = A \cos(kx + b)$ uchun: $T = \frac{2\pi}{|k|}.$
• $y = A \text{tg}(kx + b)$ va $y = A \text{ctg}(kx + b)$ uchun: $T = \frac{\pi}{|k|}.$

Muhim eslatma: Tashqi koeffitsiyent $A$ va siljish $b$ funksiyaning davriga ta'sir qilmaydi!

5. Yig'indi va ayirmaning davri

Agar $f(x) = f_1(x) \pm f_2(x)$ bo'lib, $f_1$ ning davri $T_1$, $f_2$ ning davri $T_2$ bo'lsa, umumiy funksiyaning davri $T_1$ va $T_2$ ning eng kichik umumiy karralisi (EKUK) ga teng: $T = \text{EKUK}(T_1, T_2).$

Kasr ko'rinishidagi sonlar uchun: $\text{EKUK}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{EKUK}(a, c)}{\text{EKUB}(b, d)}.$

Yechilgan misollar

1-misol. Chiziqli argumentli funksiyalar davri

Quyidagi funksiyalarning eng kichik musbat davrini toping: a) $y = 4\cos(3x - 5)$ b) $y = 2\text{tg}\left(\dfrac{2x}{3} + \dfrac{\pi}{4}\right)$

Yechish: a) $\cos x$ ning asosiy davri $2\pi$. $x$ oldidagi koeffitsiyent $k = 3$. $T = \frac{2\pi}{|k|} = \frac{2\pi}{3}.$

b) $\text{tg}\,x$ ning asosiy davri $\pi$. $x$ oldidagi koeffitsiyent $k = \dfrac{2}{3}$. $T = \frac{\pi}{|k|} = \frac{\pi}{2/3} = \frac{3\pi}{2}.$

Javob: a) $\dfrac{2\pi}{3}$; b) $\dfrac{3\pi}{2}$.

2-misol. Darajali trigonometrik funksiya davri

$y = 6\cos^2(2x) - 1$ funksiyaning asosiy davrini toping.

Yechish: Kvadrat darajani pasaytirish formulasidan foydalanamiz: $\cos^2 \alpha = \frac{1 + \cos 2\alpha}{2}.$ Bunda: $y = 6 \cdot \frac{1 + \cos(4x)}{2} - 1 = 3(1 + \cos(4x)) - 1 = 3\cos(4x) + 2.$ Hosil bo'lgan ifodada kosinusning argumenti $4x$, ya'ni $k = 4$. $T = \frac{2\pi}{4} = \frac{\pi}{2}.$

Javob: $\dfrac{\pi}{2}$.

3-misol. Funksiyalar yig'indisining davri

$y = \sin\left(\dfrac{3x}{2}\right) + \cos\left(\dfrac{5x}{4}\right)$ funksiyaning eng kichik musbat davrini toping.

Yechish:

1. $\sin\left(\dfrac{3x}{2}\right)$ uchun $k_1 = \dfrac{3}{2}$: $T_1 = \frac{2\pi}{3/2} = \frac{4\pi}{3}.$
2. $\cos\left(\dfrac{5x}{4}\right)$ uchun $k_2 = \dfrac{5}{4}$: $T_2 = \frac{2\pi}{5/4} = \frac{8\pi}{5}.$
3. Umumiy davr $T = \text{EKUK}\left(\dfrac{4\pi}{3}, \dfrac{8\pi}{5}\right)$: $T = \pi \cdot \frac{\text{EKUK}(4, 8)}{\text{EKUB}(3, 5)} = \pi \cdot \frac{8}{1} = 8\pi.$

Tekshirib ko'ramiz: $f(x + 8\pi) = \sin\left(\frac{3(x+8\pi)}{2}\right) + \cos\left(\frac{5(x+8\pi)}{4}\right) = \sin\left(\frac{3x}{2} + 12\pi\right) + \cos\left(\frac{5x}{4} + 10\pi\right) = f(x)$.

Javob: $8\pi$.

4-misol. Davriy funksiya qiymatini hisoblash

$f(x)$ davriy funksiyaning eng kichik musbat davri $T = 6$ ga teng bo'lib, $[-3; 3]$ kesmada $f(x) = x^2 - 2$ formula bilan aniqlangan. $f(23)$ ning qiymatini toping.

Yechish: Argument $23$ sonidan davr $T = 6$ ning karralilarini ayirib, berilgan $[-3; 3]$ oraliqqa tushiramiz: $23 = 4 \cdot 6 - 1 = 24 - 1.$ Demak: $f(23) = f(-1 + 4 \cdot 6) = f(-1).$ $x = -1$ nuqta $[-3; 3]$ oraliqqa tegishli. Berilgan formulaga qo'yamiz: $f(-1) = (-1)^2 - 2 = 1 - 2 = -1.$

Javob: $-1$.

Mashq

1. $y = \sin(5x + 3)$ funksiyaning eng kichik musbat davrini toping.
2. $y = \cos\left(\dfrac{x}{4}\right)$ funksiyaning asosiy davrini toping.
3. $y = \text{tg}(6x - 1)$ funksiyaning asosiy davrini toping.
4. $y = \text{ctg}\left(\dfrac{2x}{5}\right)$ funksiyaning asosiy davrini toping.
5. $y = \sin^2(5x)$ funksiyaning eng kichik musbat davrini toping.
6. $y = \cos^4 x - \sin^4 x$ funksiyaning asosiy davrini toping.
7. $y = \sin(3x) + \cos(2x)$ funksiyaning eng kichik musbat davrini toping.
8. $y = \text{tg}(3x) + \sin(6x)$ funksiyaning asosiy davrini toping.
9. $f(x)$ davriy funksiyaning davri $T = 7$ bo'lib, $f(3) = 11$ bo'lsa, $3f(17) - 2f(-11)$ ning qiymatini hisoblang.
10. $y = \{4x\}$ (sonning kasr qismi) funksiyasining asosiy davrini toping.

Javoblar va yechim yo'llari

1. $T = \dfrac{2\pi}{5}$. Javob: $\dfrac{2\pi}{5}$.

2. $k = \dfrac{1}{4} \implies T = \dfrac{2\pi}{1/4} = 8\pi$. Javob: $8\pi$.

3. Tangens uchun $T_0 = \pi \implies T = \dfrac{\pi}{6}$. Javob: $\dfrac{\pi}{6}$.

4. Kotangens uchun $T_0 = \pi \implies T = \dfrac{\pi}{2/5} = \dfrac{5\pi}{2}$. Javob: $\dfrac{5\pi}{2}$.

5. $\sin^2(5x) = \dfrac{1 - \cos(10x)}{2} \implies T = \dfrac{2\pi}{10} = \dfrac{\pi}{5}$. Javob: $\dfrac{\pi}{5}$.

6. $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos(2x) \cdot 1 = \cos(2x)$. $T = \dfrac{2\pi}{2} = \pi$. Javob: $\pi$.

7. $T_1 = \dfrac{2\pi}{3}$, $T_2 = \dfrac{2\pi}{2} = \pi$. $T = \text{EKUK}\left(\dfrac{2\pi}{3}, \pi\right) = \pi \cdot \dfrac{\text{EKUK}(2, 1)}{\text{EKUB}(3, 1)} = 2\pi$. Javob: $2\pi$.

8. $\text{tg}(3x)$ uchun $T_1 = \dfrac{\pi}{3}$; $\sin(6x)$ uchun $T_2 = \dfrac{2\pi}{6} = \dfrac{\pi}{3}$. $T = \dfrac{\pi}{3}$. Javob: $\dfrac{\pi}{3}$.

9. $f(17) = f(3 + 2 \cdot 7) = f(3) = 11$. $f(-11) = f(-11 + 2 \cdot 7) = f(3) = 11$. $3 \cdot 11 - 2 \cdot 11 = 11$. Javob: $11$.

10. $y = \{x\}$ ning asosiy davri $T_0 = 1$. Argument $4$ ga ko'paytirilgan: $T = \dfrac{1}{4}$. Javob: $\dfrac{1}{4}$ (yoki $0{,}25$).

Keng tarqalgan xatolar

1. Tangens va kotangens davrini $2\pi$ deb hisoblash. ✗ $\text{tg}(2x)$ ning davrini $T = \dfrac{2\pi}{2} = \pi$ deb topish. ✓ Tangens va kotangensning asosiy davri $\pi$ ga teng: $T = \dfrac{\pi}{k} = \dfrac{\pi}{2}$.

2. Koeffitsiyent $k$ ga bo'lish o'rniga ko'paytirib yuborish. ✗ $\sin(4x)$ ning davri $2\pi \cdot 4 = 8\pi$ deb yanglishish. ✓ Argument $k$ marta kattalashsa, davr $k$ marta kichrayadi: $T = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$.

3. Darajali trigonometrik funksiyalarni darajasini pasaytirmay hisoblash. ✗ $\sin^2 x$ ning davri $2\pi$ deb xulosa qilish. ✓ Daraja pasaytirilganda argument 2 marta ortadi: $\sin^2 x = \dfrac{1 - \cos 2x}{2}$, shuning uchun davr $\pi$ bo'ladi.

4. Funksiyalar yig'indisida davrlarni shunchaki qo'shib qo'yish. ✗ $f(x) = \sin(2x) + \cos(3x)$ ning davri $\pi + \dfrac{2\pi}{3}$ deb yozish. ✓ Yig'indining davri qismlar davrlarining eng kichik umumiy karralisi (EKUK) orqali topiladi.

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Keyingi qadam

Funksiyalarning juft-toqligi va davriyligini o'rgandik. Keyingi darsda berilgan funksiya ta'sirini teskarisiga aylantiruvchi amallar — teskari funksiya va uning mavjudlik shartlari bilan tanishamiz.