Sampla de Cheisteanna Plé Suim Riemann
Pendahuuan
Is coincheap bunúsach sa chalcalas é suim Riemann a úsáidtear chun slánuimhir chinnte feidhme a shainiú. Úsáideann an modh seo roinnt eatraimh agus suim achar dronuilleog chun an slánuimhir a mheas go garbh. Pléifidh an t-alt seo coincheap shuim Riemann go críochnúil, lena n-áirítear samplaí agus plé chun tuiscint a éascú.
Coincheap Bunúsach Suim Riemannian
Sula ndéanaimid plé ar shamplaí, tá sé tábhachtach bunchoincheap na suimeanna Riemannian a thuiscint. Is féidir suimeanna Riemannian a roinnt ina thrí phríomhchineál:
1. Suim Riemann ar chlé
2. Suim Riemann ar dheis
3. Suim Riemann lárphointe
Bristear an modh seo eatramh na feidhme atá le comhtháthú ina fho-eatramh níos lú den fhad céanna. Úsáidtear gach ceann de na fo-eatramh seo ansin chun dronuilleog a chruthú a bhfuil a airde á chinneadh ag luach na feidhme ag pointe áirithe laistigh den fho-eatramh (ar chlé, ar dheis, nó sa lár).
Foirmle Ghinearálta do Shuim Riemann
Abair gur mian linn an fheidhm f(x) a chomhtháthú ó a go b. Roinnimid an t-eatramh [a, b] ina fho-eatramh comhionanna (n) de fhad Δx = (frac{ba}{n)). Is féidir suimeanna Riemann na dtrí chineál thuasluaite a scríobh mar seo a leanas:
1. Riemann ar chlé:
[L_n = \sum_{i=0}^{n-1} f(x_i) Δx]
2. Riemann ar dheis:
[R_n = \sum_{i=1}^{n} f(x_i) Δx]
3. Riemann Láir:
[M_n = \sum_{i=0}^{n-1} f\left(\frac{x_i + x_{i+1}}{2}\right) \Deltax \]
An mana:
– Is é \( \Deltax \) leithead gach fo-eatramh.
– Is é \(x_i \) pointe tosaigh an i-ú fo-eatramh don tsuim Riemann ar chlé.
– Is é \(x_i \) críochphointe an i-ú fo-eatramh don tsuim Riemann ar dheis.
– Is é \( \frac{x_i + x_{i+1}}{2} \) lárphointe an i-ú fo-eatramh don tsuim Riemann lár.
Ceisteanna Samplacha agus Plé
Pléifimid fadhbanna samplacha do gach cineál Suim Riemann chun ár dtuiscint a dhoimhniú.
Sampla 1: Suim Riemann ar Chlé
Ríomh suim Riemann ar chlé le haghaidh \(f(x) = x^2 \) ar an eatramh \([0, 2] \) le \(n = 4 \).
Plé:
1. Leithead Fo-eatramh (Δx):
[Δx = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Pointe Roinnte Eatramh (ar chlé):
[x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5]
3. Luach Feidhme ag an bPointe Deighilte:
[f(x_0) = f(0) = 0^2 = 0]
[f(x_1) = f(0.5) = (0.5)^2 = 0.25]
[f(x_2) = f(1.0) = (1.0)^2 = 1]
[f(x_3) = f(1.5) = (1.5)^2 = 2.25]
4. Suim Riemann ar Chlé (Ln):
[L_n = \sum_{i=0}^{n-1} f(x_i) Δx = (0) ⋅0.5 + (0.25) ⋅0.5 + (1) ⋅0.5 + (2.25) ⋅0.5]
\[ L_n = 0 + 0.125 + 0.5 + 1.125 \]
\[L_n = 1.75 \]
Sampla 2: Suim Riemann ar dheis
Ríomh an tsuim Riemann cheart le haghaidh \(f(x) = x^2 \) ar an eatramh \([0, 2] \) le \(n = 4 \).
Plé:
1. Leithead Fo-eatramh (Δx):
[Δx = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Pointe Roinnte Eatramh (ar dheis):
[x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, x_4 = 2.0]
3. Luach Feidhme ag an bPointe Deighilte:
[f(x_1) = f(0.5) = (0.5)^2 = 0.25]
[f(x_2) = f(1.0) = (1.0)^2 = 1]
[f(x_3) = f(1.5) = (1.5)^2 = 2.25]
[f(x_4) = f(2.0) = (2.0)^2 = 4]
4. Suim Riemann ar Dheis (Rn):
[R_n = \sum_{i=1}^{n} f(x_i) Δx = (0.25) ⋅0.5 + (1) ⋅0.5 + (2.25) ⋅0.5 + (4) ⋅0.5]
\[R_n = 0.125 + 0.5 + 1.125 + 2 \]
\[R_n = 3.75 \]
Sampla 3: Suim Riemann Láir
Ríomh suim lár Riemann le haghaidh \(f(x) = x^2 \) ar an eatramh \([0, 2] \) le \(n = 4 \).
Plé:
1. Leithead Fo-eatramh (Δx):
[Δx = \frac{ba}{n} = \frac{2-0}{4} = 0.5 \]
2. Lárphointe an Fho-eatramh:
[x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, agus x_{n-1}=2.0]
Lárphointe an fho-eatramh:
\[tm_0 = \left(\frac{0 + 0.5}{2}\right) = 0.25 \]
\[tm_1 = \left(\frac{0.5 + 1.0}{2}\right) = 0.75 \]
\[tm_2 = \left(\frac{1.0 + 1.5}{2}\right) = 1.25 \]
\[tm_3 = \left(\frac{1.5 + 2.0}{2}\right) = 1.75 \]
3. Luach Feidhme ag an Lárphointe:
[f(0.25) = (0.25)^2 = 0.0625]
[f(0.75) = (0.75)^2 = 0.5625]
[f(1.25) = (1.25)^2 = 1.5625]
[f(1.75) = (1.75)^2 = 3.0625]
4. Suim Riemann Láir (Mn):
[M_n = \sum_{i=0}^{n-1} f(tm_i) Δx = (0.0625) ⋅0.5 + (0.5625) ⋅0.5 + (1.5625) ⋅0.5 + (3.0625) ⋅0.5]
\[M_n = 0.03125 + 0.28125 + 0.78125 + 1.53125 \]
\[M_n = 2.625 \]
Conclúid
Pléadh san alt seo conas suimeanna Riemann ar chlé, ar dheis agus sa lár a ríomh, mar aon le samplaí mionsonraithe. Soláthraíonn modh suime Riemann bealach éifeachtach chun slánuimhir feidhme a mheas trína eatramh a roinnt ina fho-eatraimh bheaga agus achar iomlán gach fo-eatraimh a ríomh. Tá tuiscint mhaith ar shuim Riemann riachtanach dóibh siúd atá ag déanamh staidéir ar chalcalas nó ag obair le feidhmeanna casta i réimsí eolaíochta éagsúla.
