Difference between revisions of "Bagus Fadhlurrohman"
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nanti akan mendapatkan hail | nanti akan mendapatkan hail | ||
+ | |||
+ | |||
+ | ==''' Metode Numerik ''' == | ||
+ | |||
+ | [[Hiburan 1]] | ||
+ | |||
+ | Mencari limit (x)=1 pada persamaan dengan pyhton | ||
+ | |||
+ | (x**2-1) / ((x-1) | ||
+ | |||
+ | step 1 | ||
+ | |||
+ | Membuka python idle lulu mengetik program seperti berikut | ||
+ | |||
+ | def limit (x) : | ||
+ | try: | ||
+ | a = (x**2-1) | ||
+ | b = (x-1) | ||
+ | result = a / b | ||
+ | print (result) | ||
+ | except ZeroDivisionError: | ||
+ | c = ((x+(1/99))**2-1) / ((x+(1/99))-1) | ||
+ | print (c) | ||
+ | d = ((x+(1/999))**2-1) / ((x+(1/999))-1) | ||
+ | print (d) | ||
+ | e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) | ||
+ | print (e) | ||
+ | f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) | ||
+ | print (f) | ||
+ | g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) | ||
+ | print (g) | ||
+ | h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) | ||
+ | print (h) | ||
+ | print ("mendekati angka 2") | ||
+ | else: | ||
+ | print ("hasilnya", resut) | ||
+ | finally: | ||
+ | print ("sudah") | ||
+ | |||
+ | step 2 | ||
+ | |||
+ | simpan program denna bents file.py | ||
+ | |||
+ | step 3 | ||
+ | |||
+ | jalankan program pada python | ||
+ | |||
+ | step 4 | ||
+ | |||
+ | nanti akan mendapatkan hail | ||
+ | |||
+ | ==''' Metode Numerik ''' == | ||
+ | |||
+ | [[Hiburan 1]] | ||
+ | |||
+ | Mencari limit (x)=1 pada persamaan dengan pyhton | ||
+ | |||
+ | (x**2-1) / ((x-1) | ||
+ | |||
+ | step 1 | ||
+ | |||
+ | Membuka python idle lulu mengetik program seperti berikut | ||
+ | |||
+ | def limit (x) : | ||
+ | try: | ||
+ | a = (x**2-1) | ||
+ | b = (x-1) | ||
+ | result = a / b | ||
+ | print (result) | ||
+ | except ZeroDivisionError: | ||
+ | c = ((x+(1/99))**2-1) / ((x+(1/99))-1) | ||
+ | print (c) | ||
+ | d = ((x+(1/999))**2-1) / ((x+(1/999))-1) | ||
+ | print (d) | ||
+ | e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) | ||
+ | print (e) | ||
+ | f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) | ||
+ | print (f) | ||
+ | g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) | ||
+ | print (g) | ||
+ | h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) | ||
+ | print (h) | ||
+ | print ("mendekati angka 2") | ||
+ | else: | ||
+ | print ("hasilnya", resut) | ||
+ | finally: | ||
+ | print ("sudah") | ||
+ | |||
+ | step 2 | ||
+ | |||
+ | simpan program denna bents file.py | ||
+ | |||
+ | step 3 | ||
+ | |||
+ | jalankan program pada python | ||
+ | |||
+ | step 4 | ||
+ | |||
+ | nanti akan mendapatkan hail | ||
+ | |||
+ | |||
+ | ==''' Metode Numerik ''' == | ||
+ | |||
+ | [[Hiburan 1]] | ||
+ | |||
+ | Mencari limit (x)=1 pada persamaan dengan pyhton | ||
+ | |||
+ | (x**2-1) / ((x-1) | ||
+ | |||
+ | step 1 | ||
+ | |||
+ | Membuka python idle lulu mengetik program seperti berikut | ||
+ | |||
+ | def limit (x) : | ||
+ | try: | ||
+ | a = (x**2-1) | ||
+ | b = (x-1) | ||
+ | result = a / b | ||
+ | print (result) | ||
+ | except ZeroDivisionError: | ||
+ | c = ((x+(1/99))**2-1) / ((x+(1/99))-1) | ||
+ | print (c) | ||
+ | d = ((x+(1/999))**2-1) / ((x+(1/999))-1) | ||
+ | print (d) | ||
+ | e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) | ||
+ | print (e) | ||
+ | f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) | ||
+ | print (f) | ||
+ | g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) | ||
+ | print (g) | ||
+ | h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) | ||
+ | print (h) | ||
+ | print ("mendekati angka 2") | ||
+ | else: | ||
+ | print ("hasilnya", resut) | ||
+ | finally: | ||
+ | print ("sudah") | ||
+ | |||
+ | step 2 | ||
+ | |||
+ | simpan program dalam bentik file.py | ||
+ | |||
+ | step 3 | ||
+ | |||
+ | jalankan program pada python | ||
+ | |||
+ | step 4 | ||
+ | |||
+ | nanti akan mendapatkan hasil | ||
+ | |||
+ | |||
+ | [[Hiburan 2]] | ||
+ | |||
+ | Mencari nilai x pada persamaan dengan python | ||
+ | |||
+ | 8x**4 + 2x**3 + x**2 - x = 0 | ||
+ | |||
+ | step 1 | ||
+ | |||
+ | Membuka python idle lulu mengetik program seperti berikut | ||
+ | |||
+ | def f(x): | ||
+ | return 8*x**3 + 2*x**2 + x - 1 | ||
+ | def fprime(x): | ||
+ | return 24*x**2 + 4*x +1 | ||
+ | |||
+ | ep = 0.001 | ||
+ | |||
+ | gu = -10 | ||
+ | i = 0 | ||
+ | |||
+ | print('8*x**3 + 2*x**2 + x - 1') | ||
+ | |||
+ | print('Results by Python 3.7') | ||
+ | |||
+ | while abs(f(gu)) >= ep: | ||
+ | gu = gu - (f(gu)/fprime(gu)) | ||
+ | i += 1 | ||
+ | print(' ' + str(i) + ' ' + str(round(gu,7))) | ||
+ | |||
+ | print('The root approach is ' + str(round(gu,2)) + | ||
+ | '| failed to calculate: ' + str(i) + ' times' ) | ||
+ | |||
+ | step 2 | ||
+ | |||
+ | simpan program dalam bentik file.py | ||
+ | |||
+ | step 3 | ||
+ | |||
+ | jalankan program pada python | ||
+ | |||
+ | step 4 | ||
+ | |||
+ | nanti akan mendapatkan hasil | ||
==''' Tugas Mekanika Fluida''' == | ==''' Tugas Mekanika Fluida''' == |
Revision as of 20:19, 16 September 2019
Contents
Profil
Nama: Bagus Fadhlurrohman
NPM: 1706070633
Fakultas: Teknik
Jurusan: Teknik Mesin
BIOGRAFI
Nama saya Bagus Fadhlurrohman lahir pada tanggal 15 Juli 1999 di kota Jakarta. Saya merupakan anak pertama dari 2 bersaudara. Ayah saya seorang karyawan dan ibu saya seorang ibu rumah tangga. Saya memiliki 1 adik laki-laki yang sudah menjadi mahasiswa.
Riwayat pendidikan
2003-2005 TK Aisiyah 04 Tebet Timur
2005-2011 SDN Tebet Barat 05 Pagi
2011-2014 SMPN 115 Jakarta
2014-2017 SMAN 8 Jakarta
2017-.... S1 Teknik Mesin Universitas Indonesia
Metode Numerik
Mencari limit (x)=1 pada persamaan dengan pyhton
(x**2-1) / ((x-1)
step 1
Membuka python idle lulu mengetik program seperti berikut
def limit (x) :
try: a = (x**2-1) b = (x-1) result = a / b print (result) except ZeroDivisionError: c = ((x+(1/99))**2-1) / ((x+(1/99))-1) print (c) d = ((x+(1/999))**2-1) / ((x+(1/999))-1) print (d) e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) print (e) f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) print (f) g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) print (g) h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) print (h) print ("mendekati angka 2") else: print ("hasilnya", resut) finally: print ("sudah")
step 2
simpan program denna bents file.py
step 3
jalankan program pada python
step 4
nanti akan mendapatkan hail
Metode Numerik
Mencari limit (x)=1 pada persamaan dengan pyhton
(x**2-1) / ((x-1)
step 1
Membuka python idle lulu mengetik program seperti berikut
def limit (x) :
try: a = (x**2-1) b = (x-1) result = a / b print (result) except ZeroDivisionError: c = ((x+(1/99))**2-1) / ((x+(1/99))-1) print (c) d = ((x+(1/999))**2-1) / ((x+(1/999))-1) print (d) e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) print (e) f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) print (f) g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) print (g) h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) print (h) print ("mendekati angka 2") else: print ("hasilnya", resut) finally: print ("sudah")
step 2
simpan program denna bents file.py
step 3
jalankan program pada python
step 4
nanti akan mendapatkan hail
Metode Numerik
Mencari limit (x)=1 pada persamaan dengan pyhton
(x**2-1) / ((x-1)
step 1
Membuka python idle lulu mengetik program seperti berikut
def limit (x) :
try: a = (x**2-1) b = (x-1) result = a / b print (result) except ZeroDivisionError: c = ((x+(1/99))**2-1) / ((x+(1/99))-1) print (c) d = ((x+(1/999))**2-1) / ((x+(1/999))-1) print (d) e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) print (e) f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) print (f) g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) print (g) h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) print (h) print ("mendekati angka 2") else: print ("hasilnya", resut) finally: print ("sudah")
step 2
simpan program denna bents file.py
step 3
jalankan program pada python
step 4
nanti akan mendapatkan hail
Metode Numerik
Mencari limit (x)=1 pada persamaan dengan pyhton
(x**2-1) / ((x-1)
step 1
Membuka python idle lulu mengetik program seperti berikut
def limit (x) :
try: a = (x**2-1) b = (x-1) result = a / b print (result) except ZeroDivisionError: c = ((x+(1/99))**2-1) / ((x+(1/99))-1) print (c) d = ((x+(1/999))**2-1) / ((x+(1/999))-1) print (d) e = ((x+(1/9999))**2-1) / ((x+(1/9999))-1) print (e) f = ((x+(1/99999))**2-1) / ((x+(1/99999))-1) print (f) g = ((x+(1/999999))**2-1) / ((x+(1/999999))-1) print (g) h = ((x+(1/9999999))**2-1) / ((x+(1/9999999))-1) print (h) print ("mendekati angka 2") else: print ("hasilnya", resut) finally: print ("sudah")
step 2
simpan program dalam bentik file.py
step 3
jalankan program pada python
step 4
nanti akan mendapatkan hasil
Mencari nilai x pada persamaan dengan python
8x**4 + 2x**3 + x**2 - x = 0
step 1
Membuka python idle lulu mengetik program seperti berikut
def f(x):
return 8*x**3 + 2*x**2 + x - 1
def fprime(x):
return 24*x**2 + 4*x +1
ep = 0.001
gu = -10 i = 0
print('8*x**3 + 2*x**2 + x - 1')
print('Results by Python 3.7')
while abs(f(gu)) >= ep:
gu = gu - (f(gu)/fprime(gu)) i += 1 print(' ' + str(i) + ' ' + str(round(gu,7)))
print('The root approach is ' + str(round(gu,2)) +
'| failed to calculate: ' + str(i) + ' times' )
step 2
simpan program dalam bentik file.py
step 3
jalankan program pada python
step 4
nanti akan mendapatkan hasil