- #1
TSN79
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Is there any way of solving [tex]\int_{0}^{1} \frac{1}{\sqrt{100-x^5}} dx[/tex] by some regular method? If not, how does one go about solving such integrals numerically?
Solving such integrals numerically?
- Thread starter TSN79
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In summary, the conversation discusses the difficulty of solving the integral \int_{0}^{1} \frac{1}{\sqrt{100-x^5}} dx using regular methods and the potential use of numerical methods. The conversation also mentions the different results from using Maple and Mathematica, and the frustration of not having a unified theory for solving integrals.
- #2
arildno
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1. I don't know about any method to crack this integral analytically.
2. As for numerical methods:
These typically start with some fancy finite Riemann sum to approximate the integral.
You could check up on the Trapezoid rule or Simpson's rule or some other rule..
2. As for numerical methods:
These typically start with some fancy finite Riemann sum to approximate the integral.
You could check up on the Trapezoid rule or Simpson's rule or some other rule..
- #3
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[tex] \int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} =\frac{1}{10} \ _{2}F_{1}\left(\frac{1}{2},\frac{1}{5},\frac{6}{5},\frac{1}{100}\right) [/tex]
according to my friend,Maple.
Daniel.
according to my friend,Maple.
Daniel.
- #4
Zurtex
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According to Mathematica:
[tex] \int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} = \frac{1}{10} \text{Hypergeometric2F1} \left[ \frac{1}{5},\frac{1}{2},\frac{6}{5},\frac{1}{100} \right][/tex]
Hmm wonder why the fractions are different, does it make a difference?
Anyway to 100 s.f:
0.1000836762086654607634453194543998535914445494139945308954127536431450557575506273354667189780577481
[tex] \int_{0}^{1} \frac{dx}{\sqrt{100-x^{5}}} = \frac{1}{10} \text{Hypergeometric2F1} \left[ \frac{1}{5},\frac{1}{2},\frac{6}{5},\frac{1}{100} \right][/tex]
Hmm wonder why the fractions are different, does it make a difference?
Anyway to 100 s.f:
0.1000836762086654607634453194543998535914445494139945308954127536431450557575506273354667189780577481
Last edited:
- #5
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Conventions,on normal basis they should be the same...
Gauss's function is [itex] _{2}F_{1}\left(a,b,c,z\right) [/itex]
Daniel.
P.S.Here are 125 sig.digs.
[tex]\allowbreak .\,10008\,36762\,08665\,46076\,34453\,19454\,39985\,35914\,44549\,41399\,45308\,95412\,[/tex]
[tex]75364\,31450\,55757\,55062\,73354\,66718\,97805\,77480\,65696\,66047\,63945\,64052\,7984 [/tex]
Gauss's function is [itex] _{2}F_{1}\left(a,b,c,z\right) [/itex]
Daniel.
P.S.Here are 125 sig.digs.
[tex]\allowbreak .\,10008\,36762\,08665\,46076\,34453\,19454\,39985\,35914\,44549\,41399\,45308\,95412\,[/tex]
[tex]75364\,31450\,55757\,55062\,73354\,66718\,97805\,77480\,65696\,66047\,63945\,64052\,7984 [/tex]
Last edited:
- #6
Zurtex
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Ahh so it's going to be like that is it, well in that case, to 2000 sf:dextercioby said:
0.1000836762086654607634453194543998535914445494139945308954127536431450557575\
506273354667189780577480656966604763945640527983993080671953023089297911515248\
996013282637274784238797044064349870623646876674724483462693995830072965962405\
190019049666775160232759330205668592578946075338958508523904278007888848833088\
060488556739935166104773509141768174230265551654691550416849059337448757791416\
602630567642386763728600302390370505747803022415391215722818064179823297750442\
434640467815965016423148168846638958221953928239081869470007493153876431418021\
157217678724106639004941740270189408736496188721513628469894013953107138449253\
791957534610058071313307798195976366400486448615041008866156544661543944553681\
807447545511569915551726649431123256963503190869480487316815727122678140209122\
738110692425545202808759680890972292418663402273431929145662980727173557540648\
152030821410027920261779677495993071187486068413623093242779260125263946810946\
539669010663042017697120941744353678783323221747296966263595416822115595315497\
544773344894826763667430935304194135002414853743041716586627006645641572144003\
038754961014790401073888263046103372397829209944018459246323194799313562020937\
625258141676770207971455031095503662945181184143286220172118052176868976940819\
363006570343110478540840140496393128738996412270357183445395020573388989278555\
813556144036563102455032850874625946333692746942852322508699926140544281120662\
532525325402041764035161169228790566576239939498698256797892476673581793593341\
565591033055741801465791277510653241163913554713152060127185501516632153831744\
084485276203982058024033779697637618294711132918471464969716668181939944429129\
590221571580066039538720668949128150132064065578982106579514989803518960288881\
581003421562857862205496312407174042287815539537348909650284394685560662915042\
685107335399595953096758002153118745564316408031460221673554730734019949375781\
303013799317984319741189620338837708872894884720821818310366605811188532339923\
2084126619627384173536035029500741646467671020765461
:tongue: :tongue: :tongue:
- #7
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Heh,i use a 10 y.o.Maple.It can't more than 1000 decimals.
Daniel.
Daniel.
- #8
marlon
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It always breaks my heart if an integral has no analytical solution. I hate it when those damned Maple or Mathematica-programs are needed in order to crack the problem. Doesn't it disturbe you that we don't have a TOE when it come to integrals. You know, how about a String Theory-variant for integral-solving...i dunno, just wondering...
regards
marlon
regards
marlon
- #9
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I'm not pissed off.I like when the results come out in terms of "fancy functions" (i call them:"common special functions").
This is not really elementary mathematics...
Daniel.
This is not really elementary mathematics...
Daniel.
1. How do you determine the accuracy of a numerical integration method?
The accuracy of a numerical integration method can be determined by comparing the results with the exact solution, if available. Additionally, the error can be calculated by using a known function with an exact solution and comparing the difference between the exact and numerical results.
2. What are the main steps in solving an integral numerically?
The main steps in solving an integral numerically include: choosing an appropriate numerical integration method, determining the limits of integration, dividing the integral into smaller subintervals, calculating the function values at each subinterval, and finally, summing up the results to approximate the integral.
3. What are the advantages of using numerical integration over analytical integration?
Numerical integration allows for the solution of integrals that cannot be solved analytically or are too complex to solve by hand. It also provides a more accurate solution compared to approximations used in analytical integration methods.
4. How do you choose the appropriate numerical integration method for a specific integral?
The choice of numerical integration method depends on the type of integral (e.g. definite or indefinite), the function being integrated, and the desired level of accuracy. Some common methods include the trapezoidal rule, Simpson's rule, and Gaussian quadrature.
5. Can numerical integration be used for any type of function?
Yes, numerical integration can be used for any type of function, including trigonometric, exponential, and logarithmic functions. However, the accuracy of the results may vary depending on the complexity of the function and the chosen numerical integration method.
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