Timeline for Binomial formula for $(x+1)^{1/3}$ (related to Newton's binomial theorem)
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2014 at 3:50 | answer | added | apnorton | timeline score: 1 | |
| Dec 11, 2013 at 15:18 | review | Close votes | |||
| Dec 11, 2013 at 15:30 | |||||
| Mar 19, 2013 at 13:01 | comment | added | Mhenni Benghorbal | @Xentius: It is the gamma function $\Gamma(n)=(n-1)!.$ | |
| Mar 19, 2013 at 7:53 | comment | added | Amadeus | @Xentius it's called gamma function. | |
| Mar 19, 2013 at 7:50 | comment | added | Xentius | @MhenniBenghorbal Could you please explain me the meaning of that symbol looks like a sickle? | |
| Mar 19, 2013 at 1:05 | comment | added | Mhenni Benghorbal | $$ 1.2.5 \ldots (3j-4)= {\frac {{3}^{j-1}\Gamma \left( j-1/3 \right) }{\Gamma \left( 2 /3 \right) }}.$$ | |
| Mar 19, 2013 at 0:48 | history | edited | Mhenni Benghorbal | CC BY-SA 3.0 | added 9 characters in body |
| Mar 19, 2013 at 0:42 | history | edited | Greg Martin | CC BY-SA 3.0 | added 2 characters in body |
| Mar 19, 2013 at 0:16 | history | edited | Xentius | CC BY-SA 3.0 | added 73 characters in body |
| Mar 18, 2013 at 23:40 | comment | added | Américo Tavares | This has nothing to do with your main question but there is a typo in the numerator: $-1^{j+1}$ instead of $(-1)^{j+1}$. | |
| Mar 18, 2013 at 23:34 | comment | added | Xentius | @GerryMyerson But the person who asked me this question told me that he saw this on a book as an exercise problem. So I guess there must be a decent expression? | |
| Mar 18, 2013 at 23:31 | comment | added | Gerry Myerson | If there isn't a solution, there isn't a problem. Not every complex mathematical expression has a simpler form. I think you have to live with what you have. | |
| Mar 18, 2013 at 23:29 | comment | added | Xentius | @GerryMyerson while solving for $\sqrt{1+x}$, there was 1*3*5*7*..*(2j-3) at the nominator, then I multiplied both the nominator and the demoniator with 2*4*6*...*(2j-2) and found the answer. However in this case (cube root), it seems not to be working (multiplying nominator and dominator with 3*4*6*7*9*11*...). So I thought maybe there can be a way to write 2*5*...*(3j-4) in another way. But you are right, I am not sure about what to do in order to solve this problem. Do you have an idea? | |
| Mar 18, 2013 at 23:23 | history | edited | Xentius | CC BY-SA 3.0 | deleted 1 characters in body |
| Mar 18, 2013 at 23:23 | comment | added | Gerry Myerson | I don't think you can write anything for $(2)(5)\cdots(3j-4)$. Why do you feel you need to? | |
| Mar 18, 2013 at 23:14 | history | edited | Américo Tavares | CC BY-SA 3.0 | TeX in title |
| Mar 18, 2013 at 23:09 | comment | added | Xentius | @GitGud thank you so much! | |
| Mar 18, 2013 at 23:08 | history | edited | Xentius | CC BY-SA 3.0 | added 18 characters in body |
| Mar 18, 2013 at 23:02 | comment | added | Git Gud | I slightly improved the $\LaTeX$ in your question. Please check that I kept the meaning of the question. | |
| Mar 18, 2013 at 23:01 | history | edited | Git Gud | CC BY-SA 3.0 | Refined TeX |
| Mar 18, 2013 at 22:56 | history | asked | Xentius | CC BY-SA 3.0 |