Timeline for Upper bound over $[0,1] $ for strange family of polynomials
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2020 at 2:38 | answer | added | Max Alekseyev | timeline score: 1 | |
| Jun 1, 2019 at 0:37 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 31, 2019 at 4:46 | comment | added | mamiladi | thanks for your help | |
| May 31, 2019 at 4:45 | comment | added | mamiladi | the dominant coefficient of $ p_n$ is the term on $x^{n-1}$, it's equal to $binomial(2n-1,n)*sum(binomial(n, 1-p)*\frac{(-1)^{p+1}}{p}, p = 1 .. 1)=binomial(2n-1,n), $ and in case n=5, you must have binomial(2*5-1,5)=126 as a coefficient of $x^4 $and not $630$, so please check the formulea write on mapple (correct it) , and you have to find $p[5](x) := 137/60-(77/2)*x+(329/2)*x^2-252*x^3+126*x^4$ as expression for $p_5(x)$ | |
| May 30, 2019 at 10:42 | comment | added | user64494 | According to your definitions, the command of Maple n:=5:sum(binomial(n+k, n)*(-1)^k*x^k*(sum(binomial(n, n-k-p)*(-1)^(p+1)/p, p = 1 .. n-k)), k = 0 .. n-1) p produces $$ {\frac{137}{60}}-{\frac {125\,x}{2}}+385\,{x}^{2}-840\,{x}^{3}+630\,{x }^{4}. $$ | |
| May 30, 2019 at 10:01 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 9:51 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 9:49 | comment | added | mamiladi | for this $p_5$ of degre $4$ you have exactly 4 distincts zeros on $]0,1[$ ( can be groupped 2 by two $x_0$ and $1-x_0$) | |
| May 30, 2019 at 9:45 | comment | added | mamiladi | have you found $p_5(x) := 137/60-(77/2)*x+(329/2)*x^2-252*x^3+126*x^4$ ? | |
| May 30, 2019 at 9:41 | comment | added | user64494 | Your statement "$P_n$ have exactly $ n−1$ zeros over $]0,1[$" does not correspond to reality in the case $n=5$ up to Maple code n:=5:fsolve(sum(binomial(n+k, n)*(-1)^k*x^k*(sum(binomial(n, n-k-p)*(-1)^(p+1)/p, p = 1 .. n-k)), k = 0 .. n-1)) which produces $0.5066328932e-1, .2452340931 $. | |
| May 30, 2019 at 7:21 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 6:48 | comment | added | mamiladi | $$ C_{n}^k= \displaystyle \frac{n!}{k!(n-k)!}$$ | |
| May 30, 2019 at 6:45 | comment | added | mamiladi | yes ; binomial coefiicient and $C_n^0=1$ | |
| May 30, 2019 at 6:42 | comment | added | Sahil Kumar | In definition of $a_{n,k}$ you have used $C^{n-k-p}_n$ which i don't understand as for example at $p=n-k$ becomes $C^0_n$ ? | |
| May 30, 2019 at 6:34 | history | edited | YCor | CC BY-SA 4.0 |
fixed many typos; edited tags
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| May 30, 2019 at 6:26 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 6:20 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 5:01 | history | edited | mamiladi | CC BY-SA 4.0 |
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| May 30, 2019 at 4:49 | history | asked | mamiladi | CC BY-SA 4.0 |