On the integrals $int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Show $lim_nto inftynint_0^fracpi2(1-sqrt[n]sin(x)),mathrmdx = fracpi ln(2)2$Calculating value of $pi$ independently using integrals.Expressing the integral $int_0^1fracmathrmdxsqrtleft(1-x^3right)left(1-a^6x^3right)$ in terms of elliptic integralsEvaluation of integral $int_0^infty frac(k-frac12)^u+j(k+frac12)^u+j+2 sqrtk J^2_ell(k) sin(tausqrtk)dk$Estimating the sum of a series within arbitrary certainty.Evaluate $lim_n to infty int_0^1 [x^n + (1-x)^n ]^1/n mathrmdx$How may we evaluate the closed form for $int_0^pi/4x^nsqrt1+sin(2x)mathrm dx=n!+F(n)?$How to approximate the numbers after finding the linear approximation?Evaluate the $int_0^infty xe^-xdx$.Evaluation of Integrals involving Legendre Functions
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On the integrals $int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Show $lim_nto inftynint_0^fracpi2(1-sqrt[n]sin(x)),mathrmdx = fracpi ln(2)2$Calculating value of $pi$ independently using integrals.Expressing the integral $int_0^1fracmathrmdxsqrtleft(1-x^3right)left(1-a^6x^3right)$ in terms of elliptic integralsEvaluation of integral $int_0^infty frac(k-frac12)^u+j(k+frac12)^u+j+2 sqrtk J^2_ell(k) sin(tausqrtk)dk$Estimating the sum of a series within arbitrary certainty.Evaluate $lim_n to infty int_0^1 [x^n + (1-x)^n ]^1/n mathrmdx$How may we evaluate the closed form for $int_0^pi/4x^nsqrt1+sin(2x)mathrm dx=n!+F(n)?$How to approximate the numbers after finding the linear approximation?Evaluate the $int_0^infty xe^-xdx$.Evaluation of Integrals involving Legendre Functions
$begingroup$
Playing with integrals of type,
$$
I(n)=int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx,
$$
$$
n in mathbbN
$$
I got two interesting results for the limiting cases $n=1$ and $n to infty$:
$$
lim_n to infty I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx approx frac pisqrt 27
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.
calculus integration definite-integrals
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
add a comment |
$begingroup$
Playing with integrals of type,
$$
I(n)=int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx,
$$
$$
n in mathbbN
$$
I got two interesting results for the limiting cases $n=1$ and $n to infty$:
$$
lim_n to infty I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx approx frac pisqrt 27
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.
calculus integration definite-integrals
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38
add a comment |
$begingroup$
Playing with integrals of type,
$$
I(n)=int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx,
$$
$$
n in mathbbN
$$
I got two interesting results for the limiting cases $n=1$ and $n to infty$:
$$
lim_n to infty I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx approx frac pisqrt 27
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.
calculus integration definite-integrals
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
Playing with integrals of type,
$$
I(n)=int_-1^0 sqrt[2n+1]x-sqrt[2n+1]x mathrm dx,
$$
$$
n in mathbbN
$$
I got two interesting results for the limiting cases $n=1$ and $n to infty$:
$$
lim_n to infty I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx approx frac pisqrt 27
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.
calculus integration definite-integrals
calculus integration definite-integrals
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
asked Apr 1 at 8:40
Awe Kumar JhaAwe Kumar Jha
633113
633113
$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38
add a comment |
$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38
$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38
$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
$endgroup$
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
add a comment |
$begingroup$
Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.
For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
add a comment |
$begingroup$
I found the solution of your interesting integral by try:
$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$
I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!
answered Apr 1 at 10:16
stochastocha
31138
$endgroup$
add a comment |
$begingroup$
I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.
I guess that the trouble comes from a transformation such as $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.
The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
$endgroup$
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
add a comment |
$begingroup$
For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
$endgroup$
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
add a comment |
$begingroup$
For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
$endgroup$
For $n in mathbbN$ we have
beginalign
I (n) &= int limits_-1^0 left[x - x^frac12n+1right]^frac12n+1 mathrmd x stackrelx = -y= int limits_0^1 left[y^frac12n+1 - yright]^frac12n+1 mathrmd y = int limits_0^1 y^frac1(2n+1)^2left[1 - y^frac2n2n+1right]^frac12n+1 mathrmd y \
&hspace-10ptstackrely = t^frac2n+12n= frac2n+12n int limits_0^1 t^fracn+1n (2n+1) (1-t)^frac12n+1 mathrmdt = frac2n+12n operatornameBleft(fracn+1n(2n+1)+1,frac12n+1 + 1right) , .
endalign
Using $Gamma(x+1) = x Gamma(x)$ we can rewrite this result to find
$$ I (n) = fracoperatornameB left(frac1n - frac12n+1, frac12n+1right)2 (2n+1) = frac12(2n+1) fracoperatornameGammaleft(frac1n - frac12n+1right) operatornameGammaleft(frac12n+1right)operatornameGammaleft(frac1nright)$$
for $n in mathbbN$. In particular,
$$ I(1) = fracoperatornameGammaleft(frac23right) operatornameGammaleft(frac13right)6 = fracpi6 sin left(fracpi3right) = fracpi3 sqrt3 ,.$$
Moreover, we obtain
$$ lim_n to infty I (n) stackrelGamma(x) , stackrelx to 0sim , frac1x= lim_n to infty frac12(2n+1) fracfracn(2n+1)n+1 (2n+1)n = lim_n to infty frac2n+12(n+1) = 1 , .$$
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
edited Apr 1 at 11:05
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
answered Apr 1 at 10:25
ComplexYetTrivialComplexYetTrivial
4,9832631
4,9832631
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
add a comment |
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
I think that your calculus is not correct. At first line you supposed $x^frac2n2n+1= (-x)^frac2n2n+1=y^frac2n2n+1$ which is false. It is true that $((-x)^2)^fracn2n+1=y^frac2n2n+1$ but $ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$. The first is complex. The second is real.
$endgroup$
– JJacquelin
Apr 1 at 11:02
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
$begingroup$
Interesting because I get another solution but I can not reply the second result
$endgroup$
– stocha
Apr 1 at 11:05
1
1
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@JJacquelin I have added an intermediate step to illustrate what I have done here. The only assumption is that $x^1/(2n+1) = - (-x)^1/(2n+1)$ holds for $x < 0$, which (as confirmed by the results) seems to agree with the OP's definition of odd roots of negative numbers.
$endgroup$
– ComplexYetTrivial
Apr 1 at 11:10
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@ComplexYetTrivial. I agree that the key point is the definition of odd roots of negative numbers. $sqrt[3]-1$ has three roots, $−1$ and two complex. I would agree with the OP results if the integral was $$int_0^1sqrt[2n+1]x+sqrt[2n+1]x dx $$
$endgroup$
– JJacquelin
Apr 1 at 11:54
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
$begingroup$
@JJacquelin, I am truly sorry if it is sarcastic, but I think at this point Wolfram Alpha has got a bug, for the integrand is indeed defined for all real $x$, whether positive or negative . In fact you can manually plot it on both sides of the y-axis and find that there is indeed a positive area bound by the curve and the y-axis between -1 and. 0.
$endgroup$
– Awe Kumar Jha
Apr 1 at 13:14
add a comment |
$begingroup$
Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.
For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
add a comment |
$begingroup$
Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.
For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
add a comment |
$begingroup$
Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.
For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=fracGamma left(frac2 (n+1)2 n+1right) Gamma left(fracn+1n(2
n+1)right)2 Gamma left(frac1nright)$$ Expanding as series
$$I_n=1-frac12 n+frac12-pi ^224 n^2+Oleft(frac1n^3right)$$ which is not "too bad" even for $n=1$; this would give $1-fracpi ^224approx 0.588766$ while $fracpi3 sqrt3approx 0.604600$.
For a few values of $n$
$$left(
beginarrayccc
n & textapproximation & textexact \
1 & 0.588766 & 0.604600 \
2 & 0.772192 & 0.774848 \
3 & 0.843196 & 0.843965 \
4 & 0.880548 & 0.880851 \
5 & 0.903551 & 0.903695 \
6 & 0.919132 & 0.919210 \
7 & 0.930383 & 0.930429 \
8 & 0.938887 & 0.938916 \
9 & 0.945540 & 0.945560
endarray
right)$$
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 3 at 6:40
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
add a comment |
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
well I didn't think that the 2nd degree Taylor approximation would be so accurate, +1.
$endgroup$
– Awe Kumar Jha
Apr 3 at 11:31
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Me neither, be sure ! In fact, we can easily go further. Cheers :-)
$endgroup$
– Claude Leibovici
Apr 3 at 16:19
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
$begingroup$
@AweKumarJha. Still better if we use $$I_n=1-frac12 n+frac12-pi ^224 n^2+frac12 (zeta (3)-2)+pi ^248 n^3+Oleft(frac1n^4right)$$ The numbers would become $$0.594897,0.772958,0.843423,0.880644,0.903600,0.919161,0.930401,0.938899,0.945549$$
$endgroup$
– Claude Leibovici
Apr 4 at 4:42
add a comment |
$begingroup$
I found the solution of your interesting integral by try:
$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$
I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!
answered Apr 1 at 10:16
stochastocha
31138
$endgroup$
add a comment |
$begingroup$
I found the solution of your interesting integral by try:
$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$
I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!
answered Apr 1 at 10:16
stochastocha
31138
$endgroup$
add a comment |
$begingroup$
I found the solution of your interesting integral by try:
$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$
I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!
answered Apr 1 at 10:16
stochastocha
31138
$endgroup$
I found the solution of your interesting integral by try:
$$int_-1^0 left(x-x^frac12 n+1right)^frac12 n+1 ,dx = frac(-1)^frac2 n1+2 n left(-1+(-1)^1+frac11+2 nright)^1+frac11+2 n (1+2 n)^2 textHypergeometric2F1left[1,2+frac1n,2+frac1n-frac11+2n,(-1)^frac2 n1+2 nright]2 (1+2 n (1+n))$$
I checked the result for several values, but since I can't get your second result, the numerical solution is complex, I guess you did a mistake in your post! Please check your second result!
answered Apr 1 at 10:16
stochastocha
31138
answered Apr 1 at 10:16
stochastocha
31138
answered Apr 1 at 10:16
stochastocha
31138
answered Apr 1 at 10:16
stochastocha
31138
31138
add a comment |
add a comment |
$begingroup$
I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.
I guess that the trouble comes from a transformation such as $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.
The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
add a comment |
$begingroup$
I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.
I guess that the trouble comes from a transformation such as $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.
The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
add a comment |
$begingroup$
I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.
I guess that the trouble comes from a transformation such as $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.
The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
$endgroup$
I am afraid that your calculus is not correct. Unfortunately the calculus for $I(1)$ is not detailed. One cannot say where exactly is the mistake.
I guess that the trouble comes from a transformation such as $x^frac2n2n+1= ((-x)^2)^fracn2n+1$ which is false. $$ x^frac2n2n+1neq ((-x)^2)^fracn2n+1$$ for $-1<x<0$ the first term is complex. The second is real.
The integral is not real $simeq fracpisqrt27$.
$$I(1)=int_-1^0 sqrt[3]x-sqrt[3]x mathrm dx simeq 0.68575-0.242772,i$$
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
answered Apr 1 at 11:18
JJacquelinJJacquelin
45.7k21858
45.7k21858
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
add a comment |
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
$begingroup$
Yes your right, see my post below, I recognize the same
$endgroup$
– stocha
Apr 1 at 11:30
add a comment |
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$begingroup$
You should avoid the ambiguity due to complex roots in using the next integral : $$int_0^1 sqrt[2n+1]x+sqrt[2n+1]x mathrm dx.$$
$endgroup$
– JJacquelin
Apr 1 at 11:38