Mixing
Geometry problem boils down to finding a closed form for $sum_n=1^karctanleft(frac1nright)$
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I was solving the following problem:
"Find $angle A + angle B + angle C$ in the figure below, assuming the three shapes are squares."![1]](https://i.stack.imgur.com/DgX67.gif)
And I found a beautiful one-liner using complex numbers:
$(1+i)(2+i)(3+i)=10i$, so $angle A + angle B + angle C = fracpi2$
Now, I thought, what if I want to generalize? What if, instead of three squares, there were $2018$ squares? What would the sum of the angles be then? Could I make a formula for $k$ squares?
Essentially, the question boiled down to finding a closed form for the argument of the complex number
$$prod_n=1^k(n+i)=prod_n=1^kleft(sqrtn^2+1right)e^icot^-1n$$
This we can break into two parts, finding a closed form for
$$prod_n=1^k(n^2+1)$$
and
$$sum_n=1^kcot^-1n$$
This I don't know how to solve.
complex-numbers summation products
asked Jul 18 at 23:32
Shrey Joshi
1389
add a comment |Â
up vote
9
down vote
favorite
I was solving the following problem:
"Find $angle A + angle B + angle C$ in the figure below, assuming the three shapes are squares."
And I found a beautiful one-liner using complex numbers:
$(1+i)(2+i)(3+i)=10i$, so $angle A + angle B + angle C = fracpi2$
Now, I thought, what if I want to generalize? What if, instead of three squares, there were $2018$ squares? What would the sum of the angles be then? Could I make a formula for $k$ squares?
Essentially, the question boiled down to finding a closed form for the argument of the complex number
$$prod_n=1^k(n+i)=prod_n=1^kleft(sqrtn^2+1right)e^icot^-1n$$
This we can break into two parts, finding a closed form for
$$prod_n=1^k(n^2+1)$$
and
$$sum_n=1^kcot^-1n$$
This I don't know how to solve.
complex-numbers summation products
asked Jul 18 at 23:32
Shrey Joshi
1389
1
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I was solving the following problem:
"Find $angle A + angle B + angle C$ in the figure below, assuming the three shapes are squares."
And I found a beautiful one-liner using complex numbers:
$(1+i)(2+i)(3+i)=10i$, so $angle A + angle B + angle C = fracpi2$
Now, I thought, what if I want to generalize? What if, instead of three squares, there were $2018$ squares? What would the sum of the angles be then? Could I make a formula for $k$ squares?
Essentially, the question boiled down to finding a closed form for the argument of the complex number
$$prod_n=1^k(n+i)=prod_n=1^kleft(sqrtn^2+1right)e^icot^-1n$$
This we can break into two parts, finding a closed form for
$$prod_n=1^k(n^2+1)$$
and
$$sum_n=1^kcot^-1n$$
This I don't know how to solve.
complex-numbers summation products
asked Jul 18 at 23:32
Shrey Joshi
1389
I was solving the following problem:
"Find $angle A + angle B + angle C$ in the figure below, assuming the three shapes are squares."
And I found a beautiful one-liner using complex numbers:
$(1+i)(2+i)(3+i)=10i$, so $angle A + angle B + angle C = fracpi2$
Now, I thought, what if I want to generalize? What if, instead of three squares, there were $2018$ squares? What would the sum of the angles be then? Could I make a formula for $k$ squares?
Essentially, the question boiled down to finding a closed form for the argument of the complex number
$$prod_n=1^k(n+i)=prod_n=1^kleft(sqrtn^2+1right)e^icot^-1n$$
This we can break into two parts, finding a closed form for
$$prod_n=1^k(n^2+1)$$
and
$$sum_n=1^kcot^-1n$$
This I don't know how to solve.
complex-numbers summation products
asked Jul 18 at 23:32
Shrey Joshi
1389
edited Jul 19 at 0:31
asked Jul 18 at 23:32
Shrey Joshi
1389
asked Jul 18 at 23:32
Shrey Joshi
1389
asked Jul 18 at 23:32
Shrey Joshi
1389
1389
1
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31
add a comment |Â
1
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31
1
1
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
With CAS help:
$$sum _n=1^k tan ^-1left(frac1nright)=\int left(sum _n=1^k fracpartial partial atan ^-1left(fracanright)right) , da=\int left(sum _n=1^k fracna^2+n^2right) ,
da=\int frac12 left(-H_-i a-H_i a+H_-i a+k+H_i a+kright) ,
da=\-frac12 i (textlog$Gamma $(1-i a)-textlog$Gamma $(1+i a)-textlog$Gamma
$(1-i a+k)+textlog$Gamma $(1+i a+k))+C$$
where $a=1$ and $C=0$ then:
$$colorbluesum _n=1^k tan ^-1left(frac1nright)=-frac12 i (textlog$Gamma
$(1-i)-textlog$Gamma $(1+i)-textlog$Gamma $((1-i)+k)+textlog$Gamma
$((1+i)+k))$$
Where: $H_i a$ is harmonic number and $textlog$Gamma $(1-i)$ is logarithm of the gamma function
MMA code:
HoldForm[Sum[ArcTan[1/n], n, 1, k] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.:)
– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
add a comment |Â
up vote
3
down vote
Playing around with Wolfy suggests that
$s(n)
=sum_k=1^n arctan(1/k)
=frac3pi4-frac12arctan(g(n))
$
where
$g(n)$ is an increasingly complicated fraction.
Some values are
$g(4) = 15/8,
g(5) = 140/71,
g(6) = 2848/7665,
g(7) = 14697/203896
$.
To get a recurrence for
$g(n)$,
$beginarray\
s(n+1)-s(n)
&=arctan(1/(n+1))\
&=(frac3pi4-frac12arctan(g(n+1)))-(frac3pi4-frac12arctan(g(n)))\
&=frac12(arctan(g(n))-arctan(g(n+1))\
endarray
$
so,
using
$arctan(x)pmarctan(y)
=arctan(fracxpm y1mp xy)
$,
$beginarray\
arctan(g(n+1))
&=arctan(g(n))-2arctan(1/(n+1))\
&=arctan(g(n))-arctan(frac2/(n+1)1-1/(n+1)^2)\
&=arctan(g(n))-arctan(frac2(n+1)(n+1)^2-1)\
&=arctan(g(n))-arctan(frac2(n+1)n^2+2n)\
&=arctan(fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n)\
&<arctan(g(n)-frac2n+1)\
endarray
$
so,
assuming that
the proper branch of
arctan is taken,
$g(n+1)
=fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n
lt g(n)-frac2n+1
$.
Since $sum 1/n$ diverges,
this shows that
exentually
$g(n) < 0$.
At this point the
next branch of
arctan has to be taken.
For example,
Wolfy calculates that
$s(20)
=frac5 À4 - frac12 arctan(frac4718365039332113702517864397263976449928)
$
and
$s(40)
=frac5À4 + frac12 arctan(frac41279370979134545450499387615832498927444174194743607269197868658553203529942799672226208517623565372926024)
$.
I'll leave it at this.
answered Jul 19 at 2:40
marty cohen
69.2k446122
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
add a comment |Â
up vote
2
down vote
We have that
$$
prodlimits_1, le ,n, le ,k left( i + n right) = 1 over iprodlimits_0, le ,n, le ,k left( i + n right)
= 1 over ii^,overline ,k + 1, = left( 1 + i right)^,overline ,k,
= Gamma left( 1 + i + k right) over Gamma left( 1 + i right) = k!left( matrix
i + k cr
k cr right)
$$
where $x^,overline ,k, = Gamma (x + k) over Gamma (x)$ denotes the Rising Factorial
and $x^,underline ,k, = left( x - k + 1 right)^,overline ,k, $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic
functions even for complex $x$ and $k$.
Then
$$
ln left( z^,overline ,k, right) = ln left( right) + iarg left( z^,overline ,k, right)
= ln Gamma (z + k) over Gamma (z)
$$
The above tells us that your question is related to the absolute value and argument of the Gamma Function,
which unfortunately do not have a closed expression, better than the above.
answered Jul 19 at 1:37
G Cab
15.1k31136
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
 |Â
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up vote
0
down vote
Note that the addition formula for the arctangent gives $arctan(x)+arctan(y)=arctan(dfracx+y1-xy)$; thus, if we define $alpha_n$ by $arctan(alpha_n)=sum_i=1^narctan(frac1i)$, then we have $alpha_n=dfracalpha_n-1+frac1n1-fracalpha_n-1n$ $=dfracnalpha_n-1+1n-alpha_n-1$; alternately, setting $alpha_n=fraca_nb_n$, we can write this as $a_n=na_n-1+b_n-1, b_n = -a_n-1+nb_n-1$ - or $beginpmatrixa_n \ b_n endpmatrix = beginpmatrixn & 1 \ -1 & nendpmatrixbeginpmatrixa_n-1 \ b_n-1 endpmatrix$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1leq nleq 3) =langle(1, 1), (3, 1), (10, 0)rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With CAS help:
$$sum _n=1^k tan ^-1left(frac1nright)=\int left(sum _n=1^k fracpartial partial atan ^-1left(fracanright)right) , da=\int left(sum _n=1^k fracna^2+n^2right) ,
da=\int frac12 left(-H_-i a-H_i a+H_-i a+k+H_i a+kright) ,
da=\-frac12 i (textlog$Gamma $(1-i a)-textlog$Gamma $(1+i a)-textlog$Gamma
$(1-i a+k)+textlog$Gamma $(1+i a+k))+C$$
where $a=1$ and $C=0$ then:
$$colorbluesum _n=1^k tan ^-1left(frac1nright)=-frac12 i (textlog$Gamma
$(1-i)-textlog$Gamma $(1+i)-textlog$Gamma $((1-i)+k)+textlog$Gamma
$((1+i)+k))$$
Where: $H_i a$ is harmonic number and $textlog$Gamma $(1-i)$ is logarithm of the gamma function
MMA code:
HoldForm[Sum[ArcTan[1/n], n, 1, k] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.:)
– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
add a comment |Â
up vote
4
down vote
accepted
With CAS help:
$$sum _n=1^k tan ^-1left(frac1nright)=\int left(sum _n=1^k fracpartial partial atan ^-1left(fracanright)right) , da=\int left(sum _n=1^k fracna^2+n^2right) ,
da=\int frac12 left(-H_-i a-H_i a+H_-i a+k+H_i a+kright) ,
da=\-frac12 i (textlog$Gamma $(1-i a)-textlog$Gamma $(1+i a)-textlog$Gamma
$(1-i a+k)+textlog$Gamma $(1+i a+k))+C$$
where $a=1$ and $C=0$ then:
$$colorbluesum _n=1^k tan ^-1left(frac1nright)=-frac12 i (textlog$Gamma
$(1-i)-textlog$Gamma $(1+i)-textlog$Gamma $((1-i)+k)+textlog$Gamma
$((1+i)+k))$$
Where: $H_i a$ is harmonic number and $textlog$Gamma $(1-i)$ is logarithm of the gamma function
MMA code:
HoldForm[Sum[ArcTan[1/n], n, 1, k] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.:)
– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With CAS help:
$$sum _n=1^k tan ^-1left(frac1nright)=\int left(sum _n=1^k fracpartial partial atan ^-1left(fracanright)right) , da=\int left(sum _n=1^k fracna^2+n^2right) ,
da=\int frac12 left(-H_-i a-H_i a+H_-i a+k+H_i a+kright) ,
da=\-frac12 i (textlog$Gamma $(1-i a)-textlog$Gamma $(1+i a)-textlog$Gamma
$(1-i a+k)+textlog$Gamma $(1+i a+k))+C$$
where $a=1$ and $C=0$ then:
$$colorbluesum _n=1^k tan ^-1left(frac1nright)=-frac12 i (textlog$Gamma
$(1-i)-textlog$Gamma $(1+i)-textlog$Gamma $((1-i)+k)+textlog$Gamma
$((1+i)+k))$$
Where: $H_i a$ is harmonic number and $textlog$Gamma $(1-i)$ is logarithm of the gamma function
MMA code:
HoldForm[Sum[ArcTan[1/n], n, 1, k] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
With CAS help:
$$sum _n=1^k tan ^-1left(frac1nright)=\int left(sum _n=1^k fracpartial partial atan ^-1left(fracanright)right) , da=\int left(sum _n=1^k fracna^2+n^2right) ,
da=\int frac12 left(-H_-i a-H_i a+H_-i a+k+H_i a+kright) ,
da=\-frac12 i (textlog$Gamma $(1-i a)-textlog$Gamma $(1+i a)-textlog$Gamma
$(1-i a+k)+textlog$Gamma $(1+i a+k))+C$$
where $a=1$ and $C=0$ then:
$$colorbluesum _n=1^k tan ^-1left(frac1nright)=-frac12 i (textlog$Gamma
$(1-i)-textlog$Gamma $(1+i)-textlog$Gamma $((1-i)+k)+textlog$Gamma
$((1+i)+k))$$
Where: $H_i a$ is harmonic number and $textlog$Gamma $(1-i)$ is logarithm of the gamma function
MMA code:
HoldForm[Sum[ArcTan[1/n], n, 1, k] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
edited Jul 20 at 7:13
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
answered Jul 19 at 7:11
Mariusz Iwaniuk
1,6611615
1,6611615
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.:)
– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
add a comment |Â
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.:)
– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
An integral sign is missing on the second line. By the way, great answer!
– Szeto
Jul 20 at 3:28
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
Moreover, I don’t think that $C=0$. Consider the case $a=0$.
– Szeto
Jul 20 at 3:30
@Szeto. Case is only $a=1$,there are No others cases.
:)– Mariusz Iwaniuk
Jul 20 at 7:17
@Szeto. Case is only $a=1$,there are No others cases.
:)– Mariusz Iwaniuk
Jul 20 at 7:17
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
cute starting point ! (+1). Now use the fact that $Gamma (tilde z) = tilde Gamma (z)$ to simplify a further step and reach to the same conclusion as in my answer : $Im(ln(Gamma(1+i+k) / Gamma(1+i)))$
– G Cab
Jul 20 at 15:52
add a comment |Â
up vote
3
down vote
Playing around with Wolfy suggests that
$s(n)
=sum_k=1^n arctan(1/k)
=frac3pi4-frac12arctan(g(n))
$
where
$g(n)$ is an increasingly complicated fraction.
Some values are
$g(4) = 15/8,
g(5) = 140/71,
g(6) = 2848/7665,
g(7) = 14697/203896
$.
To get a recurrence for
$g(n)$,
$beginarray\
s(n+1)-s(n)
&=arctan(1/(n+1))\
&=(frac3pi4-frac12arctan(g(n+1)))-(frac3pi4-frac12arctan(g(n)))\
&=frac12(arctan(g(n))-arctan(g(n+1))\
endarray
$
so,
using
$arctan(x)pmarctan(y)
=arctan(fracxpm y1mp xy)
$,
$beginarray\
arctan(g(n+1))
&=arctan(g(n))-2arctan(1/(n+1))\
&=arctan(g(n))-arctan(frac2/(n+1)1-1/(n+1)^2)\
&=arctan(g(n))-arctan(frac2(n+1)(n+1)^2-1)\
&=arctan(g(n))-arctan(frac2(n+1)n^2+2n)\
&=arctan(fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n)\
&<arctan(g(n)-frac2n+1)\
endarray
$
so,
assuming that
the proper branch of
arctan is taken,
$g(n+1)
=fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n
lt g(n)-frac2n+1
$.
Since $sum 1/n$ diverges,
this shows that
exentually
$g(n) < 0$.
At this point the
next branch of
arctan has to be taken.
For example,
Wolfy calculates that
$s(20)
=frac5 À4 - frac12 arctan(frac4718365039332113702517864397263976449928)
$
and
$s(40)
=frac5À4 + frac12 arctan(frac41279370979134545450499387615832498927444174194743607269197868658553203529942799672226208517623565372926024)
$.
I'll leave it at this.
answered Jul 19 at 2:40
marty cohen
69.2k446122
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
add a comment |Â
up vote
3
down vote
Playing around with Wolfy suggests that
$s(n)
=sum_k=1^n arctan(1/k)
=frac3pi4-frac12arctan(g(n))
$
where
$g(n)$ is an increasingly complicated fraction.
Some values are
$g(4) = 15/8,
g(5) = 140/71,
g(6) = 2848/7665,
g(7) = 14697/203896
$.
To get a recurrence for
$g(n)$,
$beginarray\
s(n+1)-s(n)
&=arctan(1/(n+1))\
&=(frac3pi4-frac12arctan(g(n+1)))-(frac3pi4-frac12arctan(g(n)))\
&=frac12(arctan(g(n))-arctan(g(n+1))\
endarray
$
so,
using
$arctan(x)pmarctan(y)
=arctan(fracxpm y1mp xy)
$,
$beginarray\
arctan(g(n+1))
&=arctan(g(n))-2arctan(1/(n+1))\
&=arctan(g(n))-arctan(frac2/(n+1)1-1/(n+1)^2)\
&=arctan(g(n))-arctan(frac2(n+1)(n+1)^2-1)\
&=arctan(g(n))-arctan(frac2(n+1)n^2+2n)\
&=arctan(fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n)\
&<arctan(g(n)-frac2n+1)\
endarray
$
so,
assuming that
the proper branch of
arctan is taken,
$g(n+1)
=fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n
lt g(n)-frac2n+1
$.
Since $sum 1/n$ diverges,
this shows that
exentually
$g(n) < 0$.
At this point the
next branch of
arctan has to be taken.
For example,
Wolfy calculates that
$s(20)
=frac5 À4 - frac12 arctan(frac4718365039332113702517864397263976449928)
$
and
$s(40)
=frac5À4 + frac12 arctan(frac41279370979134545450499387615832498927444174194743607269197868658553203529942799672226208517623565372926024)
$.
I'll leave it at this.
answered Jul 19 at 2:40
marty cohen
69.2k446122
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Playing around with Wolfy suggests that
$s(n)
=sum_k=1^n arctan(1/k)
=frac3pi4-frac12arctan(g(n))
$
where
$g(n)$ is an increasingly complicated fraction.
Some values are
$g(4) = 15/8,
g(5) = 140/71,
g(6) = 2848/7665,
g(7) = 14697/203896
$.
To get a recurrence for
$g(n)$,
$beginarray\
s(n+1)-s(n)
&=arctan(1/(n+1))\
&=(frac3pi4-frac12arctan(g(n+1)))-(frac3pi4-frac12arctan(g(n)))\
&=frac12(arctan(g(n))-arctan(g(n+1))\
endarray
$
so,
using
$arctan(x)pmarctan(y)
=arctan(fracxpm y1mp xy)
$,
$beginarray\
arctan(g(n+1))
&=arctan(g(n))-2arctan(1/(n+1))\
&=arctan(g(n))-arctan(frac2/(n+1)1-1/(n+1)^2)\
&=arctan(g(n))-arctan(frac2(n+1)(n+1)^2-1)\
&=arctan(g(n))-arctan(frac2(n+1)n^2+2n)\
&=arctan(fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n)\
&<arctan(g(n)-frac2n+1)\
endarray
$
so,
assuming that
the proper branch of
arctan is taken,
$g(n+1)
=fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n
lt g(n)-frac2n+1
$.
Since $sum 1/n$ diverges,
this shows that
exentually
$g(n) < 0$.
At this point the
next branch of
arctan has to be taken.
For example,
Wolfy calculates that
$s(20)
=frac5 À4 - frac12 arctan(frac4718365039332113702517864397263976449928)
$
and
$s(40)
=frac5À4 + frac12 arctan(frac41279370979134545450499387615832498927444174194743607269197868658553203529942799672226208517623565372926024)
$.
I'll leave it at this.
answered Jul 19 at 2:40
marty cohen
69.2k446122
Playing around with Wolfy suggests that
$s(n)
=sum_k=1^n arctan(1/k)
=frac3pi4-frac12arctan(g(n))
$
where
$g(n)$ is an increasingly complicated fraction.
Some values are
$g(4) = 15/8,
g(5) = 140/71,
g(6) = 2848/7665,
g(7) = 14697/203896
$.
To get a recurrence for
$g(n)$,
$beginarray\
s(n+1)-s(n)
&=arctan(1/(n+1))\
&=(frac3pi4-frac12arctan(g(n+1)))-(frac3pi4-frac12arctan(g(n)))\
&=frac12(arctan(g(n))-arctan(g(n+1))\
endarray
$
so,
using
$arctan(x)pmarctan(y)
=arctan(fracxpm y1mp xy)
$,
$beginarray\
arctan(g(n+1))
&=arctan(g(n))-2arctan(1/(n+1))\
&=arctan(g(n))-arctan(frac2/(n+1)1-1/(n+1)^2)\
&=arctan(g(n))-arctan(frac2(n+1)(n+1)^2-1)\
&=arctan(g(n))-arctan(frac2(n+1)n^2+2n)\
&=arctan(fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n)\
&<arctan(g(n)-frac2n+1)\
endarray
$
so,
assuming that
the proper branch of
arctan is taken,
$g(n+1)
=fracg(n)-frac2(n+1)n^2+2n1+g(n)frac2(n+1)n^2+2n
lt g(n)-frac2n+1
$.
Since $sum 1/n$ diverges,
this shows that
exentually
$g(n) < 0$.
At this point the
next branch of
arctan has to be taken.
For example,
Wolfy calculates that
$s(20)
=frac5 À4 - frac12 arctan(frac4718365039332113702517864397263976449928)
$
and
$s(40)
=frac5À4 + frac12 arctan(frac41279370979134545450499387615832498927444174194743607269197868658553203529942799672226208517623565372926024)
$.
I'll leave it at this.
answered Jul 19 at 2:40
marty cohen
69.2k446122
answered Jul 19 at 2:40
marty cohen
69.2k446122
answered Jul 19 at 2:40
marty cohen
69.2k446122
answered Jul 19 at 2:40
marty cohen
69.2k446122
69.2k446122
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
add a comment |Â
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
Sadly no matches for fraction on OEIS.
– qwr
Jul 19 at 2:55
add a comment |Â
up vote
2
down vote
We have that
$$
prodlimits_1, le ,n, le ,k left( i + n right) = 1 over iprodlimits_0, le ,n, le ,k left( i + n right)
= 1 over ii^,overline ,k + 1, = left( 1 + i right)^,overline ,k,
= Gamma left( 1 + i + k right) over Gamma left( 1 + i right) = k!left( matrix
i + k cr
k cr right)
$$
where $x^,overline ,k, = Gamma (x + k) over Gamma (x)$ denotes the Rising Factorial
and $x^,underline ,k, = left( x - k + 1 right)^,overline ,k, $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic
functions even for complex $x$ and $k$.
Then
$$
ln left( z^,overline ,k, right) = ln left( right) + iarg left( z^,overline ,k, right)
= ln Gamma (z + k) over Gamma (z)
$$
The above tells us that your question is related to the absolute value and argument of the Gamma Function,
which unfortunately do not have a closed expression, better than the above.
answered Jul 19 at 1:37
G Cab
15.1k31136
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
 |Â
show 4 more comments
up vote
2
down vote
We have that
$$
prodlimits_1, le ,n, le ,k left( i + n right) = 1 over iprodlimits_0, le ,n, le ,k left( i + n right)
= 1 over ii^,overline ,k + 1, = left( 1 + i right)^,overline ,k,
= Gamma left( 1 + i + k right) over Gamma left( 1 + i right) = k!left( matrix
i + k cr
k cr right)
$$
where $x^,overline ,k, = Gamma (x + k) over Gamma (x)$ denotes the Rising Factorial
and $x^,underline ,k, = left( x - k + 1 right)^,overline ,k, $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic
functions even for complex $x$ and $k$.
Then
$$
ln left( z^,overline ,k, right) = ln left( right) + iarg left( z^,overline ,k, right)
= ln Gamma (z + k) over Gamma (z)
$$
The above tells us that your question is related to the absolute value and argument of the Gamma Function,
which unfortunately do not have a closed expression, better than the above.
answered Jul 19 at 1:37
G Cab
15.1k31136
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
We have that
$$
prodlimits_1, le ,n, le ,k left( i + n right) = 1 over iprodlimits_0, le ,n, le ,k left( i + n right)
= 1 over ii^,overline ,k + 1, = left( 1 + i right)^,overline ,k,
= Gamma left( 1 + i + k right) over Gamma left( 1 + i right) = k!left( matrix
i + k cr
k cr right)
$$
where $x^,overline ,k, = Gamma (x + k) over Gamma (x)$ denotes the Rising Factorial
and $x^,underline ,k, = left( x - k + 1 right)^,overline ,k, $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic
functions even for complex $x$ and $k$.
Then
$$
ln left( z^,overline ,k, right) = ln left( right) + iarg left( z^,overline ,k, right)
= ln Gamma (z + k) over Gamma (z)
$$
The above tells us that your question is related to the absolute value and argument of the Gamma Function,
which unfortunately do not have a closed expression, better than the above.
answered Jul 19 at 1:37
G Cab
15.1k31136
We have that
$$
prodlimits_1, le ,n, le ,k left( i + n right) = 1 over iprodlimits_0, le ,n, le ,k left( i + n right)
= 1 over ii^,overline ,k + 1, = left( 1 + i right)^,overline ,k,
= Gamma left( 1 + i + k right) over Gamma left( 1 + i right) = k!left( matrix
i + k cr
k cr right)
$$
where $x^,overline ,k, = Gamma (x + k) over Gamma (x)$ denotes the Rising Factorial
and $x^,underline ,k, = left( x - k + 1 right)^,overline ,k, $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic
functions even for complex $x$ and $k$.
Then
$$
ln left( z^,overline ,k, right) = ln left( right) + iarg left( z^,overline ,k, right)
= ln Gamma (z + k) over Gamma (z)
$$
The above tells us that your question is related to the absolute value and argument of the Gamma Function,
which unfortunately do not have a closed expression, better than the above.
answered Jul 19 at 1:37
G Cab
15.1k31136
edited Jul 23 at 9:11
answered Jul 19 at 1:37
G Cab
15.1k31136
answered Jul 19 at 1:37
G Cab
15.1k31136
answered Jul 19 at 1:37
G Cab
15.1k31136
15.1k31136
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
 |Â
show 4 more comments
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
I get the first and second steps but I don't understand what $overlinek+1$ and $Gamma$ are. Could you please explain?
– Shrey Joshi
Jul 19 at 1:40
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
@ShreyJoshi From context I assume it's a strange factorial variant: $a^overlineb = (a+b)!/a!$?
– user7530
Jul 19 at 1:43
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
Actually the third, fourth and fifth steps are not necessary. This can be derived directly from the definition of generalized binomial coefficients.
– Szeto
Jul 19 at 1:45
1
1
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
The Gamma function ($Gamma (x)$) is the same as an offset factorial ($(x - 1)!$), but defined in such a way that it is usable for non-integer values of $x$. $i^,overline ,k + 1,$ in this context is the rising factorial: $i(i + 1)(i + 2) cdots (i + k - 1)(i + k)$.
– Bladewood
Jul 19 at 2:21
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
@above, thank you for clearing it all up, I see how it fits together much better now. One thing I find funny is how we can define factorials and choose functions for imaginary numbers. It would also be interesting to see what would happen for fractional $k$.
– Shrey Joshi
Jul 19 at 2:53
 |Â
show 4 more comments
up vote
0
down vote
Note that the addition formula for the arctangent gives $arctan(x)+arctan(y)=arctan(dfracx+y1-xy)$; thus, if we define $alpha_n$ by $arctan(alpha_n)=sum_i=1^narctan(frac1i)$, then we have $alpha_n=dfracalpha_n-1+frac1n1-fracalpha_n-1n$ $=dfracnalpha_n-1+1n-alpha_n-1$; alternately, setting $alpha_n=fraca_nb_n$, we can write this as $a_n=na_n-1+b_n-1, b_n = -a_n-1+nb_n-1$ - or $beginpmatrixa_n \ b_n endpmatrix = beginpmatrixn & 1 \ -1 & nendpmatrixbeginpmatrixa_n-1 \ b_n-1 endpmatrix$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1leq nleq 3) =langle(1, 1), (3, 1), (10, 0)rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
0
down vote
Note that the addition formula for the arctangent gives $arctan(x)+arctan(y)=arctan(dfracx+y1-xy)$; thus, if we define $alpha_n$ by $arctan(alpha_n)=sum_i=1^narctan(frac1i)$, then we have $alpha_n=dfracalpha_n-1+frac1n1-fracalpha_n-1n$ $=dfracnalpha_n-1+1n-alpha_n-1$; alternately, setting $alpha_n=fraca_nb_n$, we can write this as $a_n=na_n-1+b_n-1, b_n = -a_n-1+nb_n-1$ - or $beginpmatrixa_n \ b_n endpmatrix = beginpmatrixn & 1 \ -1 & nendpmatrixbeginpmatrixa_n-1 \ b_n-1 endpmatrix$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1leq nleq 3) =langle(1, 1), (3, 1), (10, 0)rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that the addition formula for the arctangent gives $arctan(x)+arctan(y)=arctan(dfracx+y1-xy)$; thus, if we define $alpha_n$ by $arctan(alpha_n)=sum_i=1^narctan(frac1i)$, then we have $alpha_n=dfracalpha_n-1+frac1n1-fracalpha_n-1n$ $=dfracnalpha_n-1+1n-alpha_n-1$; alternately, setting $alpha_n=fraca_nb_n$, we can write this as $a_n=na_n-1+b_n-1, b_n = -a_n-1+nb_n-1$ - or $beginpmatrixa_n \ b_n endpmatrix = beginpmatrixn & 1 \ -1 & nendpmatrixbeginpmatrixa_n-1 \ b_n-1 endpmatrix$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1leq nleq 3) =langle(1, 1), (3, 1), (10, 0)rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
Note that the addition formula for the arctangent gives $arctan(x)+arctan(y)=arctan(dfracx+y1-xy)$; thus, if we define $alpha_n$ by $arctan(alpha_n)=sum_i=1^narctan(frac1i)$, then we have $alpha_n=dfracalpha_n-1+frac1n1-fracalpha_n-1n$ $=dfracnalpha_n-1+1n-alpha_n-1$; alternately, setting $alpha_n=fraca_nb_n$, we can write this as $a_n=na_n-1+b_n-1, b_n = -a_n-1+nb_n-1$ - or $beginpmatrixa_n \ b_n endpmatrix = beginpmatrixn & 1 \ -1 & nendpmatrixbeginpmatrixa_n-1 \ b_n-1 endpmatrix$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1leq nleq 3) =langle(1, 1), (3, 1), (10, 0)rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
answered Jul 19 at 22:28
Steven Stadnicki
40.1k765119
40.1k765119
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1
It seems you summed under the square root -- that would work if the square root were a logarithm, but it doesn't work for square roots. You can check that your result doesn't give the right values for the first few values of $k$. (Speaking of which, the products and summations run to $k$, whereas $n$ is just the summation index, so the free variable in the results should be $k$, not $n$.)
– joriki
Jul 19 at 0:15
Oops, it seems that I mixed up my products and summations.
– Shrey Joshi
Jul 19 at 0:31