nyuuzyou/wikis huggingface dataset:
Greek Anthology Book XIV: Metrodorus: 124

== Arithmetical Epigram of Metrodorus ==
:The sun, the moon, and the planets of the revolving zodiac spun such a nativity for thee;
::for a sixth part of thy life to remain an orphan with thy dear mother,
::for an eighth part to perform forced labour for thy enemies.
::For a third part the gods shall grant thee home-coming, and likewise a wife and a late-born son by her.
::Then thy son and wife shall perish by the spears of the Scythians,
::and then having shed tears for them thou shalt reach the end of thy life in twenty-seven years.
== Solution ==
Let $n$ be the length in years of the subject's life.
:$\dfrac n 6$ is spent as an orphan
:$\dfrac n 8$ is spent as a slave
:$\dfrac n 3$ is spent at home with a wife and son
:$27$ years remain.
We have:
So the subject lived $72$ years, of which:
:$12$ years were spent as an orphan
:$9$ years were spent as a slave
:$24$ years were spent as a family man at home
and $27$ years remained before his death.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $124$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 125

== Arithmetical Epigram of Metrodorus ==
:I am a tomb and I cover the lamented children of Philinna, containing fruit of her vainly-travailing womb such as I describe.
::Philinna gave me my fifth portion of young men,
::my third of maidens,
::and three newly married daughters;
::the other four descended to Hades from her womb without participating at all in the sunlight and in speech.
== Solution ==
Let $n$ be the number of children of Philinna.
:$\dfrac n 5$ were young men
:$\dfrac n 3$ were maidens
:$3$ were newly married daughters
:$4$ were still-born.
We have:
So Philinna had $15$ children, of which:
:$3$ were young men
:$5$ were maidens
and as we know:
:$3$ were newly married daughters
:$4$ were still-born.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $125$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 127

== Arithmetical Epigram of Metrodorus ==
:Demochares lived for a quarter of his whole life as a boy,
::for a fifth part of it a young man,
::and for a third as a man,
::and when he reached grey old age he lived thirteen years more on the threshold of eld.
== Solution ==
Let $n$ be the age in years of Demochares at his death.
:$\dfrac n 4$ was spent as a boy
:$\dfrac n 5$ was spent as a young man
:$\dfrac n 3$ was spent as a man
:$13$ more years were spent as an old man.
We have:
So Demochares lived for $60$ years, of which:
:$15$ years were spent as a boy
:$12$ years were spent as a young man
:$20$ years were spent as a man
and as we know:
:$13$ more years were spent as an old man.
== Also see ==
* Henry Ernest Dudeney/Puzzles and Curious Problems/41 - Ancient Problem
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $127$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 128

== Arithmetical Epigram of Metrodorus ==
:What violence my brother has done me, dividing our father's fortune of five talents unjustly!
:Poor tearful I have this fifth part of the seven-elevenths of my brother's share.
:Zeus, thou sleepest sound.
== Solution ==
Let $n$ be the narrator's share of the fortune.
We have that the brother's share is therefore $5 - n$ .
Hence:
So the narrator has $\dfrac {35} {62}$ of a .
The brother, meanwhile, has $4 \frac {27} {62}$ .
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $128$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 128/Historical Note

== Historical Note on Metrodorus' Arithmetical Epigram no. 128 ==
The translator plaintively remarks on the solution:
:The one offered is that the one brother had $4 \frac 4 {11}$ of a , the other $\frac 7 {11}$, but I cannot work it out.
Doubtless, as is apparent in the calculation given here on .
This would have been correct had the question been worded:
:... I have this fifth part of the seven-elevenths of the total inheritance
but it isn't so it's not.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $128$
Category:Historical Notes
Greek Anthology Book XIV: Metrodorus: 129

== Arithmetical Epigram of Metrodorus ==
:A traveller, ploughing with his ship the broad gulf of the Adriatic, said to the captain,
::"How much sea have we still to traverse?"
:And he answered him,
::"Voyager, between Cretan Ram's Head and Sicilian Peloris are six thousand stades,
::and twice two-fifths of the distance we have traversed remains till the Sicilian strait.
== Solution ==
It is assumed that "Sicilian Peloris" and "the Sicilian strait" are the same place.
That is, the total distance is $6000$ .
Let $d$ be the distance left to travel.
Hence:
So the narrator has $2666 \frac 2 3$ remaining to travel.
He has already travelled $3333 \frac 1 3$ .
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $129$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 130

== Arithmetical Epigram of Metrodorus ==
:Of the four spouts one filled the whole tank in a day,
::the second in two days,
::the third in three days,
::and the fourth in four days.
:What time will all four take to fill it?
== Solution ==
Let $t$ be the number of days it takes to fill the tank.
Let $a, b, c, d$ be the flow rate in numbers of tanks per day of (respectively) the first, second, third and fourth spouts.
In $t$ days, the various contributions of each of the spouts is $a t$, $b t$, $c t$ and $d t$ respectively.
So for the total contribution to be $1$ tank, we have:
We have:
and so:
So the tank will be filled in $\dfrac {12} {25}$ of a day.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $130$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Category:Cistern Problems
Greek Anthology Book XIV: Metrodorus: 131

== Arithmetical Epigram of Metrodorus ==
:Open me and I, a spout with abundant flow, will fill the present cistern in four hours;
::the one on my right requires four more hours to fill it,
::and the third twice as much.
:But if you bid them both join me in pouring forth a stream of water, we will fill it in a small part of the day.
== Solution ==
Let $t$ be the number of hours it takes to fill the cistern.
Let $a, b, c$ be the flow rate in numbers of cisterns per hour of (respectively) the first, second and third spouts.
In $t$ hours, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.
So for the total contribution to be $1$ cistern, we have:
We have:
and so:
So the cistern will be filled in $2 \frac 2 7$ hours.
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $131$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Category:Cistern Problems
Greek Anthology Book XIV: Metrodorus: 131/Historical Note

== Historical Note on Metrodorus' Arithmetical Epigram no. $131$ ==
In 's $1918$ translation of , he gives the answer as $2 \frac 2 {11}$ hours.
The discrepancy between this and the $2 \frac 2 7$ reported in the calculation is explained by the imprecision of the wording of the epigram.
If we take:
:the third twice as much
to mean:
:the third spout takes twice as much time to fill a cistern as the second spout
then we deduce that the third spout takes $16$ hours (that is, twice four plus four hours) to fill the cistern.
Hence we arrive at the solution $2 \frac 2 7$ hours.
However, if we interpret:
:the third twice as much
to mean:
:the third spout takes twice as much longer (that is, eight hours) than the first spout than the second one does to fill a cistern
that would mean the third spout takes $4$ hours plus twice $4$ hours, that is $12$ hours, to fill a cistern.
If this assumption is made, then $2 \frac 2 {11}$ is correct.
== Proof ==
Thus we have:
and so:
So the cistern will be filled in $2 \frac 2 {11}$ hours.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $131$
Category:Historical Notes
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 132

== Arithmetical Epigram of Metrodorus ==
:This is Polyphemus the brazen Cyclops, and as if on him someone made an eye, a mouth, and a hand, connecting them with pipes.
:He looks quite as if he were dripping water and seems also to be spouting it from his mouth.
:None of the spouts are irregular;
::that from his hand when running will fill the cistern in three days only,
::that from his eye in one day,
::and his mouth in two-fifths of a day.
:Who will tell me the time it takes when all three are running?
== Solution ==
Let $t$ be the number of hours it takes to fill the cistern.
Let $a, b, c$ be the flow rate in numbers of cisterns per hour of (respectively) the eye, the mouth and the hand.
In $t$ hours, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.
So for the total contribution to be $1$ cistern, we have:
We have:
and so:
So the cistern will be filled in $\dfrac 6 {23}$ of a day.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $132$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Category:Cistern Problems
Greek Anthology Book XIV: Metrodorus: 133

== Arithmetical Epigram of Metrodorus ==
:What a fine stream do these two river-gods and beautiful Bacchus pour into the bowl.
:The current of the streams of all is not the same.
::Nile flowing alone will fill it up in a day, so much water does he spout from his paps,
::and the thyrsus of Bacchus, sending forth wine, will fill it in three days,
::and thy horn, Achelous, in two days.
:Now run all together and you will fill it in a few hours.
== Solution ==
Let $t$ be the number of days it takes to fill the bowl.
Let $a, b, c$ be the flow rate in numbers of bowls per hour of (respectively) Nile, Bacchus and Achelous.
In $t$ days, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.
So for the total contribution to be $1$ bowl, we have:
We have:
and so:
So the bowl will be filled in $\dfrac 6 {11}$ of a day.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $133$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Category:Cistern Problems
Greek Anthology Book XIV: Metrodorus: 134

== Arithmetical Epigram of Metrodorus ==
:O woman, how hast thou forgotten Poverty?
:But she presses hard on thee, goading thee ever by force to labour.
:Thou didst use to spin a mina's weight of wool in a day,
::but thy eldest daughter spun a mina and one-third of thread,
::while thy younger daughter contributed a half-mina's weight.
:Now thou providest them all with supper, weighing out one mina only of wool.
== Solution ==
The exercise appears to be to calculate the contribution of each of the women during the time (less than a day) they spent spinning.
Let $a$, $b$ and $c$ (respectively) be the contribution of the mother, the older daughter and the younger daughter of the $1$ of wool.
Let $w$ be the weight of wool in they would produce together if all three were to spin all day.
Then:
Thus if all $3$ worked all day, they would produce $\dfrac {17} 6$ .
It follows to produce $1$ , they worked for $\dfrac 6 {17}$ of a day.
Thus their individual contributions are
So:
:the mother contributed $\dfrac 6 {17}$ of a of wool
:the older daughter contributed $\dfrac 8 {17}$ of a of wool
:the younger daughter contributed $\dfrac 3 {17}$ of a of wool.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $134$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 135

== Arithmetical Epigram of Metrodorus ==
:We three Loves stand here pouring out water for the bath, sending streams into the fair-flowing tank.
::I on the right, from my long-winged feet, fill it full in the sixth part of a day;
::I on the left, from my jar, fill it in four hours;
::and I in the middle, from my bow, in just half a day.
:Tell me in what a short time we should fill it, pouring water from wings, bow and jar all at once.
== Solution ==
First note that the day in this context is taken to be $12$ hours long.
Let $t$ be the number of days it takes to fill the tank.
Let $a, b, c$ be the flow rate in numbers of tanks per day of (respectively) the Loves on the right, the left and the middle.
In $t$ days, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.
So for the total contribution to be $1$ tank, we have:
We have:
and so:
So the tank will be filled in $\dfrac 1 {11}$ of a day.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $135$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Category:Cistern Problems
Greek Anthology Book XIV: Metrodorus: 136

== Arithmetical Epigram of Metrodorus ==
:Brick-makers, I am in a great hurry to erect this house.
:To-day is cloudless, and I do not require many more bricks, but I have all I want but three hundred.
::Thou alone in one day couldst make as many,
::but thy son left off working when he had finished two hundred,
::and thy son-in-law when he had made two hundred and fifty.
:Working all together, in how many hours can you make these?
== Solution ==
Let $b$ be the number of bricks they can produce together in one day.
Then:
Thus it takes $\dfrac {300} {750} = \dfrac 2 5$ of a day to produce $300$ bricks when all working together.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $136$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 137

== Arithmetical Epigram of Metrodorus ==
:Let fall a tear as you pass by; for we are those guests of Antiochus whom his house slew when it fell, and God gave us in equal shares this place for a banquet and a tomb.
::Four of us from Tegea lie here,
::twelve from Messene,
::five from Argos,
::and half of the banqueters were from Sparta,
::and Antiochus himself.
::A fifth of the fifth part of those who perished were from Athens,
::and do thou, Corinth, weep for Hylas alone.
== Solution ==
Let $n$ be the number of guests who were killed in the accident.
:$4$ came from Tegea
:$12$ came from Messene
:$5$ came from Argos
:$\dfrac n 2$ came from Sparta
:$1$ was Antiochus
:$\dfrac n {5 \times 5}$ came from Athens
:$1$ (that is, Hylas) came from Corinth.
We have:
So there were $50$ guests.
:$25$ came from Sparta
:$2$ came from Athens
and the numbers of the remainder are specified explicitly.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $137$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 138

== Arithmetical Epigram of Metrodorus ==
:Nicarete, playing with five companions of her own age,
::gave a third of the nuts she had to Cleis,
::the quarter to Sappho,
::and the fifth to Aristodice,
::the twentieth and again the twelfth to Theano,
::and the twenty-fourth to Philinnis.
:Fifty nuts were left for herself.
== Solution ==
Let $n$ be the number of nuts in total.
:$\dfrac n 3$ went to Cleis
:$\dfrac n 4$ went to Sappho
:$\dfrac n 5$ went to Aristodice
:$\dfrac n {20} + \dfrac n {12}$ went to Theano
:$\dfrac n {24}$ went to Philinnis
:$50$ stayed with Nicarete.
We have:
So there were $1200$ nuts.
:$400$ went to Cleis
:$300$ went to Sappho
:$240$ went to Aristodice
:$60 + 100 = 160$ went to Theano
:$50$ went to Philinnis
and we know that Nicarete kept $50$.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $138$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 139

== Arithmetical Epigram of Metrodorus ==
:Diodorus, great glory of dial-makers, tell me the hour since when the golden wheels of the sun leapt up from the east to the pole.
:Four times three-fifths of the distance he has traversed remain until he sinks to the western sea.
== Solution ==
Let $t$ be the time since dawn.
It is assumed that the day is $12$ hours long.
We have:
So $3 \frac 9 {17}$ hours have passed since dawn.
That leaves $12 - 3 \frac 9 {17} = 8 \frac 8 {17}$ hours remaining till sunset.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $139$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 140

== Arithmetical Epigram of Metrodorus ==
:Blessed Zeus, are these deeds pleasing in thy sight that the Thessalian women do in play?
:The eye of the moon is blighted by mortals; I saw it myself.
:The night still wanted till morning twice two-sixths and twice one-seventh of what was past.
== Solution ==
Let $t$ be the time since sunset.
It is assumed that the night is $12$ hours long.
We have:
So $6 \frac 6 {41}$ hours have passed since sunset.
That leaves $12 - 6 \frac 6 {41} = 5 \frac {35} {41}$ hours remaining till sunrise.
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $140$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 141

== Arithmetical Epigram of Metrodorus ==
:Tell me the transits of the fixed stars and planets when my wife gave birth to a child yesterday.
:It was day, and till the sun set in the western sea it wanted six times two-sevenths of the time since dawn.
== Solution ==
Let $t$ be the time since dawn that the birth happened.
It is assumed that the day is $12$ hours long.
We have:
So $7 \frac {11} {19}$ hours had passed since dawn when the baby was born.
That leaves $12 - 7 \frac {11} {19} = 4 \frac 8 {19}$ hours remaining till sunset.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $141$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 142

== Arithmetical Epigram of Metrodorus ==
:Arise, work-women, it is past dawn;
:a fifth part of three-eighths of what remains is gone by.
== Solution ==
Let $t$ be the time in hours since dawn.
It is assumed that the day is $12$ hours long.
We have:
So $\dfrac {36} {43}$ of an hour has passed since dawn.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $142$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 143

== Arithmetical Epigram of Metrodorus ==
:The father perished in the shoals of the Syrtis, and this, the eldest of the brothers, came back from that voyage with five talents.
::To me he gave twice two-thirds of his share,
::on our mother he bestowed two-eighths of my share,
:nor did he sin against divine justice.
== Solution ==
Let $a$ be the elder brother's share.
Let $b$ be the narrator's share.
Let $c$ be the mother's share.
We have:
So:
:the elder brother takes $1 \frac 7 8$ 
:the narrator takes $2 \frac 1 2$ 
:the mother takes $\dfrac 5 8$ of a .
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $143$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 143/Historical Note

== Historical Note on Metrodorus' Arithmetical Epigram no. $143$ ==
In 's $1918$ translation of , he gives the answer as:
:the elder brother had $1 \frac 5 7$ 
:the narrator had $2 \frac 2 7$ 
:the mother had $1$ .
The discrepancy between this and the shares reported in the calculation appears to be due to a misinterpretation of the wording.
If we replace:
:on our mother he bestowed two-eighths of my share
with:
:on our mother he bestowed two-eighths of our combined share
then we arrive at the solution given by .
== Proof ==
Let $a$ be the elder brother's share.
Let $b$ be the narrator's share.
Let $c$ be the mother's share.
We have:
So:
:the elder brother takes $1 \frac 5 7$ 
:the narrator takes $2 \frac 2 7$ 
:the mother takes $1$ .
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $143$
Category:Historical Notes
Greek Anthology Book XIV: Metrodorus: 144

== Arithmetical Epigram of Metrodorus ==
:$A$. How heavy is the base I have to stand on together with myself!
:$B$. And my base together together with myself weighs the same number of .
:$A$. But I alone weigh twice as much as your base.
:$B$: And I alone weigh three times the weight of yours.
== Solution ==
Presumably $A$ and $B$ are talking statues.
Let $a$ be the weight of $A$.
Let $b$ be the weight of $B$.
Let $c$ be the weight of $A$'s base.
Let $d$ be the weight of $B$'s base.
We have:
We have $3$ simultaneous equations in $4$ unknowns.
Hence we cannot calculate the actual weights, but the proportions only.
So:
So, taking the base of $A$ as a unit, we have:
:$A$ weighs $4$ times its base
:$B$ weighs $3$ times the base of $A$
:the base of $B$ weighs twice the base of $B$.
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $144$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 144/Historical Note

== Historical Note on Metrodorus' Arithmetical Epigram no. $144$ ==
In 's $1918$ translation of , he gives the answer as:
:From these data not the actual weights but the proportions alone can be determined.
:The statue $A$ was a third part heavier than $B$, and $B$ only weighed $\dfrac 3 4$ of the statue $A$.
:The base of $B$ weighed thrice as much as the base of $A$.
This appears to be incorrect.
== Proof ==
Let $a$ be the weight of $A$.
Let $b$ be the weight of $B$.
Let $c$ be the weight of $A$'s base.
Let $d$ be the weight of $B$'s base.
Let us assume the solution given by .
Let us assume that $B$'s first statement is true.
:$B$. And my base together together with myself weighs the same number of .
We have:
That is, $B$ weighed $6$ times as much as the base of $A$, which deviates from the initial statement of the problem:
:$B$: And I alone weigh three times the weight of yours.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $144$
Category:Historical Notes
Greek Anthology Book XIV: Metrodorus: 145

== Arithmetical Epigram of Metrodorus ==
:$A$. Give me ten and I become three times as much as you.
:$B$. And if I got the same from you I am five times as much as you.
== Solution ==
Let $a$ be the quantity of $A$.
Let $b$ be the quantity of $B$.
We have:
So:
:$A$ is $15 \dfrac 5 7$
:$B$ is $18 \dfrac 4 7$.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $145$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 146

== Arithmetical Epigram of Metrodorus ==
:$A$. Give me two and I become twice as much as you.
:$B$. And if I got the same from you I am four times as much as you.
== Solution ==
Let $a$ be the quantity of $A$.
Let $b$ be the quantity of $B$.
We have:
So:
:$A$ is $3 \frac 5 7$
:$B$ is $4 \frac 6 7$.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $146$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Green's Identities

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian manifold with or without boundary $\partial M$.
Let $\hat g$ be the induced metric on $\partial M$.
Let $\map {C^\infty} M$ be the smooth function space.
Let $u, v \in \map {C^\infty} M$ be smooth real functions on $M$.
Let $\rd V_g$ be the Riemannian volume form.
Let $\nabla^2$ be the Laplace-Beltrami operator.
Let $\grad$ be the gradient operator.
Let $N$ be the outward-pointing unit normal vector field on $\partial M$.
Let $\innerprod \cdot \cdot_g$ be the Riemannian metric the metric $g$.
Then:
:$\ds \int_M u \nabla^2 v \rd V_g = \int_{\partial M} u N v \rd V_{\hat g} - \int_M \innerprod {\grad u} {\grad v}_g \rd V_g$
:$\ds \int_M \paren {u \nabla^2 v - v \nabla^2 u} \rd V_g = \int_{\partial M} \paren {u N v - v N u} \rd V_{\hat g}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry