How to prove that $intlimits_0^{pi} e^{sin^2(x)}dx > {3over2}pi$? [closed]
How to prove that $intlimits_0^{pi} e^{sin^2(x)} dx > {3 over 2}pi$?
definite-integrals integral-inequality
edited Nov 29 at 21:56
Bernard
118k638111
asked Nov 29 at 21:52
TBox
233
closed as off-topic by Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF Dec 4 at 22:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How to prove that $intlimits_0^{pi} e^{sin^2(x)} dx > {3 over 2}pi$?
definite-integrals integral-inequality
edited Nov 29 at 21:56
Bernard
118k638111
asked Nov 29 at 21:52
TBox
233
closed as off-topic by Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF Dec 4 at 22:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
1
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
5
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
1
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44
add a comment |
How to prove that $intlimits_0^{pi} e^{sin^2(x)} dx > {3 over 2}pi$?
definite-integrals integral-inequality
edited Nov 29 at 21:56
Bernard
118k638111
asked Nov 29 at 21:52
TBox
233
How to prove that $intlimits_0^{pi} e^{sin^2(x)} dx > {3 over 2}pi$?
definite-integrals integral-inequality
definite-integrals integral-inequality
edited Nov 29 at 21:56
Bernard
118k638111
asked Nov 29 at 21:52
TBox
233
edited Nov 29 at 21:56
Bernard
118k638111
asked Nov 29 at 21:52
TBox
233
edited Nov 29 at 21:56
Bernard
118k638111
edited Nov 29 at 21:56
Bernard
118k638111
edited Nov 29 at 21:56
Bernard
118k638111
118k638111
asked Nov 29 at 21:52
TBox
233
asked Nov 29 at 21:52
TBox
233
asked Nov 29 at 21:52
TBox
233
233
closed as off-topic by Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF Dec 4 at 22:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF Dec 4 at 22:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Rebellos, Jyrki Lahtonen, amWhy, José Carlos Santos, DRF
If this question can be reworded to fit the rules in the help center, please edit the question.
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
1
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
5
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
1
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44
add a comment |
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
1
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
5
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
1
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
1
1
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
5
5
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
1
1
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44
add a comment |
3 Answers
3
active
oldest
votes
Use the inequality $e^tge 1+t$, valid for all $t$, to get:
$$
int_0^pi e^{sin^2x}dxgeint_0^pi(1+sin^2x)dx=int_0^pileft(frac32 + frac12cos 2xright),dx
$$
You should be able to take it from here. The inequality is in fact strict, because the difference $e^{sin^2x}-(1+sin^2x)$ is continuous, non-negative and not identically zero.
answered Nov 29 at 22:10
grand_chat
20k11225
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
add a comment |
We can also use Taylor expansion: $e^{sin^2(x)}=1+sin^2(x)+sin^4(x)/2+...$ and integrate it from $0$ to $pi$
$int_0^pi(1+sin^2(x)+(sin^4(x))/2)dx=(27pi)/16 ≈ 5.3014 > 3pi/2$
edited Dec 4 at 22:45
amWhy
191k28224439
answered Dec 4 at 19:13
TBox
233
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
add a comment |
The integral is a Bessel function:
$$sqrt{e} pi I_0left(frac{1}{2}right)$$
which has numerical value $5.50843$, which is indeed less than $3 pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.
answered Nov 29 at 21:56
David G. Stork
9,65121232
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use the inequality $e^tge 1+t$, valid for all $t$, to get:
$$
int_0^pi e^{sin^2x}dxgeint_0^pi(1+sin^2x)dx=int_0^pileft(frac32 + frac12cos 2xright),dx
$$
You should be able to take it from here. The inequality is in fact strict, because the difference $e^{sin^2x}-(1+sin^2x)$ is continuous, non-negative and not identically zero.
answered Nov 29 at 22:10
grand_chat
20k11225
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
add a comment |
Use the inequality $e^tge 1+t$, valid for all $t$, to get:
$$
int_0^pi e^{sin^2x}dxgeint_0^pi(1+sin^2x)dx=int_0^pileft(frac32 + frac12cos 2xright),dx
$$
You should be able to take it from here. The inequality is in fact strict, because the difference $e^{sin^2x}-(1+sin^2x)$ is continuous, non-negative and not identically zero.
answered Nov 29 at 22:10
grand_chat
20k11225
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
add a comment |
Use the inequality $e^tge 1+t$, valid for all $t$, to get:
$$
int_0^pi e^{sin^2x}dxgeint_0^pi(1+sin^2x)dx=int_0^pileft(frac32 + frac12cos 2xright),dx
$$
You should be able to take it from here. The inequality is in fact strict, because the difference $e^{sin^2x}-(1+sin^2x)$ is continuous, non-negative and not identically zero.
answered Nov 29 at 22:10
grand_chat
20k11225
Use the inequality $e^tge 1+t$, valid for all $t$, to get:
$$
int_0^pi e^{sin^2x}dxgeint_0^pi(1+sin^2x)dx=int_0^pileft(frac32 + frac12cos 2xright),dx
$$
You should be able to take it from here. The inequality is in fact strict, because the difference $e^{sin^2x}-(1+sin^2x)$ is continuous, non-negative and not identically zero.
answered Nov 29 at 22:10
grand_chat
20k11225
edited Nov 29 at 23:42
answered Nov 29 at 22:10
grand_chat
20k11225
answered Nov 29 at 22:10
grand_chat
20k11225
answered Nov 29 at 22:10
grand_chat
20k11225
20k11225
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
add a comment |
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
I've taken and it equals to $int_0^pi(1+sin^2x)dx = {3over2}pi$ But is it really right to use $e^tge 1+t$?
– TBox
Nov 29 at 22:20
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
$$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion.
– Crostul
Nov 29 at 22:22
got it, thank you!
– TBox
Nov 29 at 22:28
got it, thank you!
– TBox
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@ArseniyBakaev The inequality $e^tge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011
– grand_chat
Nov 29 at 22:28
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
@Crostul Right, and the justification that $int_0^pi sin^4 (x) dx>0$ will equally apply to $int_0^pi e^{sin^2x}dx > int_0^pi (1+sin^2x)dx$.
– grand_chat
Nov 29 at 22:55
add a comment |
We can also use Taylor expansion: $e^{sin^2(x)}=1+sin^2(x)+sin^4(x)/2+...$ and integrate it from $0$ to $pi$
$int_0^pi(1+sin^2(x)+(sin^4(x))/2)dx=(27pi)/16 ≈ 5.3014 > 3pi/2$
edited Dec 4 at 22:45
amWhy
191k28224439
answered Dec 4 at 19:13
TBox
233
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
add a comment |
We can also use Taylor expansion: $e^{sin^2(x)}=1+sin^2(x)+sin^4(x)/2+...$ and integrate it from $0$ to $pi$
$int_0^pi(1+sin^2(x)+(sin^4(x))/2)dx=(27pi)/16 ≈ 5.3014 > 3pi/2$
edited Dec 4 at 22:45
amWhy
191k28224439
answered Dec 4 at 19:13
TBox
233
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
add a comment |
We can also use Taylor expansion: $e^{sin^2(x)}=1+sin^2(x)+sin^4(x)/2+...$ and integrate it from $0$ to $pi$
$int_0^pi(1+sin^2(x)+(sin^4(x))/2)dx=(27pi)/16 ≈ 5.3014 > 3pi/2$
edited Dec 4 at 22:45
amWhy
191k28224439
answered Dec 4 at 19:13
TBox
233
We can also use Taylor expansion: $e^{sin^2(x)}=1+sin^2(x)+sin^4(x)/2+...$ and integrate it from $0$ to $pi$
$int_0^pi(1+sin^2(x)+(sin^4(x))/2)dx=(27pi)/16 ≈ 5.3014 > 3pi/2$
edited Dec 4 at 22:45
amWhy
191k28224439
answered Dec 4 at 19:13
TBox
233
edited Dec 4 at 22:45
amWhy
191k28224439
edited Dec 4 at 22:45
amWhy
191k28224439
edited Dec 4 at 22:45
amWhy
191k28224439
191k28224439
answered Dec 4 at 19:13
TBox
233
answered Dec 4 at 19:13
TBox
233
answered Dec 4 at 19:13
TBox
233
233
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
add a comment |
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
Why post this, which copies the idea of a previous answer, only more clumsily?
– Did
Dec 5 at 0:06
add a comment |
The integral is a Bessel function:
$$sqrt{e} pi I_0left(frac{1}{2}right)$$
which has numerical value $5.50843$, which is indeed less than $3 pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.
answered Nov 29 at 21:56
David G. Stork
9,65121232
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
add a comment |
The integral is a Bessel function:
$$sqrt{e} pi I_0left(frac{1}{2}right)$$
which has numerical value $5.50843$, which is indeed less than $3 pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.
answered Nov 29 at 21:56
David G. Stork
9,65121232
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
add a comment |
The integral is a Bessel function:
$$sqrt{e} pi I_0left(frac{1}{2}right)$$
which has numerical value $5.50843$, which is indeed less than $3 pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.
answered Nov 29 at 21:56
David G. Stork
9,65121232
The integral is a Bessel function:
$$sqrt{e} pi I_0left(frac{1}{2}right)$$
which has numerical value $5.50843$, which is indeed less than $3 pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.
answered Nov 29 at 21:56
David G. Stork
9,65121232
edited Dec 5 at 0:44
answered Nov 29 at 21:56
David G. Stork
9,65121232
answered Nov 29 at 21:56
David G. Stork
9,65121232
answered Nov 29 at 21:56
David G. Stork
9,65121232
9,65121232
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
add a comment |
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
Is it possible to compare without Bessel function?
– TBox
Nov 29 at 21:57
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
The only other way to compare would be a numerical integration.
– David G. Stork
Nov 29 at 21:58
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
@DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below.
– Rebellos
Nov 29 at 22:03
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
"The only other way to compare would be a numerical integration" Not true.
– Did
Dec 5 at 0:05
add a comment |
Could you please edit the question, the math notations are not showing.
– John_Wick
Nov 29 at 21:53
1
I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page.
– DavidG
Nov 30 at 3:48
5
@DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort".
– epimorphic
Dec 4 at 17:52
1
@DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service.
– amWhy
Dec 4 at 22:42
@amWhy - will do. Thanks for the link.
– DavidG
Dec 4 at 22:44