Fdtxjr

How to answer pointed "are you quitting" questioning when I don't want them to suspect

Creating a loop after a break using Markov Chain in Tikz

Does the average primeness of natural numbers tend to zero?

extract characters between two commas?

Symmetry in quantum mechanics

Can the Produce Flame cantrip be used to grapple, or as an unarmed strike, in the right circumstances?

Ideas for 3rd eye abilities

What is the meaning of "of trouble" in the following sentence?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Why doesn't a const reference extend the life of a temporary object passed via a function?

A poker game description that does not feel gimmicky

Was there ever an axiom rendered a theorem?

Why do UK politicians seemingly ignore opinion polls on Brexit?

COUNT(*) or MAX(id) - which is faster?

What happens when a metallic dragon and a chromatic dragon mate?

Is ipsum/ipsa/ipse a third person pronoun, or can it serve other functions?

How to move the player while also allowing forces to affect it

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

Doomsday-clock for my fantasy planet

Re-submission of rejected manuscript without informing co-authors

Is there a name of the flying bionic bird?

How could a lack of term limits lead to a "dictatorship?"

Where to refill my bottle in India?

Lied on resume at previous job

Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.

Finding parametric equations for the tangent line at a point on a curveIntegral in spherical coordinates, $Omega$ is the unit sphere, of $iiint_Omega 1/(2+z)^2dx dy dz$Proving injectivity and surjectivityAnalysing/Visualising shape of multi-variate function.Minimize and maximize the function $g(a,b) = a + b$ given the constraint $h(a,b) = frac1a + frac1b = 1$.How to integrate an equation with multiple non-independent variablesHow is “expressing” a differential operator “in cylindrical coordinates” rigorously defined?Why is this relation logical?How do you integrate $e^x^2$ over a closed interval?Area of fourth quadrant in unit circle

1

$begingroup$

Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.

I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.

My Solution After Tips:

Let $u = x + y$ and $ v = x - y$. Then $ x = u - y $ and $ y = x - v $. From these equations, we have that

$ x = u - y = u - (x - v) rightarrow x = u - x + v rightarrow 2x = u + v Rightarrow x = (u + v)/2 $.

Similarly, $u - v = (x + y) - (x - y) = x + y - x + y = 2y Rightarrow y = (u - v)/2$.

Substituting, we get

$$ f(u, v) = (fracu + v2)(fracu - v2) + (fracu + v2)^2 $$

$$ f(u, v) = fracu^2 - uv2 $$

multivariable-calculus

share|cite|improve this question

edited Mar 23 at 0:33

Victor S.

asked Mar 22 at 16:09

Victor S.Victor S.

32819

$endgroup$

add a comment | 

1

$begingroup$

Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.

I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.

My Solution After Tips:

Let $u = x + y$ and $ v = x - y$. Then $ x = u - y $ and $ y = x - v $. From these equations, we have that

$ x = u - y = u - (x - v) rightarrow x = u - x + v rightarrow 2x = u + v Rightarrow x = (u + v)/2 $.

Similarly, $u - v = (x + y) - (x - y) = x + y - x + y = 2y Rightarrow y = (u - v)/2$.

Substituting, we get

$$ f(u, v) = (fracu + v2)(fracu - v2) + (fracu + v2)^2 $$

$$ f(u, v) = fracu^2 - uv2 $$

multivariable-calculus

share|cite|improve this question

edited Mar 23 at 0:33

Victor S.

asked Mar 22 at 16:09

Victor S.Victor S.

32819

$endgroup$

add a comment | 

$begingroup$

Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.

I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.

My Solution After Tips:

Let $u = x + y$ and $ v = x - y$. Then $ x = u - y $ and $ y = x - v $. From these equations, we have that

$ x = u - y = u - (x - v) rightarrow x = u - x + v rightarrow 2x = u + v Rightarrow x = (u + v)/2 $.

Similarly, $u - v = (x + y) - (x - y) = x + y - x + y = 2y Rightarrow y = (u - v)/2$.

Substituting, we get

$$ f(u, v) = (fracu + v2)(fracu - v2) + (fracu + v2)^2 $$

$$ f(u, v) = fracu^2 - uv2 $$

multivariable-calculus

share|cite|improve this question

edited Mar 23 at 0:33

Victor S.

asked Mar 22 at 16:09

Victor S.Victor S.

32819

$endgroup$

share|cite|improve this question

edited Mar 23 at 0:33

Victor S.

asked Mar 22 at 16:09

Victor S.Victor S.

32819

share|cite|improve this question

edited Mar 23 at 0:33

Victor S.

asked Mar 22 at 16:09

Victor S.Victor S.

32819

asked Mar 22 at 16:09

Victor S.Victor S.

32819

asked Mar 22 at 16:09

Victor S.Victor S.

32819

3 Answers
3

active

oldest

votes

1

$begingroup$

You can solve it directly as follows:

• $f(x+y,x-y) = xy+y^2 = (x+y)y$

Now, write

• $y = frac12((x+y)-(x-y))$

It follows:

$$f(x+y,x-y)= (x+y)cdot frac12((x+y)-(x-y)) = frac12left( (x+y)^2 - (x+y)(x-y)right)$$ $$Rightarrow f(x,y) = frac12(x^2-xy)$$

share|cite|improve this answer

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you think about this substitution for y in the first place? Just intuition?
  $endgroup$
  – Victor S.
  Mar 22 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @VictorS. When looking at the expression $(x+y)y$, the next question is, how to express $y$ in terms of $x+y$ and $x-y$. So, you either just "see" ("intuite") or know it or you may set up $a(x+y) +b(x-y) =0cdot x +1cdot y$ and find $a$ and $b$.
  $endgroup$
  – trancelocation
  Mar 22 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just understood your reasoning for why one would want this substitution. Thanks
  $endgroup$
  – Victor S.
  Mar 23 at 0:30
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $u = x+y, v = x-y$.

Now express each of $x$ and $y$ in terms of only $u$ and $v$.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 16:10

avsavs

3,944515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure I follow your process. But what I do gives the same result as you got: $$ f(u,v) = f(x+y, x-y) = xy + y^2 = [(u+v)/2] ; [(u-v)/2] + [(u-v)/2]^2 = (u^2 - v^2)/4 + (u^2 - 2uv + v^2)/4, $$ and the like terms can be collected.
  $endgroup$
  – avs
  Mar 22 at 18:30
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: $displaystyle(x,y)=left(fracx+y2+fracx-y2,fracx+y2-fracx-y2right)$

share|cite|improve this answer

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, that is not correct. The correct answer is one quarter of that: $f(x,y)=fracx^2-xy2$. You can check that, with this function, you indeed have $f(x+y,x-y)=xy+y^2$.
  $endgroup$
  – José Carlos Santos
  Mar 22 at 17:16
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158334%2fdetermine-fx-y-if-fx-y-x-y-xy-y2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

You can solve it directly as follows:

• $f(x+y,x-y) = xy+y^2 = (x+y)y$

Now, write

• $y = frac12((x+y)-(x-y))$

It follows:

$$f(x+y,x-y)= (x+y)cdot frac12((x+y)-(x-y)) = frac12left( (x+y)^2 - (x+y)(x-y)right)$$ $$Rightarrow f(x,y) = frac12(x^2-xy)$$

share|cite|improve this answer

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you think about this substitution for y in the first place? Just intuition?
  $endgroup$
  – Victor S.
  Mar 22 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @VictorS. When looking at the expression $(x+y)y$, the next question is, how to express $y$ in terms of $x+y$ and $x-y$. So, you either just "see" ("intuite") or know it or you may set up $a(x+y) +b(x-y) =0cdot x +1cdot y$ and find $a$ and $b$.
  $endgroup$
  – trancelocation
  Mar 22 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just understood your reasoning for why one would want this substitution. Thanks
  $endgroup$
  – Victor S.
  Mar 23 at 0:30
  

  
  
  
  

add a comment | 

1

$begingroup$

You can solve it directly as follows:

• $f(x+y,x-y) = xy+y^2 = (x+y)y$

Now, write

• $y = frac12((x+y)-(x-y))$

It follows:

$$f(x+y,x-y)= (x+y)cdot frac12((x+y)-(x-y)) = frac12left( (x+y)^2 - (x+y)(x-y)right)$$ $$Rightarrow f(x,y) = frac12(x^2-xy)$$

share|cite|improve this answer

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would you think about this substitution for y in the first place? Just intuition?
  $endgroup$
  – Victor S.
  Mar 22 at 19:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @VictorS. When looking at the expression $(x+y)y$, the next question is, how to express $y$ in terms of $x+y$ and $x-y$. So, you either just "see" ("intuite") or know it or you may set up $a(x+y) +b(x-y) =0cdot x +1cdot y$ and find $a$ and $b$.
  $endgroup$
  – trancelocation
  Mar 22 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just understood your reasoning for why one would want this substitution. Thanks
  $endgroup$
  – Victor S.
  Mar 23 at 0:30
  

  
  
  
  

add a comment | 

$begingroup$

You can solve it directly as follows:

• $f(x+y,x-y) = xy+y^2 = (x+y)y$

Now, write

• $y = frac12((x+y)-(x-y))$

It follows:

$$f(x+y,x-y)= (x+y)cdot frac12((x+y)-(x-y)) = frac12left( (x+y)^2 - (x+y)(x-y)right)$$ $$Rightarrow f(x,y) = frac12(x^2-xy)$$

share|cite|improve this answer

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

$endgroup$

share|cite|improve this answer

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

answered Mar 22 at 17:39

trancelocationtrancelocation

13.7k1829

4

$begingroup$

Let $u = x+y, v = x-y$.

Now express each of $x$ and $y$ in terms of only $u$ and $v$.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 16:10

avsavs

3,944515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure I follow your process. But what I do gives the same result as you got: $$ f(u,v) = f(x+y, x-y) = xy + y^2 = [(u+v)/2] ; [(u-v)/2] + [(u-v)/2]^2 = (u^2 - v^2)/4 + (u^2 - 2uv + v^2)/4, $$ and the like terms can be collected.
  $endgroup$
  – avs
  Mar 22 at 18:30
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $u = x+y, v = x-y$.

Now express each of $x$ and $y$ in terms of only $u$ and $v$.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 16:10

avsavs

3,944515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure I follow your process. But what I do gives the same result as you got: $$ f(u,v) = f(x+y, x-y) = xy + y^2 = [(u+v)/2] ; [(u-v)/2] + [(u-v)/2]^2 = (u^2 - v^2)/4 + (u^2 - 2uv + v^2)/4, $$ and the like terms can be collected.
  $endgroup$
  – avs
  Mar 22 at 18:30
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $u = x+y, v = x-y$.

Now express each of $x$ and $y$ in terms of only $u$ and $v$.

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 16:10

avsavs

3,944515

$endgroup$

share|cite|improve this answer

edited Mar 22 at 16:19

answered Mar 22 at 16:10

avsavs

3,944515

answered Mar 22 at 16:10

avsavs

3,944515

answered Mar 22 at 16:10

avsavs

3,944515

1

$begingroup$

Hint: $displaystyle(x,y)=left(fracx+y2+fracx-y2,fracx+y2-fracx-y2right)$

share|cite|improve this answer

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, that is not correct. The correct answer is one quarter of that: $f(x,y)=fracx^2-xy2$. You can check that, with this function, you indeed have $f(x+y,x-y)=xy+y^2$.
  $endgroup$
  – José Carlos Santos
  Mar 22 at 17:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: $displaystyle(x,y)=left(fracx+y2+fracx-y2,fracx+y2-fracx-y2right)$

share|cite|improve this answer

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I added my solution to the original question. Could you verify, please?
  $endgroup$
  – Victor S.
  Mar 22 at 16:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, that is not correct. The correct answer is one quarter of that: $f(x,y)=fracx^2-xy2$. You can check that, with this function, you indeed have $f(x+y,x-y)=xy+y^2$.
  $endgroup$
  – José Carlos Santos
  Mar 22 at 17:16
  

  
  
  
  

add a comment | 

$begingroup$

Hint: $displaystyle(x,y)=left(fracx+y2+fracx-y2,fracx+y2-fracx-y2right)$

share|cite|improve this answer

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 22 at 16:12

José Carlos SantosJosé Carlos Santos

173k23133241