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I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.

Given the function

$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$

Prove that the taylor polynomial of degree $4$ of $f$ is equal to $5+x^4+y^4$.

First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $

$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$

Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$

Therefore: $f(x,y)=x^4+y^4+5+4R_3$

I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.

Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

New edit

I understand this now thanks to hint of Hagen von Eitzen, thanks !

The third question is:

Determine what kind of stationary point you have in $(0,0)$.

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Are you really expected to use the first problem to solve the second question? It's easy to solve the second question directly. –  Thomas Andrews May 15 '13 at 15:38
    
Downvote, explain yourself please ! –  Kasper May 15 '13 at 15:38
1  
It looks more like the second part is a step in doing something more with the result of the first part. –  Mark Bennet May 15 '13 at 15:39
    
The second question really asks you to show inequality $a^2 + b^2 \geq \frac{(a+b)^2}{2}$ which is straight-forward. –  JavaMan May 15 '13 at 17:31

3 Answers 3

Hint: The difference left minus right is $\frac12(x^4-2x^2y^2+y^4)$

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@Kasper, you'll get $\frac{1}{2}(x^4-2x^2y^2+y^4)\ge 0 $ that can be simplified as $(x^2-y^2)^2 \ge 0$. –  Alan May 15 '13 at 16:00

Write $x^4=2x^4-x^4$ and similarly $y^4=2y^4-y^4$

$$(x^2-y^2)^2 \ge 0$$

$$x^4+y^4-2x^2y^2 \ge 0$$

$$2x^4-x^4+2y^4-y^4-2x^2y^2 \ge0$$

$$2x^4+2y^4 \ge x^4+y^4+2x^2y^2$$

$$x^4+y^4 \ge \dfrac{(x^2+y^2)^2}{2}$$

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Typo: Should be $y^4$ on the third line of the inequalities. –  BU982T May 15 '13 at 16:26
    
Corrected it. Thank you. –  Inceptio May 15 '13 at 16:27

$$x^4 + y^4 \ge \frac{(x^2+y^2)^2}{2} \\ 2x^4+2y^4\ge x^4+2x^2y^2+y^4 \\ x^4-2x^2y^2+y^4 \ge0 \\ (x^2-y^2)^2\ge0$$

Notice that $x^2-y^2\in R$, so all real number with an even exponent are always greater or iqual to zero.

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