3
$\begingroup$

I am looking for some math formula

For example \begin{align} & a^2 -b^2 = (a+b)(a-b) \\ &a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \end{align} First one related with factor a+b and the second one related with factor a+b+c then

How about some formula related with a,b,c,d

i.e., is there are some equation factors into (a+b+c+d)?

$\endgroup$
  • $\begingroup$ Well if $$a^2 - b^2 = (a + b)(a - b)$$ $$\implies (a + b)^2 - (c - d)^2 = (a + b + c + d)(a + b - c - d)$$ That is a formula related with a, b, c, d for example. Is this what you are asking for? $$(a + b)^2 = a^2 + b^2 + 2ab$$ $$\implies (c - d)^2 = c^2 + d^2 - 2cd$$ $$\therefore a^2 + b^2 + c^2 + d^2 + 2(ab - cd) = (a + b + c + d)(a + b - c - d)$$ But you can also do for something like $(a - b)^2 - (c + d)^2$ as well if you want $\endgroup$ – Mr Pie Sep 12 '17 at 3:36
  • 1
    $\begingroup$ @user477343 First term on the RHS should be $a+b+c-d$, shouldn't it? EDIT: but you can modify it to $$(a+b)^2 - (c+d)^2 = (a+b+c+d)(a+b-c-d)$$ $\endgroup$ – Zubin Mukerjee Sep 12 '17 at 3:37
  • $\begingroup$ Yes sorry about that. But I fixed it up :) EDIT: well technically $x - y - z = x - (y + z)$ so... $\endgroup$ – Mr Pie Sep 12 '17 at 3:38
  • $\begingroup$ @user477343 $$(a+b) + (c-d) \neq a + b + c + d$$ $$(a+b) - (c-d) \neq a + b + c + d$$ You either want your second term $c-d$ to be instead $c+d$, or $-c-d$, as you have it, I don't think it works $\endgroup$ – Zubin Mukerjee Sep 12 '17 at 3:47
  • $\begingroup$ Next problem , what is formula for,$$a_1,a_2,a_3\cdots,a_n$$ $\endgroup$ – neonpokharkar Sep 12 '17 at 4:20
3
$\begingroup$

It will be probably $$a^4+b^4+c^4+d^4-b^2a^2-c^2a^2-d^2a^2-b^2c^2-b^2d^2-d^2c^2+16abcd=(a+b+c+d)(a^3+b^3+c^3+d^3-ba^2-ca^2-da^2-ac^2-bc^2-dc^2-ab^2-cb^2-db^2-ad^2-bd^2-cd^2+4bca+4bda+4cad+4bcd)$$

OR

$$\sum a^4 -\sum a^2b^2 +16abcd=(\sum a)(\sum a^3-\sum ab^2 +4\sum abc)$$ $$$$ $$\sum \text{ represents cyclic summation}$$

$\endgroup$
  • $\begingroup$ Then is there any other representation for $(a^3+b^3+~$ terms? For example like $a^2+b^2+c^2 - ab-bc-ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$ I want to make the rest term positive-definite $\endgroup$ – phy_math Sep 12 '17 at 4:19
  • $\begingroup$ Let's see, i will try $\endgroup$ – neonpokharkar Sep 12 '17 at 4:23
  • $\begingroup$ Is it for 3 terms? $\endgroup$ – neonpokharkar Sep 12 '17 at 4:24
1
$\begingroup$

$$\begin{align} & a^2 -b^2 = (a+b)(a-b) \\ &a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \end{align}$$

The two relations are not quite "alike" since the second one is symmetric in $\,a,b,c\,$ (i.e. stays invariant if you permute the variables), while the first one is not (both sides change sign). Maybe a better analog would be for the first relation to be written as $\,a^2+b^2+2ab=(a+b)^2\,$.

With that note, the two equalities duplicate the Newton's identities for $\,n=2\,$ and $\,n=3\,$, respectively (where $p_k$ are the $k^{th}$ power sums, and $e_k$ the elementary symmetric polynomials):

$$ \begin{align} p_2 + 2 e_2 &= e_1 p_1 \\ p_3 - 3 e_3 &= e_1 p_2 - e_2 p_1 = e_1(p_2-e_2) \end{align} $$

The next identity for $\,n=4\,$ would then be:

$$ p_4 + 4 e_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = e_1(p_3+e_3) - e_2 p_2 $$

$$ \begin{align} \iff a^4+b^4+c^4+d^4 + 4 abcd &= (a+b+c+d)(a^3+b^3+c^3+d^3+abc+abd+acd+bcd) \\ &\quad - (a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd) \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.