30
$\begingroup$

It is given that $$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?

$\endgroup$
  • 5
    $\begingroup$ These equations may be tedious, but they're not linear, assuming all variables are unknowns. $\endgroup$ – Matt Samuel Feb 9 '15 at 4:31
  • 28
    $\begingroup$ "Today class, we'll be covering how to solve systems of tedious equations, for the people in engineering that love tedious calculations." $\endgroup$ – Dair Feb 9 '15 at 4:33
  • $\begingroup$ I see someone else who uses clevermath $\endgroup$ – Faraz Masroor Feb 10 '15 at 2:31
36
$\begingroup$

Note that \begin{align} f(k)&=(a+k)(b+k)(c+k)(d+k) \\ &=k^4+(a+b+c+d)k^3+(a b+a c+a d+b c+b d+c d) k^2 +(a b c+a b d+a c d+b c d) k+a b cd \\ &=k^4 + k^3 e_1 + k^2 e_2 + k e_3 + e_4. \end{align}

Now you have a linear system of 4 equations and 4 unknowns ($e_1,e_2,e_3,e_4$). Solve it and find $f(5)$.

$\endgroup$
  • 4
    $\begingroup$ Oh, might as well. So $f(k)=k^4+4k^2+3k+7$ hence $k(5) = 747$. $\endgroup$ – Tito Piezas III Feb 9 '15 at 7:06
33
$\begingroup$

First, by expanding the four equations, we note that the system of equations is equivalent to $$ \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 \\ 1 & 3 & 3 & 9 & 3 & 9 & 9 & 27 & 3 & 9 & 9 & 27 & 9 & 27 & 27 \\ 1 & 4 & 4 & 16 & 4 & 16 & 16 & 64 & 4 & 16 & 16 & 64 & 16 & 64 & 64 \end{bmatrix} \begin{bmatrix} abcd \\ abc \\ abd \\ ab \\ acd \\ ac \\ ad \\ a \\ bcd \\ bc \\ bd \\ b \\ cd \\ c \\ d \end{bmatrix} = \begin{bmatrix} 14 \\ 29 \\ 52 \\ 83 \end{bmatrix}. $$ Using row reduction, we see that this is equivalent to the matrix equation $$ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} abcd \\ abc \\ abd \\ ab \\ acd \\ ac \\ ad \\ a \\ bcd \\ bc \\ bd \\ b \\ cd \\ c \\ d \end{bmatrix} = \begin{bmatrix} 7 \\ 3 \\ 4 \\ 0 \end{bmatrix}. $$ This resolves into the following system of equations: $$ \begin{aligned} abcd = 7 \\ abc + abd + acd + bcd = 3 \\ ab + ac + ad + bc + bd + cd = 4 \\ a + b + c + d = 0. \end{aligned} $$

Finally, we note that $(a+5)(b+5)(c+5)(d+5)$ is equal to $$abcd + 5(abc + abd + acd + bcd) + 25(ab + ac + ad + bc + bd + cd) + 125(a + b + c + d) + 625.$$ So $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.

$\endgroup$
  • 1
    $\begingroup$ Amazing! Plus One!!! $\endgroup$ – Robert Lewis Feb 10 '15 at 6:12
12
$\begingroup$

According to Maple, there are no real solutions. There are complex solutions. They have $a$ as a root of the irreducible quartic $x^4+4 x^2-3 x+7$.

But as for the value of $(a+5)(b+5)(c+5)(d+5)$: note that $(a+x)(b+x)(c+x)(d+x)$ is a monic quartic polynomial in $x$. Find a monic quartic with values $15,\;45,\;133,\;339$ at $1,2,3,4$. This part involves linear equations, or you might use a table of differences...

$\endgroup$
  • $\begingroup$ Could you please elaborate a little bit. I am unable to understand what you want me to do? $\endgroup$ – user167045 Feb 9 '15 at 5:08
  • 6
    $\begingroup$ @GarvilSinghal The problem is actually very trivial and extremely quickly solvable if you use his suggested 'table of differences'. See here about the easily provable Method of Differences. $\endgroup$ – user26486 Feb 9 '15 at 17:48
  • 10
    $\begingroup$ @GarvilSinghal $$\begin{array}{|||||||}n&f(x)& D_1& D_2&D_3&D_4\\1& 15& 30& 58& 60& 24\\2& 45&88&118&84\\3&133& 206&202\\4&339& 408\\5&\boxed{747}\end{array}$$ $\endgroup$ – user26486 Feb 9 '15 at 17:54
  • $\begingroup$ @user314: Can you just confirm with me? The given data allows you to fill in everything strictly above the diagonal. Then you use : dominant coefficient $1$ to put $24$ in the top right corner, and then you go down on the diagonal $24$, $84$, $202$, $\ldots$. I think I understand... $\endgroup$ – Orest Bucicovschi Mar 6 '15 at 16:52
  • $\begingroup$ @orangeskid yes, that is how you do it. The $24$ there is $4!$. $\endgroup$ – user26486 Apr 5 '15 at 15:08
5
$\begingroup$

Hint: Use finite differences as was suggested earlier by @user314 and Robert Israel. They behave similar to derivatives.

With $\Delta [p](x) \colon = p(x+1) - p(x)$, we have for $p(x) = a_n x^n + \cdots $ a polynomial of degree $n$,
$$\Delta^n [p(x)] \equiv n ! \cdot a_n$$ Therefore, for our monic polynomial $p(x)= (x+a)(x+b)(x+c)(x+d)$ of degree $4$ we have

$$\Delta^4 [p] (x)= p(x+4) - \binom{4}{1}p(x+3) + \binom{4}{2} p(x+2) - \binom{4}{1} p(x+1) + p(x) \equiv 4! = 24$$ and so for $x=1$ we get

$$p(5) = 4\, p(4) - 6\, p(3) + 4 \,p(2) - p(1) + 24 = \,...$$

$\endgroup$
  • $\begingroup$ This answer needs more upvotes. $\endgroup$ – abnry Mar 6 '15 at 18:43
2
$\begingroup$

I found the solution numerically using a variant of Newton's method. It is, \begin{align} a &= 0.6625 + 1.1432i \\ b &= -0.6625 + 1.8897i \\ c &= 0.6625 - 1.1432i \\ d &= -0.6625 - 1.8897i, \end{align} which can be verified by pluging into the original equations.

The desired product is, $$(a+5)(b+5)(c+5)(d+5) = 747.00$$ and indeed the zeros after the decimal point continue out to machine precision.

Perhaps these numerical results will be of use in finding an exact (non-numerical) solution through more analytic means.


For completeness, the Matlab script I used to get this answer is included below. It's worth noting that the initial guess must be complex in order to get convergence.

%Solves system of nonlinear equations found here:
%http://math.stackexchange.com/questions/1140178/system-of-4-tedious-nonlinear-equations-akbkckdk-constant-for
%using Newton's method

g = [15; 45; 133; 339];
f_fct = @(v) [prod(v+1); prod(v+2); prod(v+3); prod(v+4)] - g; %want f(v)=0

%Jacobian matrix:
J_fct = @(v) ...
    [prod(v([2,3,4])+1), prod(v([1,3,4])+1), prod(v([1,2,4])+1), prod(v([1,2,3])+1); ...
    prod(v([2,3,4])+2), prod(v([1,3,4])+2), prod(v([1,2,4])+2), prod(v([1,2,3])+2); ...
    prod(v([2,3,4])+3), prod(v([1,3,4])+3), prod(v([1,2,4])+3), prod(v([1,2,3])+3); ...
    prod(v([2,3,4])+4), prod(v([1,3,4])+4), prod(v([1,2,4])+4), prod(v([1,2,3])+4)];

v0 = 1i*[1;2;3;4]; %initial guess must be complex for convergence
v = v0;
aa = 0;
c1 = 1e-4;
for k=1:100
    f = f_fct(v);
    disp(['k= ',num2str(k),', aa= ',num2str(aa,3),', ||f||= ', num2str(norm(f),3)]);
    if (norm(f) < 1e-9)
        break;
    end
    J = J_fct(v);
    p = -J\f; %Newton search direction

    %Find step size 'aa' satisfying first Armijo linesearch condition
    aa = 1;
    if (k < 1)
        aa = 0.01;
    else
        for jj=0:5
            aa = 2^(-jj);
            armijo_lhs = norm(f_fct(v + aa*p));
            armijo_rhs = norm(f_fct(v)) + c1*aa*p'*J_fct(v)'*f_fct(v);
            if (armijo_lhs < armijo_rhs)
                break;
            end
        end
    end
    v = v + aa*p;
end
a = v(1)
b = v(2)
c = v(3)
d = v(4)
answer = prod(v+5)
$\endgroup$
  • $\begingroup$ But that's "the solution" to a different problem than is being posed. $\endgroup$ – AakashM Feb 10 '15 at 11:25
  • 1
    $\begingroup$ Err, just plug the values into the formula $(a+5)(b+5)(c+5)(d+5)$. You get 747. $\endgroup$ – Nick Alger Feb 10 '15 at 11:57

Your Answer

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