Numerical Solution to $ 7000 = \sum_{n=1}^{5} \frac{ \ 1000}{(1 + x)^n} + \sum_{n=6}^{10} \frac{\ 2000}{(1 + x)^n}. $

Question

$$ 7000 = \sum_{n=1}^{5} \frac{ \ 1000}{(1 + x)^n} + \sum_{n=6}^{10} \frac{\ 2000}{(1 + x)^n}. $$

I need to solve the equation above. Since it generates a $10th$ degree polynomial, the equation must be solved by some numerical method. However, I'm unfamiliar with numerical methods, so I need a little hand.

Can you see why there is at least one solution between say $x=0$ and $x=9$? — Carl Christian, Oct 14 '20 at 19:46
A equation $f(x)=0$ where $f$ is continuous on an interval $[a,b]$ has at least one root in $(a,b)$ if $f(a)f(b)<0$. — Carl Christian, Oct 14 '20 at 19:52
The answers below do not actually explain any numerical methods. You can see here for example. To guarantee convergence, bracketing methods can be used. A number of root-finding methods are explained here. — Simply Beautiful Art, Oct 16 '20 at 13:14

Robert Israel · Accepted Answer · 2020-10-16T15:49:48.250

2

If $y = 1/(1+x)$ the equation can be written as $$ 2\,{y}^{10}+2\,{y}^{9}+2\,{y}^{8}+2\,{y}^{7}+2\,{y}^{6}+{y}^{5}+{y}^{4 }+{y}^{3}+{y}^{2}+y-7 = 0 $$ (EDIT): There is one positive real root, approximately $y=0.8772131213$. The corresponding $x$ value is $1/y - 1 \approx 0.1399738282$. There is one negative real root, approximately $y=-1.190960306$. The corresponding $x$ value is $-1.839658547$.

edited Oct 16 '20 at 15:49

answered Oct 14 '20 at 19:49

Robert Israel

448,999

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It may be easier to find a value if one multiplies through by $y-1$ to get $2y^11 - y^6 - 8y + 7 = 0$. Or it may not. On the one hand there are less terms; on the other hand $y = 1$ is a root of this polynomial but isn't a root of the polynomial we actually care about. – Michael Lugo Oct 14 '20 at 19:53
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Clearly there are two real roots? – Simply Beautiful Art Oct 16 '20 at 04:28
Sorry: one positive real root. – Robert Israel Oct 16 '20 at 15:45

score 0 · Answer 2 · answered Oct 15 '20 at 03:14

Considering what @Robert Israel did, $$f(y)=2\,{y}^{10}+2\,{y}^{9}+2\,{y}^{8}+2\,{y}^{7}+2\,{y}^{6}+{y}^{5}+{y}^{4 }+{y}^{3}+{y}^{2}+y-7 = 0$$ it seems to me that there must be two real roots.

By inspection, $f(-2)=1379$, $f(-1)=-6$, $f(0)=-7$, $f(1)=8$. Using Newton method $$y_1=-1.19096\qquad y_2=0.877213 \implies x_1=0.139974\qquad x_2=-1.83966$$

Looking at the original problem in $x$, there is a vertical asymptote at $x=-1$ and then one solution on the left and one on the right of it.

Numerical Solution to $ 7000 = \sum_{n=1}^{5} \frac{ \ 1000}{(1 + x)^n} + \sum_{n=6}^{10} \frac{\ 2000}{(1 + x)^n}. $

2 Answers2