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Express the polynomial $x^3-4x-4$ as a linear combination of $x-2$, $(x-2)^2$ and $(x-2)^3$

I've been looking everywhere but I still don't quite understand the question. I know that a linear combination is like a matrix consisting of a specific combination of vectors multiplied by a coefficient. In the form... $$a_1v_1+a_2v_2 +a_3v_3 ,\text{for }a_1 \,to \, a_n \, real \, numbers$$

So to express it as the question asks i think i have to find the coefficients off... $$x^3-4x-4 = a(x-2)+b(x-2)^2 +c(x-2)^3$$

But i'm not too sure about it or where to go. I thought I'd had to involve vectors and matrices somehow. Please help, I really want to understand this content well, I've been having trouble picking up content from this new class.

We've also been talking about basis and span. I think a vector forms a basis for a system. If for a $R^n$ system if for 3 row (have 3 pivots showing 1 = 0, after row reducing) then the vectors/columns corresponding to those rows make a basis for the system and the system hence contains/span all of "R^3".

EDIT: You guys are right the question given to me was inconsistent. I asked the teacher who then admitted there was a typo and re-wrote the question. I think I've got plenty to work on anyway with this already.

He actually meant to ask

Express the polynomial $x^3-2x-4$ as a linear combination of $x-2$, $(x-2)^2$ and $(x-2)^3$

To wich i got C=1, B=6, a=10

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    $\begingroup$ If you expand the right hand side, you should get a polynomial in $x$ with coefficients in terms of $a,b,c$; since this polynomial is equal to $x^3 - 4x - 4$, the coefficients on the right have to equal the coefficients on the left. For instance, the coefficient of $x^3$ on the right is $c$; on the left, the coefficient of $x^3$ is $1$. This implies that $c = 1$. The same kind of logic can be used to find $a$ and $b$. $\endgroup$ – Marcus M Aug 17 '15 at 3:31
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Notice, $$x^3-4x-4=a(x-2)+b(x-2)^2+c(x-2)^3$$ $$x^3-4x-4=cx^3+(b-6c)x^2+(a-4b+12c)x-2(a-2b+4c)$$ Now, comparing the corresponding coefficients, we get $$c=1$$ $$b-6c=0\iff b=6\times 1=6$$ $$a-4b+12c=-4\iff a=-4+4(6)-12(1)=8$$ But, these values have to satisfy $$-2(a-2b+4c)=-4$$$$-2(8-2(6)+4(1))=-4$$ $$ 0\neq-4$$ Hence, the given polynomial $x^3-4x-4$ can't be expressed in the linear form of $(x-2)$

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It cannot be done. For any linear combination of $x-2$, its square, and its cube has $2$ as a root, but our given polynomial does not.

We can do it if in addition to the given polynomials we use $(x-2)^0$, that is, $1$.

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As written, it is impossible. Make the substitution $x=2.$ The right-hand side becomes $0,$ but the left-hand side does not.

Unfortunately, the three terms you're given do not comprise a basis for the space of polynomials of degree no greater than $3.$ A fourth element is needed--any non-zero constant will do.

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  • $\begingroup$ Oh no, does this mean the question is broken? $\endgroup$ – kingportable Aug 17 '15 at 3:53
  • $\begingroup$ Quite possibly, but it's hard for me to say how. $\endgroup$ – Cameron Buie Aug 17 '15 at 3:54
  • $\begingroup$ I'm just guessing now, but do you reckon i'd be able to include a constant? To balance out RHS? $\endgroup$ – kingportable Aug 17 '15 at 3:57
  • $\begingroup$ @kingportable You can include a constant, but just be aware that you are then changing the question, thus: "Express the given polynomial as a linear combination of $1, (x-2), (x-2)^2$, and $(x-2)^3$". $\endgroup$ – Théophile Aug 17 '15 at 4:04
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You also need a constant term, else when $x=2$ we have $LHS=p(2)=-4$ but for any linear combination we have $RHS=0$.

However, this is not then a linear combination of the given powers of $(x-2)$.


If you allow the constant term, then:

So $$p(x)=x^3-4x-4=a_0+a_1(x-2)+a_2(x-2)^2+a_3(x-2)^3 \tag{1}$$

Evaluate at $x=2: 8-8-4=a_0 \implies \boxed{a_0=-4}$

Taking derivative of (1):

$p'(x) = 3x^2-4 = a_1 + 2a_2(x-2) + 3a_3(x-2)^2 \tag{2}$

Evaluate (2) at $x=2$:

$3(4)-4 = a_1 \implies \boxed{a_1 = 8}$

Taking further derivatives will enable you to find $a_2,a_3$.

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You are correct, we are looking for numbers $a,b,c$ so that $x^3-4x-4=a(x-2)+b(x-2)^2+c(x-2)^3$. Let's start by comparing coefficients: On the left-hand side (LHS), the coefficient in front of $x^3$ is $1$. On the right-hand side (RHS), the only place we can get something with $x^3$ in it is from the $(x-2)^3=x^3-6x^2+24x-8$ portion. So, we know that $x^3$ (from the RHS) is equal to $cx^3$ (from the LHS), and therefore $c=1$. Then, we can continue this process to find the rest of the coefficients, by comparing the coefficients in front of $x^2$, $x$, and the parts that are constant on both sides. As Andre and Cameron have pointed out, in this particular instance, there are no solutions. However, this is a good general way to solve these sorts of problems: compare the coefficients on both sides. In this case, if we blindly continue pushing forward with our method, we should eventually arrive at a contradiction, where a coefficient has to equal two distinct numbers.

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  • $\begingroup$ It's a nice thought, but doesn't work. See my answer or Andre's. $\endgroup$ – Cameron Buie Aug 17 '15 at 3:47
  • $\begingroup$ @CameronBuie Yup...I realized it about two minutes after I posted, but I thought it might still be useful for King Portable above to see how to solve this sort of problem in general, if there were a solution. Thanks for the comment! $\endgroup$ – Ben Sheller Aug 17 '15 at 3:49
  • $\begingroup$ You're very welcome! It may be worth pointing out in your post that this is a nice way to find some conditions that a linear combination must satisfy, but in some cases (and this one in particular) won't be sufficient (or, rather, will show instead that no such linear combination exists, when we check the solution's validity). $\endgroup$ – Cameron Buie Aug 17 '15 at 3:53
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An alternate route. For all polynomials of degre 3, by the Taylor expansion, you have

$$ P(x)=P(a)+P'(a)(x-a)+\frac{P^{(2)}(a)}{2!}(x-a)^2+\frac{P^{(2)}(a)}{3!}(x-a)^3 $$

You may apply it to $P(x)=x^3-4x-4$ and $a=2$. Observe that here $P(2)\neq0$.

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Let $f(x)=x^3-4x-4$, then expanding we get $$f(y+2)={y}^{3}+6{y}^{2}+8y-4$$ Let $y=x-2$ to get $$f(x)=(x-2)^{3}+6(x-2)^{2}+8(x-2)-4$$

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To solve the problem, here are possible hints. About understanding the problem statement, Ben S. has explained.

The problem, as stated, is wrong, as pointed out by Cameron Buie below. However, it can be remedied by including the constant polynomial. That is, to write $x^3-4x-4=k\cdot 1+a(x-2)+b(x-2)^2+c(x-2)^3$.

Hint: We have $$k=\left(x^3-4x-4\right)\big|_{x=2}\,,$$ $$a=\left(\frac{x^3-4x-4-k}{x-2}\right)\Big|_{x\to 2}\,,$$ $$b=\left.\left(\frac{\frac{x^3-4x-4-k}{x-2}-a}{x-2}\right)\right|_{x\to 2}\,,$$ and, trivially, $c=1$.

Alternative Hint: $k=\left(x^3-4x-4\right)\big|_{x=2}$, $a=\frac{1}{1!}\left(x^3-4x-4\right)'\big|_{x=2}$ and $b=\frac{1}{2!}\left(x^3-4x-4\right)''\big|_{x=2}$. The prime notation $'$ denotes taking derivative with respect to $x$.

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    $\begingroup$ For something like this, I would probably just compare coefficients without mucking about with any limits. But to each their own, I guess! $\endgroup$ – Ben Sheller Aug 17 '15 at 3:44
  • $\begingroup$ It's a nice thought, but doesn't work. See my answer or Andre's. $\endgroup$ – Cameron Buie Aug 17 '15 at 3:47
  • $\begingroup$ @BenS. You don't need to take limits for my first hint. I merely suggested that the OP divide polynomials are replace $x$ by $2$. However, the second hint is a much better shortcut. $\endgroup$ – Batominovski Aug 17 '15 at 4:03
  • $\begingroup$ @CameronBuie. You are completely right. I had been aware of the missing constant term as I wrote my hints, but I decided not to bother about it (trusting that the problem statement was correct). $\endgroup$ – Batominovski Aug 17 '15 at 4:04

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