How to equate coefficients properly?

Question

$$C\cdot7^n=5C\cdot7^{n-1}-6C\cdot7^{n-2}+7^n\\ 5C\cdot7^{n-1}-6C\cdot7^{n-2}-C\cdot7^n+7^n=0\\ C(5\cdot7^{n-1}-6\cdot7^{n-2}-7^n)+7^n=0$$

I understand that the correct value of $C$ is $\dfrac{49}{20}.$

However, I am getting $C=0$ by equating the coefficients of $(5\cdot7^{n-1}-6\cdot7^{n-2}-7^n)$ on both sides of the last line.

Please explain why equating coefficients is not working here. Is it invalid or am I doing it incorrectly?

Answer 1

I think you are confused by what it means to equate the coefficients. I personally think this “fact” is something that high school math usually brushes over instead of actually explaining it (at least that was the case for me and my students). Anyway, let me show you an example of when equating the coefficients actually work.

Suppose we want to find two numbers $A$ and $B$ such that $$ Ax^2 + Bx + 2 = 3x^2 + 5x + 2 $$ for all (real) numbers $x$. Then, we may equate the coefficients and say $A = 3$ and $B = 5$. Why? Read the question again, it says that “blah blah blah … for ALL NUMBERS $x$”. That means that we are trying to find $A$ and $B$ so that NO MATTER WHAT VALUE OF $x$ we substitute, $Ax^2 + Bx + 2$ and $3x^2 + 5x + 2$ will ALWAYS be equal. So, for example, we have

$$ A(1)^2 + B(1) + 2 = 3(1)^2 + 5(1)^2 + 2, $$ $$ A(2)^2 + B(2) + 2 = 3(2)^2 + 5(2)^2 + 2, $$ $$ A(3)^2 + B(3) + 2 = 3(3)^2 + 5(3)^2 + 2, $$ $$ A(-123)^2 + B(-123) + 2 = 3(-123)^2 + 5(-123)^2 + 2, $$

etc, one equation (of $A$ and $B$) for each number(!). So, of course, $A$ and $B$ better equal $3$ and $5$ respectively. And that is indeed the case which you can see by solving the two linear equations

$$ A(1)^2 + B(1) + 2 = 3(1)^2 + 5(1)^2 + 2, $$ $$ A(2)^2 + B(2) + 2 = 3(2)^2 + 5(2)^2 + 2 $$

for $A$ and $B$. So, the sole reason why we can equate coefficients in this case, is because the equation is true for ALL values of $x$ (in fact, we only needed it to be true for two values of $x$ as we can solve for $A$ and $B$ once we get two equations). It turns out, we can do the same argument for polynomial identities. For example, if

$$ Ax^5 + Bx^3 + Cx^2 + D = 2x^5 - x^3 + 5x^2 + 1 $$

for ALL numbers $x$, then we can, with full confidence, say that $A = 2$, $B = -1$, $C = 5$ and $D = 1$ (in fact you only need four different values of $x$, check this if you’re interested). There is a deeper reason why polynomials work for us, things could go wrong with other functions. For example, suppose we have

$$ A2^{x+1} + B2^x = 2^{x+1} + 2^x $$

for ALL numbers $x$. Can we confidently say that $A = B = 1$? No, because $A = 0$ and $B = 3$ is also possible (why), and so does $A = -1$ and $B = 5$ (why). I hope this explains your confusion.

Answer 2

Given $$C\cdot7^n=5C\cdot7^{n-1}-6C\cdot7^{n-2}+7^n \tag 1$$ You say:

If I equate the coefficients of $(5\cdot7^{n-1}-6\cdot7^{n-2}-7^n)$ on both sides, I get $C=0$.

I understand you are not looking for how to solve the expression.

I think you are hinting to writing (1) as: $$C\cdot7^n=C(5\cdot7^{n-1}-6\cdot7^{n-2})+7^n \tag 2$$

We need to recognize that this expression has $2$ variables, namely $C$ and $n$ that must "together" be satisfied to claim that we have a valid solution. Not all systems derived from an expression will lead to the correct solution set, even if such a set is not empty. Another point is that arrangement of terms alone don't always lead to a solution.

Let's see some possible "systems" of equations that arise from the expression:

System $1$: $$(0)(0)+C\cdot7^n=C(5\cdot7^{n-1}-6\cdot7^{n-2})+7^n \tag 3$$ In this system $C=0$ And $C=1$ And $5\cdot7^{n-1}=6\cdot7^{n-2}+7^n$ - All at the same time. This is impossible and rejected.

System $2$: $$C(0)+C\cdot7^n=C(5\cdot7^{n-1}-6\cdot7^{n-2})+7^n \tag 4$$ This implies $C=1$ and $5\cdot7^{n-1}=6\cdot7^{n-2}+7^n$ However the later expression is impossible, so this entire system is rejected.

I guess the issue with your approach (system 1) is that you only considered the variable $C$ (partially) and not the entire "system" (consisting of $n$ and $C$).

Answer 3

1. In $$\tan(2x)\equiv\frac{2\tan (x)}{1-\tan^2(x)}\\\tan(2x)\not\equiv\tan(x),$$ the ≡ symbol means that the equation is an identity, that is, its two sides are identically equal (equal whenever both are defined), rather than merely conditionally equal (equal for only some values of its variables).

2. Your final line $$\color{red}C(5\cdot7^{n-1}-6\cdot7^{n-2}-7^n)+7^n\equiv0$$ has the form $$\color{red}Cv_1+1v_2\equiv0,$$ so “equating coefficients” can only be contrived to give $\color{red}C=0$ and $1=0,$ so the operation is not valid. Now, rewrite it as $$\color{red}{\left[C\left(\frac57-\frac6{49}-1\right)+1\right]}7^n\equiv0\tag1$$ and observe that $$\color{red}K\ne0\implies \color{red}K(7^n)\ne0;$$ thus, $\color{red}{\left[C\left(\frac57-\frac6{49}-1\right)+1\right]}$ must equal $0,$ which is what it means to equate coefficients in $(1).$

3. In general, equating coefficients means $$\color{red}{C_1}v_1+\color{red}{C_2}v_2+\color{red}{C_3}v_3\equiv0\iff \color{red}{C_1},\color{red}{C_2},\color{red}{C_3}=0,$$ which means that $v_1,$ $v_2$ and $v_3$ are linearly independent.

4. Consider the following partial-fraction decomposition. \begin{align}&\frac{-8x-8}{x^2+2x}\equiv\frac{A}{x}+\frac{B}{x+2}\\\iff{}&\forall x{\in}\mathbb R{\setminus}\{-2,0\}\quad -8x-8=A(x+2)+Bx\\\iff{}&\forall x{\in}\mathbb R{\setminus}\{-2,0\}\quad (A+B+8)x+(2A+8)=0\\\implies{}&(A+B+8)x+(2A+8) \;\text{ has more than one root}\\\implies{}&(A+B+8)x+(2A+8) \;\text{ cannot be a degree-$1$ polynomial}\\\implies{}&(A+B+8)x+(2A+8) \;\text{ must be the zero polynomial};\end{align} hence, in fact, \begin{align}&\frac{-8x-8}{x^2+2x}\equiv\frac{A}{x}+\frac{B}{x+2}\\\iff{}&(A+B+8)x+(2A+8)\equiv0\tag2\\\iff{}&(A+B+8)x+(2A+8) \;\text{ is the zero polynomial};\end{align} that last line means that all the coefficients are $0,$ that is, we can equate coefficients in $(2).$ And, by similar reasoning, for each $n\in\mathbb N,$ the indeterminates $x^0,x^1,x^2,\ldots,x^n$ are linearly independent, that is, $$\color{red}{C_n}x^n+\color{red}{C_{n-1}}x^{n-1}+\ldots+\color{red}{C_0}\equiv0\iff \color{red}{C_n},\color{red}{C_{n-1}},...,\color{red}{C_0}=0.$$

5. Another example: How can we be sure that we can equate coefficients?

6. Alternatively, we can sidestep equating coefficients: the values of $A,B$ and $C$ in identities $(1)$ and $(2)$ can be determined simply by substituting as many values as necessary into $n$ and $x$ then solving each resulting system of equations.

Answer 4

Let $7^{n-2}=x$ $\implies 7^{n-1}=7x$ and $7^n=49x$

Our equation now becomes $$49Cx=35Cx-6Cx+49x$$ $$\implies 20Cx=49x$$ $$\implies C=\frac{49}{20}$$

Answer 5

Hint: $7^{n-1} = 7^n 7^{-1}$ and $7^{n-2} = 7^n 7^{-2}$ .