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My question relates to the theorem and proof given in Erwin Kreyszig's Advanced engineering mathematics:

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Theorem 1 is as follows:

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My question is as follows: In the proof of Theorem 2, what is the justification for concluding $a_{m+1}$ = $b_{m+1}$, given that $a_{m+1}$ = $b_{m+1}$ only holds when z = 0 and the second equation has not been shown to be equivalent to the first at z = 0?

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1 Answer 1

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Lets review all the steps.

Base case: n=0

By hypothesis:

$$a_{0}+a_1z+a_2z^2+..=b_0+b_1z+b_2z^2+... \quad |z|>0 $$

Then, taking the limit as $z \to 0$.

$$a_0 = b_0$$

Inductive step. Suppose this is valir for $n\leq m$:

$$ a_0 = b_0,\; b_1=b_1, \; a_3 = b_3, ...., a_m = b_m$$

Third step. We have to prove this for $n=m+1$

By hypothesis:

$$a_{0}+a_1z+a_2z^2+..=b_0+b_1z+b_2z^2+... \quad |z|>0 $$

Then

$$(a_0+a_1z+a_2z^2+...+a_mz^m)+a_{m+1}z^{m+1}+.. = (b_0+a_1z+b_2z^2+...+b_mz^m)+b_{m+1}b^{m+1}+..$$

From the inductive step:

$$a_{m+1}z^{m+1}+a_{m+2}z^{m+2}+a_{m+3}z^{m+3}+.. = +b_{m+1}z^{m+1}+b_{m+2}z^{m+2}+b_{m+3}z^{m+3}+..$$

Given that $z\neq 0 $ we can divide both sides by $z^{m+1}$

$$a_{m+1}+a_{m+2}z+a_{m+3}z^{2}+.. = b_{m+1}+b_{m+2}z+b_{m+3}z^{2}+..$$

Taking the limit as $z\to 0$:

$$ a_{m+1} = b_{m+1}$$

Hence this is valid $\forall n\in \mathbb{N}$

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  • $\begingroup$ Thank you. But what is the motivation/justification for taking | z | > 0 in the hypothesis? $\endgroup$
    – DL1
    Apr 20 at 16:27
  • $\begingroup$ In step three you have to divide both sides by $z^{m+1}$, this is not possible if $z=0$. However, you still can calculate the limit as $z \to 0$. Perhaps you may ask: what happen with Taylor series is convergent and defined at $z=0$? In that particular case you always have one term: $a_0$, there is no need for induction. $\endgroup$
    – Bertrand87
    Apr 20 at 16:31
  • $\begingroup$ Given that we are looking at limits as z -> 0, but excluding z = 0 from our considerations, is the is the fact that the sums of the series are continuous at 0 relevant? $\endgroup$
    – DL1
    Apr 20 at 16:44
  • $\begingroup$ Yes it is relevant for taking the limit $\endgroup$
    – Bertrand87
    Apr 20 at 17:00
  • $\begingroup$ Thank you. Could you perhaps explain why? $\endgroup$
    – DL1
    Apr 20 at 17:08

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