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I want to understand why: From K.C Chang's book "Infinite Dimensional Morse Theory and Multiple Solution Problems":

if i have

enter image description here

then $(4.1)$ is formal : it means that enter image description here

EDIT1: $(4.1)$ tel us that $\displaystyle\sum_{q=0}^{\infty} (M_q-\beta_q)t^q=(1+t)Q(t)$ Whene $t=-1$ we have directly that $\displaystyle\sum_{q=0}^{\infty}(-1)^q M_q=\displaystyle\sum_{q=0}^{\infty} (-1)^q \beta_q$

and wehave also that $\displaystyle\sum_{q=0}^{\infty}(-1)^q M_q\geq\displaystyle\sum_{q=0}^{\infty} (-1)^q \beta_q$ because $Q(t)$ has nonnegative coefficient

but how to obtain that $\displaystyle\sum_{j=0}^{q}(-1)^{q-j} M_j\geq\displaystyle\sum_{j=0}^{q} (-1)^{q-j} \beta_j$ ???

I ask this question on Mathematics stackex change but they don't answer me and they told me to ask it here

i need your help please help me

thank you

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2 Answers 2

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Let us set

$$ M(t)=\sum_{q=0}^\infty M_tt^q,\;\;P(t)=\sum_{q=0}^\infty \beta_q t^q. $$

We can rewrite (4.1) as

$$ M(t)=P(t)+(1+t)Q(t), $$

where the formal power series $Q(t)$ has nonnegative coefficients. We deduce

$$ (1+t)^{-1} M(t)=(1+t)^{-1}P(t) +Q(t). \tag{1}$$

This shows that the Taylor coefficients of $(1+t)^{-1} M(t)$ are $\geq $ than the corresponding Taylor coefficients of $(1+t)^{-1}P(t)$. To compute these Taylor coefficients use the know expansion

$$(1+t)^{-1}=\sum_{q=0}^\infty(-1)^q t^q. $$

Using this in (1) you obtain all the Morse inequalities.

The coefficient of $t^q$ in $(1+t)^{-1}M(t)$ is

$$ \sum_{\substack{k+j=q\\k,j\geq 0}}(-1)^k M_j\stackrel{k=q-j}{=}\sum_{j=0}^q (-1)^{q-j} M_j $$.

This coefficient is $\geq $ than the coefficient of $t^q$ in $(1+t)^{-1}P(t)$ which is

$$ \sum_{j=0}^q (-1)^{q-j} \beta_j. $$

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  • $\begingroup$ i must find that $\displaystyle\sum_{j=0}^{q}(-1)^{q-j} M_j\geq\displaystyle\sum_{j=0}^{q} (-1)^{q-j} \beta_j$ $\endgroup$
    – Vrouvrou
    Commented Apr 2, 2014 at 15:35
  • $\begingroup$ if i understand i must replace $(1+t)^{-1}$ by $\sum_{q=0}^{\infty}(-1)^q t^q.$ in (1) so i have $(1+t)^{-1}M_t=\sum_{q=0}^\infty(-1)^q t^q \sum_{q=0}^{\infty}M_q t^q=\sum_{q=0}^{\infty}\sum_{j=0}^q (-1)^j (-1)^{q-j} M_q t^q =\sum_{q=0}^{\infty}\sum_{j=0}^q (-1)^j M_q t^q$ i dont oblaine the result ! $\endgroup$
    – Vrouvrou
    Commented Apr 2, 2014 at 16:38
  • $\begingroup$ There is an error in your above computation. In the double sum you must have $M_j$. I've updated the answer so you can see better. $\endgroup$ Commented Apr 2, 2014 at 19:12
  • $\begingroup$ $(1+t)^{-1} M(t)= (\sum_{q=0}^{\infty}(-1)^q t^q)(\sum_{q=0}^{\infty} M_q t^q)=\sum_{q=0}^{\infty} \sum_{j=0}^q (-1)^j M_{q-j}t^q$ it's right ? what happen to $\sum_{q=0}^{\infty}$ after ? thank you $\endgroup$
    – Vrouvrou
    Commented Apr 7, 2014 at 18:08
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i have search google and i find a link where you can download this book :

http://www.doc88.com/p-607135346844.html

i read your problem in this book roughly (theorem 4.3 page 37 )

i also find the solution to your question in this book , it is included in the proof of theorem 4.3 itself actually !

you should read its proof carefully, your question is a consequence of Corollary 4.1 and Theorem 4.2 , the answer is in its proof indeed !

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