176 Fourier's Theorem and Roots of Equations. 



Example (continued). 

 (Negative roots.) 

 Put x=-—y YQ.f(x) and its derived functions and we have 



</>(/ =/ + 6/+/-50/-62/ + 104?/ + 170==r0 ? 

 ${xj) = 6/ + 30/ + 4/ - 150/ - 124?/ + 104, 

 $<( y ) = 30/ + 120/ + 12/ - SOOy - 124, 

 4>'"(y) = 120/ + 360/ + 24y - 300, 

 <£ IV (/) = 360/+720y + 24, 

 0v( y ) =7203/ + 720, 

 <pQ,) =720 



y- 



<p. 



f. 



0". 



<p"'. 



<p 1 ?. 



0V. 



0VI. 



3 



+ 



+ 



+ 



+ 



+ 



+ 



+ 



2 



+ 



- 



+ 



+ 



+ 



+ 



+ 



1 



+ 



- 





+ 



+ 



+ 



4- 







+ 



+ 



- 



- 



+ 



+ 



+ 



Two changes o£ sign are lost between 2 and 3. 



Let v = 2+-. 



1 + 6 + 1-50-62 + 104 + 170 

 2 + 16 + 34 - 32 - 188 - 168 



+ 8 + 17 - 16 - 94 - 

 2 + 20 + 74 +116 + 



84 + 

 44 



2 



+10 + 37 + 58+ 22 - 

 2 + 24 +122 + 360 



40 





12 + 61 +180+382 

 2 + 28 +178 





14 + 89 +358 



2 + 32 





16 +121 



2 



18 





