x² x3 log(1 + x) = x – - 2 3 4 for -1

Number 5

x²<br>x3<br>log(1 + x) = x – -<br>2<br>3 4<br>for -1 < x < 1, by first calculating by long division<br>7<br>the series<br>1<br>=1 – x + x² – x³+ ..,<br>1+x<br>odi<br>and then integrating termwise between 0 and x.<br>5. Prove that<br>1+x<br>log<br>x7<br>+..<br>= 2 x +<br>%3D<br>5<br>laimonylog<br>for – 1 < x < 1, and hence<br>:).<br>3 1<br>+..<br>535<br>hg?-2+ )<br>1 1<br>1<br>log 2 = 2<br>3.<br>1<br>8.<br>%3D<br>3 33<br>. Supply the details of the following derivation, due to<br>Euler, of the infinite series expansion for log(1+x):<br>(a) Show that log(1 + x) can be given by the limit<br>= lim n[(1+x)/

Extracted text: x² x3 log(1 + x) = x – - 2 3 4 for -1 < x="">< 1,="" by="" first="" calculating="" by="" long="" division="" 7="" the="" series="" 1="1" –="" x="" +="" x²="" –="" x³+="" ..,="" 1+x="" odi="" and="" then="" integrating="" termwise="" between="" 0="" and="" x.="" 5.="" prove="" that="" 1+x="" log="" x7="" +..="2" x="" +="" %3d="" 5="" laimonylog="" for="" –="" 1="">< x="">< 1,="" and="" hence="" :).="" 3="" 1="" +..="" 535="" hg?-2+="" )="" 1="" 1="" 1="" log="" 2="2" 3.="" 1="" 8.="" %3d="" 3="" 33="" .="" supply="" the="" details="" of="" the="" following="" derivation,="" due="" to="" euler,="" of="" the="" infinite="" series="" expansion="" for="" log(1+x):="" (a)="" show="" that="" log(1="" +="" x)="" can="" be="" given="" by="" the="" limit="lim" n[(1+x)/"="" –="" 1].="" log(1="" +="" x)="lim" n[(1="">

Jun 05, 2022

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x² x3 log(1 + x) = x – - 2 3 4 for -1

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