Solved ex2 from dis hw3 and removed ex3

1 files changed, 10 insertions, 8 deletions

diff --git a/Year 1/Calculus/Homeworks/Homework3.tex b/Year 1/Calculus/Homeworks/Homework3.tex
index b8db2d3..bf93382 100644
--- a/Year 1/Calculus/Homeworks/Homework3.tex
+++ b/Year 1/Calculus/Homeworks/Homework3.tex
@@ -72,16 +72,18 @@
 	f(x) = (x^2-1) \cos 2x + \ln (3 - 6x^2)
 \end{align*}
 
-\paragraph{Решение:} [TODO]
-
-\paragraph{\colorbox{yellow}{Задача 3.}}
-
-Да се докаже, че за всяко $x < 0$ е изпълнено, че
+\paragraph{Решение:}
 \begin{align*}
-	\frac{1}{2(1 + \sqrt{2})}\arctg \frac{x^2 - 1}{2x} + \ln \sqrt{x^2 + 1} - \frac{(\sqrt{2} - 1)\pi}{8} \geq \frac{1}{2} \ln \left( 4 - 2 \sqrt{2} \right)
+	& \ln (3 - 6x^2) = \ln \left( 3 (1 - 2x^2) \right) = \ln 3 + \ln (1 - 2x^2) = \ln 3 + \sum_{k = 1}^n \frac{(-1)^{k-1}}{k}(-2x^2)^k + o(x^{2n}) = \\
+	& \qquad = \ln 3 - \sum_{k = 1}^n \frac{2^kx^{2k}}{k} + o(x^{2n}) = \ln 3 - \sum_{k = 2}^n \frac{2^kx^{2k}}{k} - 2x^2 + o(x^{2n}) \\
+	& (x^2 - 1)\cos 2x = (x^2 - 1) \left( \sum_{k = 0}^n \frac{(-1)^k (2x)^{2k}}{(2k)!} + o(x^{2n}) \right) = \sum_{k = 0}^n \frac{(-4)^k x^{2k+2}}{(2k)!} - \sum_{k = 0}^n \frac{(-4)^k x^{2k}}{(2k)!} + o(x^{2n}) = \\
+	& \qquad = \sum_{k = 2}^n \frac{(-4)^{k-2} x^{2(k-2)+2}}{(2(k-2))!} + \frac{(-4)^{2n+1} x^{2n+1}}{(2(n-1))!} + \frac{(-4)^{2n+2} x^{2n+2}}{(2n)!} - \sum_{k = 2}^n \frac{(-4)^k x^{2k}}{(2k)!} - 1 + 2x^2 + o(x^{2n}) = \\
+	& \qquad = -1 + 2x^2 + \sum_{k = 2}^n \left( \frac{(-4)^{k-2} x^{2k-2}}{(2k-4)!} - \frac{(-4)^k x^{2k}}{(2k)!} \right) + o(x^{2n})
+\end{align*}
+\begin{align*}
+	f(x) & = \ln 3 - \sum_{k = 2}^n \frac{2^kx^{2k}}{k} - \cancel{2x^2} - 1 + \cancel{2x^2} + \sum_{k = 2}^n \left( \frac{(-4)^{k-2} x^{2k-2}}{(2k-4)!} - \frac{(-4)^k x^{2k}}{(2k)!} \right) + o(x^{2n}) = \\
+	& = \answerbox{\ln 3 - 1 + \sum_{k = 2}^n \left( \frac{(-4)^{k-2} x^{2k-2}}{(2k-4)!} - \frac{(-4)^k x^{2k}}{(2k)!} - \frac{2^kx^{2k}}{k} \right) + o(x^{2n})}
 \end{align*}
-
-\paragraph{Решение:} [TODO]
 
 \paragraph{\colorbox{yellow}{Задача 4.}}