Proof Concerning Central Line \(X_5X_6\) of Triangle

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Hello everyone. Today we will prove theorem about central line \(X_5X_6\) of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let \(ABCD\) be a golden rectangle where \( \frac{AB}{BC} = \phi \), and construct the square \( BCQP \) inside it. Reflect \( P \) over \( D \) to obtain \( E \). Then, the line \( EB \) coincides with the central line \( X_5X_6 \) of triangle \( ABP \). Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: \( A = (0,0) \), \( B = (\phi x, 0) \), \( C = (\phi x, x) \), \( D = (0, x) \). The square \( BCPQ \) ensures \( P = (\phi x, 2x) \). The reflection of \( P \) across \( D \) is \( E = (-\phi x, 2x) \). 2. Compute the Nine-Point Center \( X_5 \) The nine-point center \( X_5 \) is the circumcenter of the medial triangle, which consists of the midpoints: \[ M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \] \[ M_2 = \left(\f...
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Proof of the formula for the nth derivative of the natural logarithm

Hello everyone. Today we will prove the formula for the nth derivative of the natural logarithm. We will use the method of mathematical induction as a proof method. 

Theorem: The nth derivative of the function \(\ln (x) \) for \(n \ge 1 \) is given by the formula: \[\frac {\mathrm {d} ^ n} { \mathrm {d} x ^ n} \ln (x) = \frac {(n-1)! (-1) ^ {n-1}} {x ^ n} \] 

Proof: 

1. Base case (n = 1) \[\frac {\mathrm {d}} {\mathrm {d} x} \ln (x) = \frac {(1-1)! (-1) ^ {1-1}} {x ^ 1} \] \[ \frac {1} {x} = \frac {(0)! (-1) ^ {0}} {x} \] \[\frac {1} {x} = \frac {1} {x} \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\frac{\mathrm{d}^m}{\mathrm{d}x^m}\ln(x)=\frac{(m-1)!(-1)^{m-1}}{x^m}\]

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\frac {\mathrm {d} ^ {m + 1}} {\mathrm {d} x ^ {m + 1}} \ln (x) = \frac {m! (-1) ^ {m}} {x ^ {m + 1}} \]

So,

\[\frac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}\ln(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}^m}{\mathrm{d}x^m}\ln(x)\right)=\]\[\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{(m-1)!(-1)^{m-1}}{x^m}\right)=\]\[\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left((m-1)!(-1)^{m-1}\right)\cdot x^m-\left((m-1)!(-1)^{m-1}\right)\cdot \frac{\mathrm{d}}{\mathrm{d}x}x^m}{x^{2m}}=\]\[\frac{-(m-1)!(-1)^{m-1}mx^{m-1}}{x^{2m}}=\]\[\frac{m(m-1)!(-1)^mx^{m-1}}{x^{2m}}=\]\[\frac{m!(-1)^m}{x^{2m-m+1}}=\]\[\frac{m!(-1)^m}{x^{m+1}}\]

\(\blacksquare\)

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