Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), 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)\ln (x) for n≥1n \ge 1 is given by the formula: dndxnln⁡(x)=(n−1)!(−1)n−1xn\frac {\mathrm {d} ^ n} { \mathrm {d} x ^ n} \ln (x) = \frac {(n-1)! (-1) ^ {n-1}} {x ^ n}  

Proof: 

1. Base case (n = 1) ddxln⁡(x)=(1−1)!(−1)1−1x1\frac {\mathrm {d}} {\mathrm {d} x} \ln (x) = \frac {(1-1)! (-1) ^ {1-1}} {x ^ 1} 1x=(0)!(−1)0x \frac {1} {x} = \frac {(0)! (-1) ^ {0}} {x} 1x=1x\frac {1} {x} = \frac {1} {x}  

2. Induction hypothesis (n = m) 

Suppose that: dmdxmln⁡(x)=(m−1)!(−1)m−1xm\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: dm+1dxm+1ln⁡(x)=m!(−1)mxm+1\frac {\mathrm {d} ^ {m + 1}} {\mathrm {d} x ^ {m + 1}} \ln (x) = \frac {m! (-1) ^ {m}} {x ^ {m + 1}}

So,

dm+1dxm+1ln⁡(x)=ddx(dmdxmln⁡(x))=\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)=ddx((m−1)!(−1)m−1xm)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{(m-1)!(-1)^{m-1}}{x^m}\right)=ddx((m−1)!(−1)m−1)⋅xm−((m−1)!(−1)m−1)⋅ddxxmx2m=\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}}=−(m−1)!(−1)m−1mxm−1x2m=\frac{-(m-1)!(-1)^{m-1}mx^{m-1}}{x^{2m}}=m(m−1)!(−1)mxm−1x2m=\frac{m(m-1)!(-1)^mx^{m-1}}{x^{2m}}=m!(−1)mx2m−m+1=\frac{m!(-1)^m}{x^{2m-m+1}}=m!(−1)mxm+1\frac{m!(-1)^m}{x^{m+1}}

■\blacksquare

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