Infinite vector

Algebra Level 3

Suppose I define two vectors as follows:

[ 1 1 2 1 3 1 4 . . . 1 m ] and [ 1 2 1 2 2 1 2 3 1 2 4 . . . 1 2 m ] . \begin{bmatrix} 1\\ \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{4} \\ . \\ . \\ . \\\frac{1}{m}\end{bmatrix} \text{and} \begin{bmatrix} \frac{1}{2}\\ \frac{1}{2^{2}} \\ \frac{1}{2^{3}} \\ \frac{1}{2^{4}} \\ . \\ . \\ . \\ \frac{1}{2^{m}}\end{bmatrix}.

If m → ∞ m \to \infty , what is the limiting angle between these two vectors in radians?


The answer is 0.359.

Amal Hari by Amal Hari , 22, Canada

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1 solution

Amal Hari
Nov 17, 2019

We can use dot product and our knowledge on infinite series to find the answer for this.

A ⋅ B = ∣ A ∣ ∣ B ∣ cos ⁡ θ \textbf{A }\cdot \textbf{B} =|\textbf{A}||\textbf{B}|\cos\theta

If we perform the dot product on these pair we will get ∑ n = 1 m 1 n 2 n \displaystyle\sum_{n=1}^{m} \frac{1}{n2^{n}} this sum is the case of x being equal to 1 for the infinite series of ln ⁡ ( 1 + x ) \ln\left(1+x\right) and will converge to ln ⁡ ( 2 ) \ln\left(2\right) when m approach ∞ \infty .

Norm of A = ∑ n = 1 m 1 n 2 \textbf{A}=\sqrt{\displaystyle\sum_{n=1}^{m} \frac{1}{n^{2}}} , we know lim ⁡ m → ∞ ∑ n = 1 m 1 n 2 = π 2 6 \displaystyle \lim_{m\to \infty}\sum_{n=1}^{m} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}

Therefore norm of A ∣ A ∣ = π 2 6 = π 6 |\textbf{A}| =\sqrt{\frac{\pi^{2}}{6}}=\frac{\pi}{\sqrt{6}}

For ∣ B ∣ = ∑ n = 1 ∞ ( 1 2 n ) 2 = ∑ n = 1 ∞ 1 ( 2 2 ) n |\textbf{B}| =\sqrt{\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2^{n}}\right)^{2}}=\sqrt{\displaystyle \sum_{n=1}^{\infty} \frac{1}{\left(2^{2}\right)^{n}}}

∑ n = 1 ∞ 1 ( 2 2 ) n = 1 3 \displaystyle \sum_{n=1}^{\infty} \frac{1}{\left(2^{2}\right)^{n}} =\frac{1}{3}

Therefore norm of B ∣ B ∣ = 1 3 = 1 3 |\textbf{B}| =\sqrt{\frac{1}{3}}=\displaystyle\frac{1}{\sqrt{3}}

Rewriting the dot product again

ln ⁡ ( 2 ) = π 6 × 1 3 cos ⁡ θ \ln\left( 2\right) =\frac{\pi}{\sqrt{6}} \times \frac{1}{\sqrt{3}} \cos \theta

cos ⁡ θ = ( ln ⁡ 2 ) 18 π \displaystyle \cos \theta =\frac{\left(\ln 2 \right) \sqrt{18}}{\pi}

θ = arccos ⁡ ( ln ⁡ 2 ) 18 π ≈ 0.359 r a d \theta =\arccos{\frac{\left(\ln 2 \right) \sqrt{18}}{\pi} } \approx 0.359 rad

0 Helpful 0 Interesting 3 Brilliant 1 Confused

For x = 1 x=1 , ln ⁡ ( 1 + x ) = 1 − 1 2 + 1 3 − 1 4 + . . . \ln {(1+x)}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+... . How does it equal the sum given above?

A Former Brilliant Member - 1 year, 6 months ago

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try integrating 1 1 + x \frac{1}{1+x} using integration by parts , we can derive an infinite series which is defined for wider range of x , in the form of ∑ n = 1 ∞ x n n ( 1 + x ) n \displaystyle \sum_{n=1} ^{\infty}\frac{x^{n}}{n\left(1+x\right)^{n}}

Amal Hari - 1 year, 6 months ago

Thank you for posting this problem. It is interesting. To evaluate the following sum:

∑ n = 1 ∞ 1 n 2 n = ∑ n = 0 ∞ 1 ( n + 1 ) 2 n + 1 = S \sum_{n=1}^{\infty} \frac{1}{n2^{n}} = \sum_{n=0}^{\infty} \frac{1}{(n+1)2^{n+1}}= S

One can start with the following series:

∑ n = 0 ∞ p n = 1 1 − p \sum_{n=0}^{\infty} p^n = \frac{1}{1-p}

Where ∣ p ∣ < 1 \mid p \mid <1 . By carrying out the following operation:

∫ 0 x ∑ n = 0 ∞ p n d p = ∫ 0 x d p 1 − p \int_{0}^{x} \sum_{n=0}^{\infty} p^n dp =\int_{0}^{x} \frac{dp}{1-p}

∑ n = 0 ∞ 1 ( n + 1 ) x n + 1 = ∫ 0 x d p 1 − p \sum_{n=0}^{\infty} \frac{1}{(n+1)}x^{n+1} = \int_{0}^{x} \frac{dp}{1-p}

Replacing x x by 0.5 0.5 one gets:

∑ n = 0 ∞ 1 ( n + 1 ) 2 n + 1 = ∑ n = 1 ∞ 1 n 2 n = ∫ 0 1 2 d p 1 − p = ln ⁡ 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)2^{n+1}} = \sum_{n=1}^{\infty} \frac{1}{n2^{n}}= \int_{0}^{\frac{1}{2}} \frac{dp}{1-p} = \ln{2}

Karan Chatrath - 1 year, 6 months ago

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This is an elegant approach. So we get a beautiful identity between the two series.

A Former Brilliant Member - 1 year, 6 months ago

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