MHB Calculating $f_n(\theta)$ for Positive Integers $n$

  • Thread starter Thread starter sbhatnagar
  • Start date Start date
  • Tags Tags
    Integers Positive
AI Thread Summary
The function \( f_n(\theta) \) is defined as \( f_n(\theta) = \tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta) \). It has been proven through mathematical induction that \( f_n(\theta) = \tan(2^n \theta) \) for positive integers \( n \). Consequently, the values for \( f_2 \left(\frac{\pi}{16}\right) \), \( f_3 \left(\frac{\pi}{32}\right) \), \( f_4 \left(\frac{\pi}{64}\right) \), and \( f_5 \left(\frac{\pi}{128}\right) \) all equal \( \tan\left(\frac{\pi}{4}\right) \), which is 1. This result simplifies the calculations for these specific cases. The discussion highlights the elegant relationship between the function and the tangent of doubled angles.
sbhatnagar
Messages
87
Reaction score
0
For a positive integer $n$, let

$$f_n(\theta)=\tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta)$$

Find the value of

(i) $f_2 \left(\dfrac{\pi}{16} \right)$

(ii) $f_3 \left(\dfrac{\pi}{32} \right)$

(iii) $f_4 \left(\dfrac{\pi}{64} \right)$

(iv) $f_5 \left(\dfrac{\pi}{128} \right)$
 
Mathematics news on Phys.org
sbhatnagar said:
For a positive integer $n$, let

$$f_n(\theta)=\tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta)$$

Find the value of

(i) $f_2 \left(\dfrac{\pi}{16} \right)$

(ii) $f_3 \left(\dfrac{\pi}{32} \right)$

(iii) $f_4 \left(\dfrac{\pi}{64} \right)$

(iv) $f_5 \left(\dfrac{\pi}{128} \right)$

Hi sbhatnagar, :)

It can be shown by mathematical induction that,

\[f_n(\theta)=\tan{2^{n}\theta}\mbox{ where }n\in\mathbb{Z}^{+}\]

Therefore,

\[f_2 \left(\dfrac{\pi}{16} \right)=f_3 \left(\dfrac{\pi}{32} \right)=f_4 \left(\dfrac{\pi}{64} \right)=f_5 \left(\dfrac{\pi}{128} \right)=\tan\left(\frac{\pi}{4}\right)=1\]

Kind Regards,
Sudharaka.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
1
Views
1K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Back
Top