Average length of a perpendicular drawn from a random point taken on the arc of a semicircle on its diameter

[Win_odd Dhamnekar]

Show that

(1)if from a random point on the diameter of a semicircle a perpendicular be erected to meet the arc, its average length will be π ÷ r of the radius.

(2)But if from a random point on the arc , a perpendicular be let fall on the diameter , its average length will be 2 ÷ π of the radius.

I computed the answer to (1) but I don't know how did author compute the answer 2 ÷ π to question (2)?

Any math help will be accepted.

 

[Steven G]

Welcome back.
Which course is this from and what topics have you recently covered?

 

[Win_odd Dhamnekar]

Welcome back.
Which course is this from and what topics have you recently covered?

This question and given answer was taken from the book titled "DCC EXERCISES IN CHOICE AND CHANCE" written by W.A. Whitworth.(late professor from Cambridge University, London, England ).

Author has given the following answer. But I have one question about it.





The question: We can write expression in ①,as [imath] \frac{\pi (AQ)^2}{4}\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}= (\pi +2) r^2[/imath] = Area of ABQFG
But while computing the next step i-e [math] \therefore (AQ)^2 = 4r^2 +\frac{8r^2}{\pi}[/math], author omitted [imath]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}[/imath]
why?

 

[blamocur]

Where did you get [imath]\sum_{n=1}^\infty \frac{1}{n}[/imath] from? I believe the sum in question is [imath]\sum_{k=1}^n \frac{1}{n}[/imath], which is equal 1.

IMHO, the author is is rather sloppy with sums and limits.

 

[Deleted member 4993]

This question and given answer was taken from the book titled "DCC EXERCISES IN CHOICE AND CHANCE" written by W.A. Whitworth.(late professor from Cambridge University, London, England ).

Author has given the following answer. But I have one question about it.

View attachment 33833
View attachment 33834


The question: We can write expression in ①,as [imath] \frac{\pi (AQ)^2}{4}\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}= (\pi +2) r^2[/imath] = Area of ABQFG
But while computing the next step i-e [math] \therefore (AQ)^2 = 4r^2 +\frac{8r^2}{\pi}[/math], author omitted [imath]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}[/imath]
why?

You should know that (1/n) creates a "harmonic series" and

[imath]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} \ = \ \infty [/imath]

 

[Win_odd Dhamnekar]

Where did you get [imath]\sum_{n=1}^\infty \frac{1}{n}[/imath] from? I believe the sum in question is [imath]\sum_{k=1}^n \frac{1}{n}[/imath], which is equal 1.

IMHO, the author is is rather sloppy with sums and limits.

Yes. You are correct. But limit should be inserted in your answer. [math] \lim_{n\to \infty}\displaystyle\sum_{k=1}^{n} \frac{1}{n}=1[/math]

 

[blamocur]

Yes. You are correct. But limit should be inserted in your answer. [math] \lim_{n\to \infty}\displaystyle\sum_{k=1}^{n} \frac{1}{n}=1[/math]

That sum is 1 for any [imath]n[/imath] without even using limits.

 

[Win_odd Dhamnekar]

That sum is 1 for any [imath]n[/imath] without even using limits.

No.
Let's say n= 100, then[math] \displaystyle\sum_{k=1}^{100} \frac{1}{n}= 5.18737751764[/math]

 

[Deleted member 4993]

No.
Let's say n= 100, then[math] \displaystyle\sum_{k=1}^{100} \frac{1}{n}= 5.18737751764[/math]

Where is "k" getting involved in the sum?

 

[Win_odd Dhamnekar]

Where is "k" getting involved in the sum?

You may replace k by n. That's a literal error.

 

[Deleted member 4993]

In that case it is a harmonic series and the sum is ∞

You may replace k by n. That's a literal error.

You are repeating that error over-&-over - yet arguing about it.

Please review your post prior to arguing.

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} \ = \ \infty \)

You do not need an explicit "limit" statement here - it is already implied.

 

[blamocur]

You may replace k by n. That's a literal error.

You may not: the author of the answer states, correctly, that each angle is [imath]\frac{2\pi}{n}[/imath], i.e., no [imath]k[/imath] is involved.

 

[Win_odd Dhamnekar]

You may not: the author of the answer states, correctly, that each angle is [imath]\frac{2\pi}{n}[/imath], i.e., no [imath]k[/imath] is involved.

each angle is [imath] \frac{\pi}{2n}[/imath]

Secondly, I agree with what you wrote [imath] \displaystyle\sum_{k=1}^{n}\frac{1}{n}=1[/imath]