How can this be an integer??

One of my friends told me that 1+21+31+41+51+...\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}} comes out to be an integer. Can anyone help me out with guessing that single digit integer?

Note by Pranjal Jain
4 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

The method to solve this kind of nested radical is to define a function f(x)=1+x1+(x+1)1+(x+2)1+f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}} Clearly f(x)21=xf(x+1)f(x)^2-1=xf(x+1) Now note that f(0)=1f(0)=1 We will use induction to show that f(x)=x+1f(x)=x+1 for all positive integers xx (which will in turn give us the value of f(2)f(2)) First note the base case f(0)=1=0+1f(0)=1=0+1 Now suppose f(k)=k+1f(k)=k+1 for some positive integer kk. Then f(k)21=kf(k+1)    (k+1)21=kf(k+1)    k2+2k=kf(k+1)    f(k+1)=k+2f(k)^2-1=kf(k+1)\implies (k+1)^2-1=kf(k+1)\implies k^2+2k=kf(k+1)\implies f(k+1)=k+2 and the induction is done.

We conclude that f(x)=x+1f(x)=x+1 for all positive integers xx.

Now we compute the value of f(2)f(2) simply as 2+1=32+1=3. :)

Nathan Ramesh - 4 years, 10 months ago

Log in to reply

Thanks dude! Wikipedia had a more general one which was more difficult to be understood! Keep it up!

Pranjal Jain - 4 years, 10 months ago

Log in to reply

Thanks! :)

There is still a slight issue with the induction though. (I think). See if you can find it.

Nathan Ramesh - 4 years, 10 months ago

Log in to reply

@Nathan Ramesh Yeah sure! I just checked till f(x)21=xf(x+1)f(x)^{2}-1=xf(x+1)

Pranjal Jain - 4 years, 10 months ago

Log in to reply

@Pranjal Jain Here is the issue (which I'm not sure if it's really an issue, maybe @Calvin Lin can clear it up)

When going from f(0)f(0) to f(1)f(1) using the recurrence relation and solving for f(1)f(1), what do we get?

Nathan Ramesh - 4 years, 10 months ago

Log in to reply

@Nathan Ramesh Oh yeah! 00\frac{0}{0} indeterminate form!!!

Pranjal Jain - 4 years, 10 months ago

Log in to reply

Not quite. In essence, you have not proven anything as yet. You need the fact that f(1)=2 f(1) = 2 to conclude that f(k)=k+1 f(k) = k+1 .

The induction step is only valid if you can divide by kk (right at the end). Hence, this need not hold true for k=0k = 0 . As such, your base case needs to be k=1k=1 .

Calvin Lin Staff - 4 years, 10 months ago

Log in to reply

@Calvin Lin does this prove the base case?

2=1+132=\sqrt{1+1\cdot 3} 1+13=1+11+24\sqrt{1+1\cdot 3}=\sqrt{1+1\sqrt{1+2\cdot 4}} 1+11+24=1+11+21+35\sqrt{1+1\sqrt{1+2\cdot 4}}=\sqrt{1+1\sqrt{1+2\sqrt{1+3\cdot 5}}}

We can continue this indefinitely

Nathan Ramesh - 4 years, 10 months ago

Log in to reply

@Nathan Ramesh This way you can directly prove mine one to be 3!!!!

3=1+2×4=1+21+3×53=\sqrt{1+2×4}=\sqrt{1+2\sqrt{1+3×5}} and so on as you did in your solution!!

Pranjal Jain - 4 years, 10 months ago

Log in to reply

@Pranjal Jain Yes, of course.

Nathan Ramesh - 4 years, 10 months ago

Log in to reply

Can you help me in proving this? I guess @Nathan Ramesh 's method is not bad and "something" can be done with induction.

Pranjal Jain - 4 years, 10 months ago

Log in to reply

It is 3! And it is reported in http://en.wikipedia.org/wiki/Nested_radical

Chew-Seong Cheong - 4 years, 10 months ago

Log in to reply

Is there any other possible approach?

Pranjal Jain - 4 years, 10 months ago

Log in to reply

Not one that I know.

Chew-Seong Cheong - 4 years, 10 months ago

Log in to reply

Its value is 3

This is very famous nested radical given by ramanujan.

Krishna Sharma - 4 years, 10 months ago

Log in to reply

yes i also approached same question on this site, the key is ramanujan

Mardokay Mosazghi - 4 years, 10 months ago

Log in to reply

From: f(x)21=xf(x+1)f(x)^2 -1 = xf(x+1) (f(x)+1)(f(x)1)=xf(x+1) ( f(x)+1 ) ( f(x)-1 ) = xf(x+1)

Suppose f(x)=x+1f(x) = x+1, then: (x+1+1)(x+11)=x(x+2)(x+1+1)(x+1-1)=x(x+2) x(x+2)x(x+2)x(x+2) \equiv x(x+2)

It appears that f(x)=x+1f(x)=x+1 satisfies the first equation for all real values of xx. The special case of x=0x=0 is also shown to be f(0)=0+1f(0) = 0 +1.

I hope this is adequate.

Chew-Seong Cheong - 4 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...