What is the 100th digit after the decimal point in the following expression:

If any one can help me regain my sanity I'll really appreciate it.

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I've been working on this for 2 days, i don't feel like I'm getting any closer.

What is the 100th digit after the decimal point in the following expression:

If any one can help me regain my sanity I'll really appreciate it.

What is the 100th digit after the decimal point in the following expression:

( \sqrt{5} + \sqrt {7} )^{2011}

If any one can help me regain my sanity I'll really appreciate it.

joekz wrote:I've been working on this for 2 days, i don't feel like I'm getting any closer.

What is the 100th digit after the decimal point in the following expression:( \sqrt{5} + \sqrt {7} )^{2011}

If any one can help me regain my sanity I'll really appreciate it.

Was this on the Putnam exam (or some other math contest)? It seems like it since they tend to incorporate the current year into their tests.

Here are my thoughts:

I have no idea if I'm on the right track, this is just stream of consciousness.

Goals:

1. Disprove something widely accepted to be true (In progress... I'm getting there.)

2. Have an action figure made of myself (With karate chop action!)

3. Win a dancing contest (COMPLETE)

1. Disprove something widely accepted to be true (In progress... I'm getting there.)

2. Have an action figure made of myself (With karate chop action!)

3. Win a dancing contest (COMPLETE)

Hint:

**Spoiler:**

- Indigo is a lie.

Which idiot decided that websites can't go within 4cm of the edge of the screen?

There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.

**Spoiler:**

Macbi wrote:Hint:Spoiler:

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.Macbi wrote:Hint:Spoiler:Spoiler:

Ah, for some reason I thought we were in 2010. I now have no idea.

(Wolfram Alpha says the answer is 4.)

- Indigo is a lie.

Which idiot decided that websites can't go within 4cm of the edge of the screen?

There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

I got 6 from W|A like the above maple poster, 4 is the 99th digit?

http://www.wolframalpha.com/input/?i=fr ... +sqrt+7%29^2011%29*10^99%29

A few things I would consider, (bearing in mind I relatively suck at this type of problem), would be to

a) look at the Taylor series of things like (sqrt (4+x) + sqrt (9-2x))^2011

b) 2011 = 2048 - 37, look at the binary decomposition that way, in a similar way to what Xias is saying

c) Can we find the Nth digit of sqrt(5) + sqrt(7) on its own? How about its square? If we can do those, can we generalize?

d) Write it as

No clue if these lead anywhere.

http://www.wolframalpha.com/input/?i=fr ... +sqrt+7%29^2011%29*10^99%29

A few things I would consider, (bearing in mind I relatively suck at this type of problem), would be to

a) look at the Taylor series of things like (sqrt (4+x) + sqrt (9-2x))^2011

b) 2011 = 2048 - 37, look at the binary decomposition that way, in a similar way to what Xias is saying

c) Can we find the Nth digit of sqrt(5) + sqrt(7) on its own? How about its square? If we can do those, can we generalize?

d) Write it as

\frac{2^{2011}}{(\sqrt 7 - \sqrt 5)^{2011}}

and use b above to get the annoying double radical term in the numerator only.No clue if these lead anywhere.

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

- Yakk
- Poster with most posts but no title.
**Posts:**10879**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

In balanced trinary 2011 is:

2457 - 819 + 273 +81 + 27 -9 + 1

Or +-+++-0+

In trinary 2011 is:

2110201

sqrt(5)+sqrt(7) cubed is (12 + 2 sqrt(35)) * (sqrt(5) + sqrt(7)) = 12 sqrt(5) + 12 sqrt(7) + 10 sqrt(7) + 14 sqrt(5) = 26 sqrt(5) + 22 sqrt (7)

a sqrt(5) + b sqrt(7) cubed is:

= (5 a^3 + 21 ab^2) sqrt(5) + (15 b a^2 + 7 b^3) sqrt(7)

hmm, that doesn't look all that useful.

In balanced trinary 2011 is:

2457 - 819 + 273 +81 + 27 -9 + 1

Or +-+++-0+

In trinary 2011 is:

2110201

sqrt(5)+sqrt(7) cubed is (12 + 2 sqrt(35)) * (sqrt(5) + sqrt(7)) = 12 sqrt(5) + 12 sqrt(7) + 10 sqrt(7) + 14 sqrt(5) = 26 sqrt(5) + 22 sqrt (7)

a sqrt(5) + b sqrt(7) cubed is:

= (5 a^3 + 21 ab^2) sqrt(5) + (15 b a^2 + 7 b^3) sqrt(7)

hmm, that doesn't look all that useful.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.

Macbi wrote:Hint:Spoiler:

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?

Goals:

1. Disprove something widely accepted to be true (In progress... I'm getting there.)

2. Have an action figure made of myself (With karate chop action!)

3. Win a dancing contest (COMPLETE)

1. Disprove something widely accepted to be true (In progress... I'm getting there.)

2. Have an action figure made of myself (With karate chop action!)

3. Win a dancing contest (COMPLETE)

Xias wrote:Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.Macbi wrote:Hint:Spoiler:

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?

- Yakk
- Poster with most posts but no title.
**Posts:**10879**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

2011 isn't even...

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Wait...

Wouldn't it just be:

(Radical 5+Radical 7) ^2

Then under distributive property:

5+7

Then to derive the answer:

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

Wouldn't it just be:

(Radical 5+Radical 7) ^2

Then under distributive property:

5+7

Then to derive the answer:

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

- t1mm01994
**Posts:**298**Joined:**Mon Feb 15, 2010 7:16 pm UTC**Location:**San Francisco.. Wait up, I'll tell you some tales!

sundalius wrote:Wait...

Wouldn't it just be:

(Radical 5+Radical 7) ^2

Then under distributive property:

5+7

Then to derive the answer:

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

http://en.wikipedia.org/wiki/Freshmans_dream

Quite relevant. Basically, following your train of thought

(2+3)^2 = 2^2 + 3^2

5^2 = 4+9

25 = 13 and that doesnt quite sound right to me.. A search for "expanding brackets" might help you as well.

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5942**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

@sundalius: the distributive property means that a(b+c) = ab+ac. From this you can prove that (a+b)(c+d) = ac + ad + bc + bd. You should be able to work out where you're going wrong from that.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Xias wrote:Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.Macbi wrote:Hint:Spoiler:

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?

Didn't think about this enough when i commented earlier...

So (\sqrt 5 + \sqrt 7)^{2010} is close enough to an integer that, for the purpose of finding the 100th digit, we can just assume that it is an integer (the difference is smaller than 10^(-500)). So we essentially have a huge 1000 digit integer times \sqrt 5 + \sqrt 7. (And we know what the integer is, it's 2*(5^1005+7^1005)+2*(2010 choose 2)*35*(5^1004 + 7^1004) + ...., so maybe something can be said for digits of (ab)^n(a^m+b^m)(\sqrt a + \sqrt b)?

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

- Yakk
- Poster with most posts but no title.
**Posts:**10879**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

Well, only the first 100 digits of that number matter.

So what if we looked at it mod 10^100?

It might end up having a pattern in its first 100 digits. So we could also look at it mod 10 and see what we get.

So what if we looked at it mod 10^100?

It might end up having a pattern in its first 100 digits. So we could also look at it mod 10 and see what we get.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Pretty sure every digit in the big integer matters, root 5 + root 7 being irrational.

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

- Yakk
- Poster with most posts but no title.
**Posts:**10879**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

... ya, that was dumb.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

sundalius wrote:Wait...

Wouldn't it just be:

(Radical 5+Radical 7) ^2

Then under distributive property:

5+7

Then to derive the answer:

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

Even if the first part worked, you'd actually get

(5 + 7)^(2010/2) * (sqrt 5 + sqrt 7) = 12^1005 * (sqrt 5 + sqrt 7),

or maybe (sqrt 12)^(2011),

and I'm not sure you're any better off.

I don’t know if this helps, but we can rewrite (√7 + √5) = 2/(√7 - √5).

wee free kings

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

I think you need a series that converges more rapidly than the binomial theorem.

Macbi wrote:joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.Macbi wrote:Hint:Spoiler:Spoiler:

Ah, for some reason I thought we were in 2010. I now have no idea.

(Wolfram Alpha says the answer is 4.)

Yes, W|A has the 100th digit as being 4, but is that not just the 100th digit and not the 100th digit after the decimal place?

EDIT: I am no mathematician, I just use W|A a lot...

I'm never sarcastic.

Kick wrote:Yes, W|A has the 100th digit as being 4, but is that not just the 100th digit and not the 100th digit after the decimal place?

EDIT: I am no mathematician, I just use W|A a lot...

Ah, yes. That was my mistake.

- Indigo is a lie.

Which idiot decided that websites can't go within 4cm of the edge of the screen?

There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

After working on this problem for a while, I'm wondering whether the OP is remembering it incorrectly. All the approaches seem to eventually lead to knowing the 100th digit after the decimal place of \sqrt5, or some other equally (in)accessable data.

Late last night, I had an inspiration. Since there are a lot of symbols, and I don't really want to mess with the math interface for that, I wrote it up and put in a PDF; the link is http://math.la.asu.edu/~checkman/extras/sqrt5sqrt7.pdf .

Late last night, I had an inspiration. Since there are a lot of symbols, and I don't really want to mess with the math interface for that, I wrote it up and put in a PDF; the link is http://math.la.asu.edu/~checkman/extras/sqrt5sqrt7.pdf .

After looking at your pdf:

**Spoiler:**

t1mm01994 wrote:sundalius wrote:Wait...

Wouldn't it just be:

(Radical 5+Radical 7) ^2

Then under distributive property:

5+7

Then to derive the answer:

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

http://en.wikipedia.org/wiki/Freshmans_dream

Quite relevant. Basically, following your train of thought

(2+3)^2 = 2^2 + 3^2

5^2 = 4+9

25 = 13 and that doesnt quite sound right to me.. A search for "expanding brackets" might help you as well.

Thanks. Interesting read, will have to show my Algebra II/Geometry teacher tomorrow... I feel so lost in the math parts here, being only a freshman/sophomore/Junior(it's bloody complicated) in High School. But yeah, that doesn't sound right... I'll go study now >.>

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

mfb wrote:After looking at your pdf:

The expression inside the "[ ]" is ~10^1071, which is a lot (~300 orders of magnitude) smaller than the original number. I wonder where this comes from.

A mistake. The 12 in the auxiliary equation should be a 24. (Also, remember that you're substituting n=1005 into the formula for b_n, not 2011.) This means

b_n = \left[ \left(12 + 2 \sqrt{35}\right)^n + \left(12 - 2 \sqrt{35}\right)^n\right]\cdot 10^{100} \cdot\sqrt 5

Unfortunately, there are still \sqrt7s lurking about. (I wonder about how \sqrt5+\sqrt7 turned into \sqrt5, though ...

Well, (\sqrt{7}+\sqrt{5})^2 = 12+2\sqrt{35}

Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your b_{n} definition and you have the original problem.

So after your correction, the whole work looks a bit pointless.

Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your b

So after your correction, the whole work looks a bit pointless.

Cheating with Wolfram Alpha

**Spoiler:**

- Proginoskes
**Posts:**313**Joined:**Mon Nov 14, 2011 7:07 am UTC**Location:**Sitting Down

mfb wrote:Well, (\sqrt{7}+\sqrt{5})^2 = 12+2\sqrt{35}

Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your b_{n}definition and you have the original problem.

So after your correction, the whole work looks a bit pointless.

Of course, it's going to evaluate to the same number (or it should, anyway). It's the rewriting-and-putting-into-a-form-we-can-deal-with that is important; if it's done properly, it will enable us to solve the problem without doing the long calculation directly. (The "simpler" problem is an example of how this can be done.)

BTW, I think I solved the problem of where the \sqrt7 went. b_1 = 52, and I imagine the original formula pops up again. (I realized this last night when I was dropping off to sleep.)

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