A few questions

Agnishom · 2013-02-10 02:05:37

bobbym wrote:

Hi;
You can also get quite close using:
$gif.latex?\large%20\bold{%205^{44}=%20\frac{10^{44}}{\left%20(%202^{10}%20\right%20)^4%20\cdot%202^4}}$

Hi bobbym,
What should I be doing after getting this form?

Hi stefy,
How? If I could have approximated the logarithm of 5^44, then there would have been no problem at all. Can you do it?

Last edited by Agnishom (2013-02-10 02:07:04)

bobbym · 2013-02-10 02:18:08

10^44 is a 45 digit number.

2^10 ≈ 1000 = 10^3

(10^3)^4 = 10^12

10^44 / (10^12 * 2^4) = 10^32 / 2^4

Now 2^4 = 1.6 *10

10^32 / ( 1.6 * 10) = 10^31 / 1.6 ≈ 10^30.

10^30 is a 31 digit number. But this is only an approximation.

Agnishom · 2013-02-10 02:23:52

Yes, yes, I get it

Well, what had stefy been talkiing about?

bobbym · 2013-02-10 02:28:35

You can use Taylor's series to get log(5), but it will require a little bit of calculus.

If you want to see how...

Last edited by bobbym (2013-02-10 03:15:30)

Agnishom · 2013-02-11 01:24:08

Indeed I want to see,

My problem is that I am not acquainted enough with calculus

bobbym · 2013-02-11 01:26:39

You are a little young for your school system to have taught you it. They prefer a gradual introduction to it.

I am using log(x) to denote the common logarithm and ln(x) as the natural log.

You start with the Mclaurin series

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\ln\left%20(\frac{1+x}{1-x}%20\right%20)=2%20x+\frac{2%20x^3}{3}+\frac{2%20x^5}{5}+\frac{2%20x^7}{7}+...+$

If we substitute x = 2 / 3 in the we get:

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\ln\left%20(\frac{1+\frac{2}{3}}{1-\frac{2}{3}}%20\right%20)=%20\ln(5)\approx%20\frac{2\%202}{3}+\frac{2}{3}%20\left(\frac{2}{3}\right)^3+\frac{2}{5}%20\left(\frac{2}{3}\right)^5+\frac{2}{7}%20\left(\frac{2}{3}\right)^7%20=%201.600261284211901$

So ln(5) ≈ 1.600261284211901

But we needed log(5) not ln(5)! How do we get it?

We use this relationship

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\log(x)=\frac{\ln(x))}{\ln(10)%20}$

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\log(5)=\frac{\ln(5))}{\ln(10)%20}$

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\log(5)=\frac{\ln(5))}{\ln(2)%20+%20\ln(5)}$

$gif.latex?\small%20\inline%20\dpi{120}%20\bg_green%20\fn_cs%20\log(5)%20\approx%20\frac{1.60026}{0.693\,%20+1.60026}%20\approx%200.698970004$

which is quite close to the true value.

Last edited by bobbym (2013-02-11 02:37:12)

Agnishom · 2013-02-11 01:54:28

Let me try it out

anonimnystefy · 2013-02-11 02:03:23

Hi bobbym

$gif.latex?\frac{\ln{\left%20(1+\frac{2}{3}%20\right%20)}}{1-\frac{2}{3}}\neq%20\ln{5}$

bobbym · 2013-02-11 02:07:24

Please hold on the latex maker is not holding up, I am aware of that and am editing the original post.

All done!

Last edited by bobbym (2013-02-11 02:27:57)

Agnishom · 2013-03-14 04:32:52

Please help me with the problem attached

Bob · 2013-03-14 05:07:38

hi Agnishom,

I'm not getting this yet. I've drawn a line L . Now it says 'on a straight line l' so I assumed that the circles had their centres on the line, and that each circle touches the ones on either side of it. That covers the 'externally tangential to circle(n-1) and circle(n+1) but apparently also tangential to L itself. How does that happen? And I haven't even got to the second sentence yet.

Bob

anonimnystefy · 2013-03-14 08:05:57

I am getting 1/2.

bobbym · 2013-03-14 10:42:50

Hi Agnishom;

bobbym wrote:

A picture is worth a thousand words.

For people like me who are geometrically challenged this problem is a nightmare. So far I have come up with 10^16 different possible drawings that to me fit those constraints.

Now I am busy looking up externally tangent, internally tangent, tangerine tangent... in the hopes of what externally tangent to the line means. Intersects at one point?

I see that bob has got his circles with their centers on the line, I have lifted my circles and it was hard to do, after all there are an infinite amount of them, to roll on the line. That takes care of externally tangent. Now this second bunch of circles, where do they go?

John McCarthy wrote:

As the Chinese say, 1001 words is worth more than a picture.

Maybe I shouldn't need the diagram...

Last edited by bobbym (2013-03-14 19:08:18)

Agnishom · 2013-03-14 13:59:39

Bob, if you want I can give a diagram

Agnishom · 2013-03-14 14:12:12

This is the only diagram I can imagine.

A rough sketch though...

1001 Words ≈ 7 KB
1 picture > 100 KB

Obviously, a picture is worth more....

Last edited by Agnishom (2013-03-14 14:14:14)

Nehushtan · 2013-03-14 14:45:43

Agnishom wrote:

Please help me with the problem attached

This is what I got.

Let be the centre of and the point of contact of with . Let be the centre of and the point of contact of with .

Let

be the point on such that is a rectangle. Applying Pythagorass theorem to gives .

Let

be the point on such that is a rectangle. Applying Pythagorass theorem to gives .

Let

be the point on such that is a rectangle. Applying Pythagorass theorem to gives .

Then

Hence:

[align=center]

[/align]

Last edited by Nehushtan (2013-03-14 14:53:34)

bobbym · 2013-03-14 18:19:41

Hi Agnishom;

Thanks for the drawing.

Last edited by bobbym (2013-03-14 21:19:25)

Agnishom · 2013-03-14 19:18:18

Thanks Nehushtan
I shall ask you if I have any doubts

Agnishom · 2013-03-15 13:54:42

Please upload a diagram, I really do not understand from "Applying Pythagoras Theorem..."

Nehushtan · 2013-03-16 04:09:30

Here you go.

Agnishom · 2013-03-16 16:06:15

Thank you

Math Is Fun Forum

#176 2013-02-10 02:05:37

Re: A few questions

#177 2013-02-10 02:18:08

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#178 2013-02-10 02:23:52

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#179 2013-02-10 02:28:35

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#180 2013-02-11 01:24:08

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#187 2013-03-14 08:05:57

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#189 2013-03-14 13:59:39

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