<![CDATA[Math Is Fun Forum / Help Me !]]> 2019-10-16T16:12:39Z FluxBB http://www.mathisfunforum.com/index.php <![CDATA[Visualizing proof]]> hi

Here's the proof that I was taught many years ago. Draw a circle radius R and centre C.  Mark points A and B as shown and D so that AD is perpendicular to CB.

Let angle ACB be x in radians.

sin(x) = AD / R

arc length AB / whole circumference = arcAB / 2.pi.R = angle of arc / whole circle angle = x / 2.pi

cancelling the 2.pi gives arcAB / R = x

So sin(x) / x = AD / arcAB

implies sin(x) < x but as x approaches zero AD approaches arcAB so sin(x) / x tends to 1.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-16T16:12:39Z http://www.mathisfunforum.com/viewtopic.php?id=25269&action=new
<![CDATA[Statistics and Economics]]> Thanks Bob. I'm most grateful.

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http://www.mathisfunforum.com/profile.php?id=210744 2019-10-16T13:23:03Z http://www.mathisfunforum.com/viewtopic.php?id=25266&action=new
<![CDATA[Using limits with complex numbers]]> hi 666 bro

Strictly speaking, the Argand diagram is not a graph.  As it has 'x' and 'y' coordinates I can see why you might think it was.  But it is just a way to represent complex numbers and there is lots of useful maths that stems from it.

A function such as y = 2x + 3 has an input (x) and an output (y).  So you can show the effect of the function by plotting a graph.

A function like z = x^2 + y^2 cannot be represented in 2 dimensions, so, if you want a visual representation you have to try and represent the z axis, perpendicular to the other two … usually rising up from the x-y plane.  Tricky to show clearly because we are only 3 dimensional beings (as far as I can tell )

So if that was the  function of complex numbers of the form x + iy and z was limited to real values that's how you'd have to graph it.  But z could also be complex in which case we run out of dimensions to show what is happening. Nevertheless there's a whole area of maths that does consider such functions.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-15T12:45:30Z http://www.mathisfunforum.com/viewtopic.php?id=25264&action=new
<![CDATA[how to defeat time while study a book?]]> There's no easy answer to this.  You need to understand the topic, and, if you skip some examples, you may be left not really understanding.  But I have sometimes discovered that looking at the next chapter helps to make sense of the earlier ones.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-14T07:59:09Z http://www.mathisfunforum.com/viewtopic.php?id=25263&action=new
<![CDATA[Family of Exponential Functions]]> I did it and the and read the example again not it's clear thank you
and I understand now how the ln is an inverse of e

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http://www.mathisfunforum.com/profile.php?id=212310 2019-10-12T23:31:06Z http://www.mathisfunforum.com/viewtopic.php?id=25237&action=new
<![CDATA[[ASK]Quadratic Inequation]]> That graph is not consistent with calculus for the minimum point so something is going wrong.  Haven't worked out what yet.  Assuming it is correct that the graph stays above the x axis, then the expression aY^2 -bY + 4 is always positive for all Y (remember that Y here is the across value not the up).  So Y = log(x) > 0 for all x which means that x > 1.  So we have a solution.  It also seems it doesn't matter which log base we choose, as x > 1 in any base.

I tried the calculations again using Excel with very different results.  Maybe I had the format wrong for Wolfram.  Anyway it shows the importance of checking whether a result seems reasonable.  This time I got a minimum where it should be ie. at Y = 1.52....

So I think This graph is correct: You can use the quadratic formula to get the exact crossing points and hence the right solution set(s).

Bob

ps.  I will amend the earlier post to clarify why that graph image no longer exists.

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-09T07:41:19Z http://www.mathisfunforum.com/viewtopic.php?id=25250&action=new
<![CDATA[Conversion of recurring decimal to fraction algebraically]]> Hi UltraV

Welcome to the forum.

I agree with 3929/33300.

But it sounds like an impossible task if the number of missing digits is unknown.
0.251 recurring ;. 0.0251251 recurring; and 0.6789251251251 recurring will have very different fractional forms.  You have to know what all the digits are.  It would
be like trying to solve an equation when you don't know what it is.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-05T17:16:56Z http://www.mathisfunforum.com/viewtopic.php?id=25252&action=new
<![CDATA[proofs!]]> hi ANon

Welcome to the forum.

Assuming that all the Ps are the same then P^P = P and PVP = P.  It should be easy to prove that in Q1 all statements become P, and in Q2 they become P=>P

Q3.  If f is invertible then it won't work so we need a non-invertible function.  One way is to construct a function that leads to a constant after being applied twice.

eg:

f: If |x| ≥ 10        then f(x) = 10
If 0 < |x| < 10  then f(x) 10.|x|
If x = 0            then f(x) = 10

f(f(x))  = 10 for all x and f(f(f(x))) = 10 so these are equal.

But when x is between 0 and 10 , f(x) is not the same as f(f(x)).

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-10-02T09:04:24Z http://www.mathisfunforum.com/viewtopic.php?id=25249&action=new
<![CDATA[does exponential function start from zero?]]> Q = S(1 − e^(−kt) )

please see the full example with its graph

my question is I thought exponential function don't start from zero (0,0) but this function say the opposite ?
and my second question is that... the example ask and talk about S and t values
but what is the number " 1 "
in the equation?

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http://www.mathisfunforum.com/profile.php?id=212310 2019-09-25T05:32:58Z http://www.mathisfunforum.com/viewtopic.php?id=25240&action=new
<![CDATA[exponent]]> Hi,

P = P0 * e^kt
where where P0 and initial quantities
e =  2.71828

We say that P and Q are growing or decaying at a continuous rate of k. (For example, k =0 .02 corresponds to a continuous rate of 2%.)

what is the meaning of continuous rate and when we say it's a constant rate ?

see this please :

P = 5(1 .07)^s

in this other example the growth rate it 0.07 which is 7%

in the first example the growth rate was an exponent ! ( k value )

what changed ? how would the growth rate changing his places in a functions

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http://www.mathisfunforum.com/profile.php?id=212310 2019-09-24T12:56:51Z http://www.mathisfunforum.com/viewtopic.php?id=25239&action=new
<![CDATA[bank interests]]> hi Phro,

Thanks for the analysis.  I had thought the calcs would be different if the sum is subtracted at the start of each month but was worried it might confuse HL.  As this is the fault of the question setter it shouldn't cost any credit.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-09-24T09:58:13Z http://www.mathisfunforum.com/viewtopic.php?id=25205&action=new
<![CDATA[converting a decimal value into a percent method]]> 1.034 is a multiplier.  New population = old population times the multiplier.

3.4% is the percentage increase.  new population = old population + 3.4% of the old population.

So we have two ways of describing the same increase.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-09-22T12:13:12Z http://www.mathisfunforum.com/viewtopic.php?id=25213&action=new
<![CDATA[calculus and cubics]]> hi alphacenturi9

Welcome to the forum.

Please post the four you have identified.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-09-22T12:06:05Z http://www.mathisfunforum.com/viewtopic.php?id=25230&action=new
<![CDATA[Calculate pi?]]> I have used Microsoft Excel to create the calculations needed. The formula view on the right shows how it works.

The first column (D) just alternates between -1 and +1 so that I can subtract and then add successive terms.

The second column (E) just counts the number of terms so far.

The third column (F) shows the odd numbers needed for the calculation

The fourth column (G) does the calculation.  The previous total has 1/[an odd number] either subtracted or added.  This is achieved by multiplying by the D column number which is either -1 or +1

Finally, in the fifth column (H) the fourth column number is multiplied by 4 to give the calculated value for pi.

This last column shows just how slow the process is.  Even at the 19th row we only have 3.19....    It takes a lot more rows to make progress.

I have jumped to show rows 262 and 263.  At last we are getting 3.14 but still not very accurate.

If you subtract the row 262 value from the row 263 value you get 0.0076....

The question asks for 6 significant digits.  That means don't stop until we see 3.14159.  If a row shows this, and the next one as well, then subtracting one from the other would give 0.00000.....  I cannot say for certain that all the digits would be zeros because the displayed numbers will have been rounded off.  By testing when this difference is smaller than 0.0000001 we can be certain we have 6 significant figures.

It may be we could have stopped sooner, but the step of multiplying by 4 means any small difference  is multiplied.  Also the figures oscillate either side of the correct value so I 'played safe' by choosing this level of test.

If the test was to stop when the difference is < 0.01 then row 263 would be the moment when the algorithm stops.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-09-19T12:43:44Z http://www.mathisfunforum.com/viewtopic.php?id=25224&action=new
<![CDATA[car’s use]]> Firstly I think it is intended that both the journey to work and the journey back home should be counted as part of "commuting to work".  Other wise you are counting the journey home as "personal use" and it isn't.

Let's make up some figures.

Miles to and from work in one week = 168 miles.

Total miles on the odometer = 300 miles.

Then the fraction for travel to work is 168/300 = 56/100

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-09-19T07:49:30Z http://www.mathisfunforum.com/viewtopic.php?id=25223&action=new