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$$1-(1-1.40*10^{-36})^x \ge 1.09*10^{-9}$$

I want to estimate $x$ such that the probability on the left becomes larger than the probability on the right.

A solution must exist because $1-(1-1.40*10^{-36})^0=0$, $\lim\limits_{n \to \infty} 1-(1-1.40*10^{-36})^n=1$ and the function is continuous.

However Wolfram|Alpha seems to return a wrong result.

Is my reasoning correct? Is there any (practical) way to estimate the result? For my problem would be sufficient to find the order of magnitude of $x$.

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3 Answers 3

up vote 7 down vote accepted

You can solve the inequality by logarithming :

We get $1-1.09*10^{-9}\ge (1-1.4*10^{-36})^x$

Taking logarithm gives $$\frac {\log(1-1.09*10^{-9})}{\log(1-1.4*10^{-36})}\le x$$

Note that we divided by a negative number!

GP/PARI gives

$$\frac{log(1-1.09*10^{-9})}{log(1-1.4*10^{-36})}\ =$$ $$778571428995750000308340238.34730623955575580298358$$

So, $x$ must be very large

An excellent approximation of the given value is $\frac {1.09*10^{-9}}{1.4*10^{-36}}$ because for $x \approx 0$ we have $log(1+x)\approx x$. And this is very easy to calculate.

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+1 Great step by step explanation: thanks! Here's the correct result following your approach –  collimarco Aug 30 at 10:10
2  
+1, for the approximation ;-) –  Jean-Claude Arbaut Aug 30 at 11:20

Mathematica input:

Solve[1 - (1 - 7/(5 10^36))^x == 109/(100 10^9), x, Reals]

Result:

$$x\to \frac{11 \log (2)-2 \log (3)+11 \log (5)-\log (11)-\log (19)-\log (4441)-\log (11971)}{36 \log (2)+37 \log (5)-\log (170207)-\log (29375995111834413390753611778599)} \\ \approx 7.7857142899575*10^{26}$$

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Is there any reason why you have used integers? –  collimarco Aug 30 at 9:40

Could you try replacing $1.4$ by $14/10$ and $1.09$ by $109/100$ ? This would make a significant difference (I hope).

Doing the same as user111187, the solution of the equation is $$x=\frac{\log \left(\frac{100000000000}{99999999891}\right)}{\log \left(\frac{5000000000000000000000000000000000000}{4999999999999999999999999999999 999993}\right)}$$ which is $$x=\frac{\log(1-109 \times 10^{-11})}{\log(1-14 \times 10^{-37})}\simeq \frac{109\times 10^{-11}} {14 \times 10^{-37}}=\frac{109}{14}10^{26}$$

Added later to this answer

The problem you posted was "simple" because it was possible to isolate the solution of the equation; it was a very nice illustration of the power of Taylor series.

But, let me take a more complex problem. What is $x$ such that $$e^{x^4}-\cos \left(x^3\right) \gt \epsilon$$ You basically could not answer this question since there is no way to find the solution of the equation. Then use Taylor to find that $$e^{x^4}-\cos \left(x^3\right) \simeq x^4+\frac{x^6}{2}+\frac{x^8}{2}+O\left(x^9\right)$$ So $x\gt \sqrt[4]\epsilon$ is the solution

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Is there any reason for using integers? Or is it only a trick with Wolfram|Alpha due to its implementation? –  collimarco Aug 30 at 9:39
1  
When non integers, CAS do not use infinite precision. Did that make any change ? –  Claude Leibovici Aug 30 at 9:45
    
+1 Thanks! The issue was that when using real numbers (float precision) instead of integers I should have used more decimal digits! Here's the correct result –  collimarco Aug 30 at 10:05

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