Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the possible value of $2·((1+\tfrac{1}{100})^{100})$?

Google will give $2·((1+\tfrac{1}{100})^{100}) = 5.40962765884$.

How can I find the possible value without Google or a calculator?

How can I solve an equation like $(x^{100})$ or $(x^{20000})$?

Or like $(1.01^{100})$?

share|improve this question
6  
A good approximation is $2e$ (but only to an error of about 5 premille) –  Hagen von Eitzen Feb 16 at 10:08
    
What is the precision requirement? –  Hagen von Eitzen Feb 16 at 10:19
    
@HagenvonEitzen min 4 –  Nifty Feb 16 at 10:19
    
You will be unlucky on Google for any x in x^20000 which results in large numbers. –  Thomas W. Feb 17 at 2:49

4 Answers 4

up vote 12 down vote accepted

Hint : Do you know about the limit that $$\lim_{x\to \infty}(1+\frac{1}{x})^x = e$$

share|improve this answer

I would go as follows:

Let $y=2(1+\frac{1}{100})^{100}$ so $\mathrm{ln}(y)=\mathrm{ln}(2)+100\mathrm{ln}(1.01)$. A first-order Taylor expansion around $\mathrm{ln}(1.01)$ is quite accurate and equal to $\mathrm{ln}(1)+1*(1.01-1)=0.01$. Therefore, $\mathrm{ln}(y)=\mathrm{ln}(2)+100*0.01=\mathrm{ln}(2)+1=\mathrm{ln}(2)+\mathrm{ln}(e)=\mathrm{ln}(2e)$. Finally, undo the transformation in both sides so $y=2e$.

share|improve this answer

When you face situations such as $(x+1)^n$ where $x$ is small, you can use the binomial theorem for a few terms. In your case where $n=100$, the first terms are $$1+100 x+4950 x^2+161700 x^3+3921225 x^4+75287520 x^5$$ If you replace $x$ by $\frac{1}{100}$, you see that the successive terms contribute less and less. Uisng these terms, the value of the above expression is $2.70344$ and twice this number leads to $5.40688$ which is very close to the exact value $5.40963$.

But, going to very large values, remember what Trafalgar Law wrote.

share|improve this answer

To compute $(1+\frac1{100})^{100}$ with an error $<10^{-4}$, expand: $$\left(1+\frac1{100}\right)^{100} =1+100\frac1{100}+{100\choose 2}\frac1{10000}+\ldots$$ until you notice that the summands fall below a suitable theshold (which should happen at ${100\choose 8}\frac1{10^{16}}$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.