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Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.

I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 \cdot 1976^7>0$. How should I do that?

Jyrki Lahtonen
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Puzzled417
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3 Answers3

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We have LHS is greater than $\frac{1}{5\cdot2000^3}-\frac{1}{10\cdot1000^7}>2.4\cdot10^{-11}$

I suspect the LHS should be the 3rd power instead of the 7th. The inequality is still true in that case, but a little more care is needed.

[For example you could approximate as $\frac{1}{4\cdot2000^3}-\frac{1}{10\cdot1500^3}=\frac{1}{32}10^{-9}-\frac{1}{3.375}10^{-10}>(\frac{1}{32}-\frac{1}{33})10^{-9}>10^{-13}$]

almagest
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$$E=\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}-\frac{1}{10^{1444}}\gt 0\large?$$ $$E=1-\frac{1}{4\cdot 1976^4}-\frac{4\cdot 1976^3}{10^{1444}}\gt 0\large?$$

The answer is YES because $$\frac{1}{4\cdot 1976^3}\lt \frac 14$$ and because the numerator $4\cdot 1976^3$ has $11$ digits.

Piquito
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Seems really simple - am I missing something?

$a = 4\times 1976^3 < 4\times 2000^3 = 32\times 10^9$

$\implies 1/a > \frac{1}{32}10^{-9} > 3\times 10^{-11}$

$b= 16\times 1976^7 > 10^{22}$ and $1/b <10^{-22}$

$\implies \frac{1}{a} - \frac{1}{b} > 3 \times 10^{-11} - 10^{-22} >10^{-19.76}$ (howsoever you interpret that exponent)

Joffan
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