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Upper estimate and lower estimate are mathematical terms for auxiliary quantities in connection with inequalities .

An upper estimate for size is a different size , if it can be shown that: . ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle B\geq A}$

Accordingly, it is called a lower estimate for when it can be shown that: . ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle B\leq A}$

This is (often also ) usually an expression that depends on other variables. It can then also be the assessment; the condition must apply independently of the other quantities in the entire domain of definition. If necessary, considerations of the type used in interval arithmetic - for example in error calculation  - are used here. ${\displaystyle A}$ ${\displaystyle B}$

In this context, the term “estimate” does not mean a loss of reliability. It only has to do with the current concept of estimating insofar as the estimation can deviate from the "estimated" value - in some cases even very far, as long as it is only in the right direction.

use

Estimation is used as a tool to prove inequalities. One makes use of the transitivity of the greater / less than relation.

If one knows for an upper estimate , one can reduce the proof of to the proof of ; one can also reduce the proof of to the proof of . Correspondingly, one can show or if a lower estimate for is and or can be shown. ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle A\leq C}$ ${\displaystyle B\leq C}$ ${\displaystyle A<C}$ ${\displaystyle B<C}$ ${\displaystyle A>C}$ ${\displaystyle A\geq C}$ ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle B>C}$ ${\displaystyle B\geq C}$

If this approach fails with a particular one, the inequality to be proven is not refuted; maybe you just need a stricter estimate because the one used deviates too far from . ${\displaystyle B}$ ${\displaystyle A}$

Examples

General sample group. In principle, every numerical “ worst-case ” figure, as it occurs in everyday life, is a kind of lower or upper estimate.

Example 1. Estimates are, for example, the basis for proofs of divergence with the minorant criterion such as that for using the minorant . Clearly presented, you consider the following: ${\displaystyle \sum _{i=2}^{\infty }{\frac {1}{i}}}$ ${\displaystyle \sum _{i=2}^{\infty }{\frac {1}{2^{\lceil log_{2}(i)\rceil }}}}$

It should be shown that with a sum of consecutive fractions one can surpass any arbitrarily large natural number by choosing a suitable one. ${\displaystyle A={\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots +{\frac {1}{k}}}$ ${\displaystyle N}$ ${\displaystyle k}$

To do this, one forms another sum of parent fractions, in which every summand in with the denominator corresponds to a summand with the next not smaller power of two. Since the denominator in the new sum is equal to or greater than the corresponding denominator in the original, the fraction is equal to or less, and thus the total sum (equal to or) less. It is a lower estimate of . ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle i}$ ${\displaystyle A}$

The estimate is now chosen so that one can always summarize finite parts of the series to ½:

${\displaystyle A}$

${\displaystyle {\frac {1}{2}}}$

${\displaystyle {\frac {1}{3}}}$

${\displaystyle {\frac {1}{4}}}$

${\displaystyle {\frac {1}{5}}}$

${\displaystyle {\frac {1}{6}}}$

${\displaystyle {\frac {1}{7}}}$

${\displaystyle {\frac {1}{8}}}$

${\displaystyle {\frac {1}{9}}}$

${\displaystyle {\frac {1}{10}}}$

${\displaystyle {\frac {1}{11}}}$

${\displaystyle {\frac {1}{12}}}$

${\displaystyle {\frac {1}{13}}}$

${\displaystyle {\frac {1}{14}}}$

${\displaystyle {\frac {1}{15}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{17}}}$

${\displaystyle {\frac {1}{18}}}$

${\displaystyle B}$

${\displaystyle {\frac {1}{2}}}$

${\displaystyle {\frac {1}{4}}}$

${\displaystyle {\frac {1}{4}}}$

${\displaystyle {\frac {1}{8}}}$

${\displaystyle {\frac {1}{8}}}$

${\displaystyle {\frac {1}{8}}}$

${\displaystyle {\frac {1}{8}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{16}}}$

${\displaystyle {\frac {1}{32}}}$

${\displaystyle {\frac {1}{32}}}$

Summary of summands

${\displaystyle {\frac {1}{2}}}$

${\displaystyle {\frac {1}{2}}}$

${\displaystyle {\frac {1}{2}}}$

${\displaystyle {\frac {1}{2}}}$

etc.

If you choose , you have such parts; So it adds up . Thus it is even greater than . ${\displaystyle k=2^{2N+1}}$ ${\displaystyle 2N+1}$ ${\displaystyle B}$ ${\displaystyle N+{\frac {1}{2}}}$ ${\displaystyle A}$ ${\displaystyle N}$

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Example 2. For arbitrary bounded and real-valued functions f and g with the same domain holds ${\displaystyle D}$

${\displaystyle A:=|\max _{x\in D}f(x)-\max _{x\in D}g(x)|\leq \max _{x\in D}|f(x)-g(x)|=:B}$,

ie the distance between the maxima of two functions can always be estimated upwards by the maximum distance between the function values. Although this estimate does not have to be immediately obvious, the proof is very simple. It is noteworthy that at f and g except for the limitations (so that the peaks are adopted) and the common domain nothing further is provided: therefore applies z. B. for continuous functions on a compact (which are always restricted), but also for discontinuous functions, such as discrete functions, as long as they are restricted. ${\displaystyle A\leq B}$

Proof . Put and ; furthermore o. B. d. A. (can be achieved by renaming). ${\displaystyle x_{1}:=\arg \max _{x\in D}f(x)}$ ${\displaystyle x_{2}:=\arg \max _{x\in D}g(x)}$ ${\displaystyle f(x)\geq g(x)}$

Is sure . So follows ${\displaystyle g(x_{1})\leq g(x_{2})}$

${\displaystyle A=|f(x_{1})-g(x_{2})|=f(x_{1})-g(x_{2})\leq f(x_{1})-g(x_{1})=|f(x_{1})-g(x_{1})|\leq \max _{x\in D}|f(x)-g(x)|=B}$.

This estimate is sharp (ie equality occurs in certain cases), like the example

${\displaystyle g(x)=f(x)+\varepsilon }$

shows: Here is (Caution: The fact that an estimate is sharp does not mean that the estimate is as fine as possible; consider the example and : both estimates are sharp, but of different fineness). ${\displaystyle A=\varepsilon =B}$ ${\displaystyle 0\leq x^{2}}$ ${\displaystyle 0\leq x^{2}+x^{4}}$