Description: Lemma for fta : Main proof. We have already shifted the minimum found in ftalem3 to zero by a change of variables, and now we show that the minimum value is zero. Expanding in a series about the minimum value, let K be the lowest term in the polynomial that is nonzero, and let T be a K -th root of -u F ( 0 ) / A ( K ) . Then an evaluation of F ( T X ) where X is a sufficiently small positive number yields F ( 0 ) for the first term and -u F ( 0 ) x. X ^ K for the K -th term, and all higher terms are bounded because X is small. Thus, abs ( F ( T X ) ) <_ abs ( F ( 0 ) ) ( 1 - X ^ K ) < abs ( F ( 0 ) ) , in contradiction to our choice of F ( 0 ) as the minimum. (Contributed by Mario Carneiro, 14-Sep-2014) (Revised by AV, 28-Sep-2020)
Ref | Expression | ||
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Hypotheses | ftalem.1 | |
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ftalem.2 | |
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ftalem.3 | |
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ftalem.4 | |
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ftalem4.5 | |
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ftalem4.6 | |
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ftalem4.7 | |
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ftalem4.8 | |
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ftalem4.9 | |
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Assertion | ftalem5 | |