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math: Fix inaccuracy of j0f for x >= 2^127 when sin(x)+cos(x) is tiny
Checked on x86_64-linux-gnu and i686-linux-gnu.
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@ -5748,6 +5748,8 @@ j0 0x1p16382
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j0 0x1p16383
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# the next value generates larger error bounds on x86_64 (binary32)
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j0 0x2.602774p+0 xfail-rounding:ibm128-libgcc
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# the next value exercises the flt-32 code path for x >= 2^127
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j0 0x8.2f4ecp+124
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j1 -1.0
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j1 0.0
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@ -55,7 +55,22 @@ __ieee754_j0f(float x)
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z = -__cosf(x+x);
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if ((s*c)<zero) cc = z/ss;
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else ss = z/cc;
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}
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} else {
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/* We subtract (exactly) a value x0 such that
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cos(x0)+sin(x0) is very near to 0, and use the identity
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sin(x-x0) = sin(x)*cos(x0)-cos(x)*sin(x0) to get
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sin(x) + cos(x) with extra accuracy. */
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float x0 = 0xe.d4108p+124f;
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float y = x - x0; /* exact */
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/* sin(y) = sin(x)*cos(x0)-cos(x)*sin(x0) */
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z = __sinf (y);
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float eps = 0x1.5f263ep-24f;
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/* cos(x0) ~ -sin(x0) + eps */
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z += eps * __cosf (x);
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/* now z ~ (sin(x)-cos(x))*cos(x0) */
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float cosx0 = -0xb.504f3p-4f;
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cc = z / cosx0;
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}
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/*
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* j0(x) = 1/sqrt(pi) * (P(0,x)*cc - Q(0,x)*ss) / sqrt(x)
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* y0(x) = 1/sqrt(pi) * (P(0,x)*ss + Q(0,x)*cc) / sqrt(x)
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