Is there any easier or more efficient way to calculate SF when there is different k?
I = [1:100]k03 = 0.3k1 = 1k3 = 3k5 = 5k8 = 8k13 = 13k15 = 15k18 = 18k23 = 23k25 = 25k28 = 28S03F = k03*log(I)S1F = k1*log(I)S3F = k3*log(I)S5F = k5*log(I)S8F = k8*log(I)S13F = k13*log(I)S15F = k15*log(I)S18F = k18*log(I)S23F = k23*log(I)S25F = k25*log(I)S28F = k28*log(I)
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