How can I prove that∑i=1nan⋅exp⁡(qn(x))=0cannot have more than two roots?

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Jesse Gates · Accepted Answer

Step 1Consider a1=1, q1(x)=1−10(x−1)2 and a2=−1, q2(x)=1−10(x−2)2We havea1eq1(1)=1e1=eand for |x−1|≥1 that q1(x)≤1−10=−9 and hence0&amp;lt;a1eq1(x)≤1e−9&amp;lt;2×10−4 under that condition.Similarly,a2eq2(2)=−1e1=−e and 0&amp;gt;a2eq2(x)≥−1e−9&amp;gt;−2×10−4In other words, for i=1,2 the term aieqi(i) is (−1)i+1e, but |x−i|≥1 we have |aieqi(x)|≤e−9, which is &quot;small&quot;.The sum a1eqi(x)+a2eq2(x) is thus positive at x=1 (first summand is e, second summand is &quot;small&quot; so that the sum is positive), while it is negative for x=2 (second summmand is -e, first summand is small). Due the the continuity of a1eqi(x)+a2eq2(x), this means there is a zero of that function in the interval (1, 2)But we can continue this idea, we set, for i=3,4⋯ai=(−1)i+1,qi(x)=1−10(x−i)2.We get the same result as above, but now for all i=1, 2, ⋯aieqi(i)=(−1)i+1e,;|aieqi(x)|≤e−9  if  |x−i|≥1If we consider the for some n the sum∑i=1naieqi(x)for x=1, 2, ⋯, n we find that one summand (namely for i=x) is either e or -e, while the others are &quot;small&quot;.Even with n=5001, the &quot;small&quot; values add up (in absolute values) to less then 5000×2×10−4=1, so less then e, so they cannot change the sign dictated by the big value of absolute value e.That means that sum has alternating signs atx=1, 2, ⋯, n, so has at least n−1 zerosSure, at some higher n the argument breaks down because the small values can add up to more than e, but if you want more zeros, just change the 10 in the definition of the qi(x) to a higher value, which guarantees that the &quot;small&quot; contributions get even smaller, so you can choose n as high as you want.

How can I prove that \(\displaystyle{\sum_{{{i}={1}}}^{{n}}}{a}_{{n}}\cdot{\exp{{\left({q}_{{n}}{\left({x}\right)}\right)}}}={0}\) cannot have more

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